def minPath(grid, k):
"""
Given a grid with N rows and N columns (N >= 2) and a positive integer k,
each cell of the grid contains a value. Every integer in the range [1, N * N]
inclusive appears exactly once on the cells of the grid.
You have to find the minimum path of length k in the grid. You can start
from any cell, and in each step you can move to any of the neighbor cells,
in other words, you can go to cells which share an edge with you current
cell.
Please note that a path of length k means visiting exactly k cells (not
necessarily distinct).
You CANNOT go off the grid.
A path A (of length k) is considered less than a path B (of length k) if
after making the ordered lists of the values on the cells that A and B go
through (let's call them lst_A and lst_B), lst_A is lexicographically less
than lst_B, in other words, there exist an integer index i (1 <= i <= k)
such that lst_A[i] < lst_B[i] and for any j (1 <= j < i) we have
lst_A[j] = lst_B[j].
It is guaranteed that the answer is unique.
Return an ordered list of the values on the cells that the minimum path go through.
Examples:
Input: grid = [ [1,2,3], [4,5,6], [7,8,9]], k = 3
Output: [1, 2, 1]
Input: grid = [ [5,9,3], [4,1,6], [7,8,2]], k = 1
Output: [1]
"""
N = len(grid)
x, y = 0, 0
for i in range(N):
for j in range(N):
if grid[i][j] == 1:
x, y = i, j
mn = N * N
if x > 0: mn = min(mn, grid[x - 1][y])
if x < N - 1: mn = min(mn, grid[x + 1][y])
if y > 0: mn = min(mn, grid[x][y - 1])
if y < N - 1: mn = min(mn, grid[x][y + 1])
return [1 if i % 2 == 0 else mn for i in range(k)]
[[[[1, 2, 3], [4, 5, 6], [7, 8, 9]], 3], [[[5, 9, 3], [4, 1, 6], [7, 8, 2]], 1], [[[1, 2, 3, 4], [5, 6, 7, 8], [9, 10, 11, 12], [13, 14, 15, 16]], 4], [[[6, 4, 13, 10], [5, 7, 12, 1], [3, 16, 11, 15], [8, 14, 9, 2]], 7], [[[8, 14, 9, 2], [6, 4, 13, 15], [5, 7, 1, 12], [3, 10, 11, 16]], 5], [[[11, 8, 7, 2], [5, 16, 14, 4], [9, 3, 15, 6], [12, 13, 10, 1]], 9], [[[12, 13, 10, 1], [9, 3, 15, 6], [5, 16, 14, 4], [11, 8, 7, 2]], 12], [[[2, 7, 4], [3, 1, 5], [6, 8, 9]], 8], [[[6, 1, 5], [3, 8, 9], [2, 7, 4]], 8], [[[1, 2], [3, 4]], 10]] [[[[1, 2, 3], [4, 5, 6], [7, 8, 9]], 1], [[[1, 2, 3], [4, 5, 6], [7, 8, 9]], 9], [[[9, 8, 7], [6, 5, 4], [3, 2, 1]], 3], [[[2, 3, 4], [5, 6, 7], [8, 9, 1]], 5], [[[5, 4, 7], [2, 1, 8], [3, 6, 9]], 4], [[[1, 2, 3], [4, 5, 6], [7, 8, 9]], 5], [[[1, 5, 3], [4, 2, 6], [7, 8, 9]], 4], [[[1, 4, 7], [2, 5, 8], [3, 6, 9]], 6], [[[1, 2, 3], [4, 5, 6], [7, 8, 9]], 15], [[[1, 5, 3], [4, 2, 6], [7, 8, 9]], 2], [[[1, 2, 3], [4, 5, 6], [7, 8, 9]], 3], [[[2, 3, 4], [5, 6, 7], [8, 9, 1]], 13], [[[1, 2, 3], [4, 5, 6], [7, 8, 9]], 6], [[[2, 3, 4], [5, 6, 7], [8, 9, 1]], 4], [[[1, 4, 7], [2, 5, 8], [3, 6, 9]], 5], [[[1, 4, 7], [2, 5, 8], [3, 6, 9]], 7], [[[2, 3, 4], [5, 6, 7], [8, 9, 1]], 18], [[[1, 4, 7], [2, 5, 8], [3, 6, 9]], 9], [[[1, 2, 3], [4, 5, 6], [7, 8, 9]], 10], [[[1, 5, 3], [4, 2, 6], [7, 8, 9]], 3], [[[5, 4, 7], [2, 1, 8], [3, 6, 9]], 12], [[[2, 3, 4], [5, 6, 7], [8, 9, 1]], 1], [[[1, 4, 7], [2, 5, 8], [3, 6, 9]], 20], [[[1, 4, 7], [2, 5, 8], [3, 6, 9]], 19], [[[1, 2, 3], [4, 5, 6], [7, 8, 9]], 20], [[[2, 3, 4], [5, 6, 7], [8, 9, 1]], 17], [[[1, 4, 7], [2, 5, 8], [3, 6, 9]], 17], [[[2, 3, 4], [5, 6, 7], [8, 9, 1]], 3], [[[2, 3, 4], [5, 6, 7], [8, 9, 1]], 6], [[[1, 4, 7], [2, 5, 8], [3, 6, 9]], 13], [[[1, 4, 7], [2, 5, 8], [3, 6, 9]], 14], [[[2, 3, 4], [5, 6, 7], [8, 9, 1]], 2], [[[1, 4, 7], [2, 5, 8], [3, 6, 9]], 21], [[[2, 3, 4], [5, 6, 7], [8, 9, 1]], 12], [[[5, 4, 7], [2, 1, 8], [3, 6, 9]], 1], [[[2, 3, 4], [5, 6, 7], [8, 9, 1]], 15], [[[1, 6, 13, 14], [9, 8, 7, 16], [12, 10, 2, 11], [5, 4, 3, 15]], 16], [[[1, 5, 9, 13], [2, 6, 10, 14], [3, 7, 11, 15], [4, 8, 12, 16]], 16], [[[2, 7, 4], [3, 1, 5], [6, 8, 9]], 1], [[[1, 3, 5, 7], [9, 11, 13, 15], [2, 4, 6, 8], [10, 12, 14, 16]], 5], [[[10, 5, 20, 14, 16], [7, 11, 12, 17, 9], [13, 19, 1, 15, 8], [3, 18, 4, 6, 2], [22, 21, 23, 24, 25]], 10], [[[1, 2, 3, 4, 5], [6, 7, 8, 9, 10], [11, 12, 13, 14, 15], [16, 17, 18, 19, 20], [21, 22, 23, 24, 25]], 20], [[[10, 7, 14, 1], [15, 9, 8, 5], [12, 6, 2, 16], [13, 11, 3, 4]], 10], [[[1, 2], [3, 4]], 2], [[[1, 2, 3, 4, 5], [6, 7, 8, 9, 10], [11, 12, 13, 14, 15], [16, 17, 18, 19, 20], [21, 22, 23, 24, 25]], 1], [[[1, 2, 3, 4, 5], [6, 7, 8, 9, 10], [11, 12, 13, 14, 15], [16, 17, 18, 19, 20], [21, 22, 23, 24, 25]], 9], [[[1, 2], [3, 4]], 3], [[[1, 3, 5, 7], [9, 11, 13, 15], [2, 4, 6, 8], [10, 12, 14, 16]], 8], [[[10, 5, 20, 14, 16], [7, 11, 12, 17, 9], [13, 19, 1, 15, 8], [3, 18, 4, 6, 2], [22, 21, 23, 24, 25]], 11], [[[1, 2], [3, 4]], 24], [[[10, 7, 14, 1], [15, 9, 8, 5], [12, 6, 2, 16], [13, 11, 3, 4]], 11], [[[1, 3, 5, 7], [9, 11, 13, 15], [2, 4, 6, 8], [10, 12, 14, 16]], 6], [[[10, 7, 14, 1], [15, 9, 8, 5], [12, 6, 2, 16], [13, 11, 3, 4]], 4], [[[1, 3, 5, 7], [9, 11, 13, 15], [2, 4, 6, 8], [10, 12, 14, 16]], 10], [[[1, 3, 5, 7], [9, 11, 13, 15], [2, 4, 6, 8], [10, 12, 14, 16]], 9], [[[1, 3, 5, 7], [9, 11, 13, 15], [2, 4, 6, 8], [10, 12, 14, 16]], 15], [[[1, 2], [3, 4]], 18], [[[1, 6, 13, 14], [9, 8, 7, 16], [12, 10, 2, 11], [5, 4, 3, 15]], 14], [[[10, 7, 14, 1], [15, 9, 8, 5], [12, 6, 2, 16], [13, 11, 3, 4]], 5], [[[10, 5, 20, 14, 16], [7, 11, 12, 17, 9], [13, 19, 1, 15, 8], [3, 18, 4, 6, 2], [22, 21, 23, 24, 25]], 23], [[[10, 5, 20, 14, 16], [7, 11, 12, 17, 9], [13, 19, 1, 15, 8], [3, 18, 4, 6, 2], [22, 21, 23, 24, 25]], 6], [[[1, 2, 3, 4, 5], [6, 7, 8, 9, 10], [11, 12, 13, 14, 15], [16, 17, 18, 19, 20], [21, 22, 23, 24, 25]], 8], [[[10, 5, 20, 14, 16], [7, 11, 12, 17, 9], [13, 19, 1, 15, 8], [3, 18, 4, 6, 2], [22, 21, 23, 24, 25]], 9], [[[1, 6, 13, 14], [9, 8, 7, 16], [12, 10, 2, 11], [5, 4, 3, 15]], 15], [[[1, 3, 5, 7], [9, 11, 13, 15], [2, 4, 6, 8], [10, 12, 14, 16]], 18], [[[1, 6, 13, 14], [9, 8, 7, 16], [12, 10, 2, 11], [5, 4, 3, 15]], 9], [[[10, 5, 20, 14, 16], [7, 11, 12, 17, 9], [13, 19, 1, 15, 8], [3, 18, 4, 6, 2], [22, 21, 23, 24, 25]], 4], [[[1, 2, 3], [4, 5, 6], [7, 8, 9]], 21], [[[1, 3, 5, 7], [9, 11, 13, 15], [2, 4, 6, 8], [10, 12, 14, 16]], 17], [[[10, 5, 20, 14, 16], [7, 11, 12, 17, 9], [13, 19, 1, 15, 8], [3, 18, 4, 6, 2], [22, 21, 23, 24, 25]], 25], [[[10, 5, 20, 14, 16], [7, 11, 12, 17, 9], [13, 19, 1, 15, 8], [3, 18, 4, 6, 2], [22, 21, 23, 24, 25]], 16], [[[1, 2, 3, 4, 5], [6, 7, 8, 9, 10], [11, 12, 13, 14, 15], [16, 17, 18, 19, 20], [21, 22, 23, 24, 25]], 7], [[[1, 3, 5, 7], [9, 11, 13, 15], [2, 4, 6, 8], [10, 12, 14, 16]], 19], [[[1, 3, 5, 7], [9, 11, 13, 15], [2, 4, 6, 8], [10, 12, 14, 16]], 3], [[[1, 6, 13, 14], [9, 8, 7, 16], [12, 10, 2, 11], [5, 4, 3, 15]], 21], [[[10, 5, 20, 14, 16], [7, 11, 12, 17, 9], [13, 19, 1, 15, 8], [3, 18, 4, 6, 2], [22, 21, 23, 24, 25]], 5], [[[10, 7, 14, 1], [15, 9, 8, 5], [12, 6, 2, 16], [13, 11, 3, 4]], 9], [[[1, 3, 5, 7], [9, 11, 13, 15], [2, 4, 6, 8], [10, 12, 14, 16]], 14], [[[10, 5, 20, 14, 16], [7, 11, 12, 17, 9], [13, 19, 1, 15, 8], [3, 18, 4, 6, 2], [22, 21, 23, 24, 25]], 7], [[[1, 3, 5, 7], [9, 11, 13, 15], [2, 4, 6, 8], [10, 12, 14, 16]], 11], [[[1, 2, 3, 4, 5], [6, 7, 8, 9, 10], [11, 12, 13, 14, 15], [16, 17, 18, 19, 20], [21, 22, 23, 24, 25]], 2], [[[1, 2, 3, 4, 5], [6, 7, 8, 9, 10], [11, 12, 13, 14, 15], [16, 17, 18, 19, 20], [21, 22, 23, 24, 25]], 22], [[[1, 2], [3, 4]], 25], [[[1, 3, 5, 7], [9, 11, 13, 15], [2, 4, 6, 8], [10, 12, 14, 16]], 7], [[[1, 2], [3, 4]], 17], [[[10, 5, 20, 14, 16], [7, 11, 12, 17, 9], [13, 19, 1, 15, 8], [3, 18, 4, 6, 2], [22, 21, 23, 24, 25]], 12], [[[2, 7, 4], [3, 1, 5], [6, 8, 9]], 20], [[[10, 5, 20, 14, 16], [7, 11, 12, 17, 9], [13, 19, 1, 15, 8], [3, 18, 4, 6, 2], [22, 21, 23, 24, 25]], 17], [[[10, 7, 14, 1], [15, 9, 8, 5], [12, 6, 2, 16], [13, 11, 3, 4]], 2], [[[1, 2, 3, 4, 5], [6, 7, 8, 9, 10], [11, 12, 13, 14, 15], [16, 17, 18, 19, 20], [21, 22, 23, 24, 25]], 13], [[[1, 3, 5, 7], [9, 11, 13, 15], [2, 4, 6, 8], [10, 12, 14, 16]], 16], [[[1, 3, 5, 7], [9, 11, 13, 15], [2, 4, 6, 8], [10, 12, 14, 16]], 24], [[[1, 2], [3, 4]], 19], [[[10, 7, 14, 1], [15, 9, 8, 5], [12, 6, 2, 16], [13, 11, 3, 4]], 13], [[[1, 3, 5, 7], [9, 11, 13, 15], [2, 4, 6, 8], [10, 12, 14, 16]], 13], [[[10, 5, 20, 14, 16], [7, 11, 12, 17, 9], [13, 19, 1, 15, 8], [3, 18, 4, 6, 2], [22, 21, 23, 24, 25]], 24], [[[2, 7, 4], [3, 1, 5], [6, 8, 9]], 9], [[[1, 5, 9, 13], [2, 6, 10, 14], [3, 7, 11, 15], [4, 8, 12, 16]], 22], [[[1, 6, 13, 14], [9, 8, 7, 16], [12, 10, 2, 11], [5, 4, 3, 15]], 12], [[[1, 2], [3, 4]], 16], [[[2, 7, 4], [3, 1, 5], [6, 8, 9]], 19], [[[1, 6, 13, 14], [9, 8, 7, 16], [12, 10, 2, 11], [5, 4, 3, 15]], 10], [[[1, 2, 3], [4, 5, 6], [7, 8, 9]], 4], [[[10, 7, 14, 1], [15, 9, 8, 5], [12, 6, 2, 16], [13, 11, 3, 4]], 3], [[[1, 2], [3, 4]], 22], [[[10, 5, 20, 14, 16], [7, 11, 12, 17, 9], [13, 19, 1, 15, 8], [3, 18, 4, 6, 2], [22, 21, 23, 24, 25]], 22], [[[1, 6, 13, 14], [9, 8, 7, 16], [12, 10, 2, 11], [5, 4, 3, 15]], 24], [[[1, 6, 13, 14], [9, 8, 7, 16], [12, 10, 2, 11], [5, 4, 3, 15]], 23], [[[10, 7, 14, 1], [15, 9, 8, 5], [12, 6, 2, 16], [13, 11, 3, 4]], 23], [[[1, 6, 13, 14], [9, 8, 7, 16], [12, 10, 2, 11], [5, 4, 3, 15]], 22], [[[1, 2, 3], [4, 5, 6], [7, 8, 9]], 19], [[[1, 2], [3, 4]], 5], [[[1, 2, 3, 4, 5], [6, 7, 8, 9, 10], [11, 12, 13, 14, 15], [16, 17, 18, 19, 20], [21, 22, 23, 24, 25]], 10], [[[1, 2, 3], [4, 5, 6], [7, 8, 9]], 12], [[[1, 3, 5, 7], [9, 11, 13, 15], [2, 4, 6, 8], [10, 12, 14, 16]], 1], [[[10, 5, 20, 14, 16], [7, 11, 12, 17, 9], [13, 19, 1, 15, 8], [3, 18, 4, 6, 2], [22, 21, 23, 24, 25]], 8], [[[1, 2, 3], [4, 5, 6], [7, 8, 9]], 25], [[[2, 7, 4], [3, 1, 5], [6, 8, 9]], 18], [[[1, 2, 3, 4, 5], [6, 7, 8, 9, 10], [11, 12, 13, 14, 15], [16, 17, 18, 19, 20], [21, 22, 23, 24, 25]], 23], [[[1, 6, 13, 14], [9, 8, 7, 16], [12, 10, 2, 11], [5, 4, 3, 15]], 11], [[[1, 6, 13, 14], [9, 8, 7, 16], [12, 10, 2, 11], [5, 4, 3, 15]], 2], [[[10, 7, 14, 1], [15, 9, 8, 5], [12, 6, 2, 16], [13, 11, 3, 4]], 12], [[[1, 6, 13, 14], [9, 8, 7, 16], [12, 10, 2, 11], [5, 4, 3, 15]], 7], [[[1, 2, 3], [4, 5, 6], [7, 8, 9]], 11], [[[1, 6, 13, 14], [9, 8, 7, 16], [12, 10, 2, 11], [5, 4, 3, 15]], 18], [[[1, 5, 9, 13], [2, 6, 10, 14], [3, 7, 11, 15], [4, 8, 12, 16]], 8], [[[1, 2, 3], [4, 5, 6], [7, 8, 9]], 18], [[[1, 2], [3, 4]], 21], [[[1, 3, 5, 7], [9, 11, 13, 15], [2, 4, 6, 8], [10, 12, 14, 16]], 21], [[[10, 7, 14, 1], [15, 9, 8, 5], [12, 6, 2, 16], [13, 11, 3, 4]], 24], [[[1, 5, 9, 13], [2, 6, 10, 14], [3, 7, 11, 15], [4, 8, 12, 16]], 2], [[[10, 5, 20, 14, 16], [7, 11, 12, 17, 9], [13, 19, 1, 15, 8], [3, 18, 4, 6, 2], [22, 21, 23, 24, 25]], 2], [[[1, 3, 5, 7], [9, 11, 13, 15], [2, 4, 6, 8], [10, 12, 14, 16]], 12], [[[1, 6, 13, 14], [9, 8, 7, 16], [12, 10, 2, 11], [5, 4, 3, 15]], 8], [[[10, 5, 20, 14, 16], [7, 11, 12, 17, 9], [13, 19, 1, 15, 8], [3, 18, 4, 6, 2], [22, 21, 23, 24, 25]], 21], [[[1, 2, 3], [4, 5, 6], [7, 8, 9]], 16], [[[10, 5, 20, 14, 16], [7, 11, 12, 17, 9], [13, 19, 1, 15, 8], [3, 18, 4, 6, 2], [22, 21, 23, 24, 25]], 19], [[[1, 2], [3, 4]], 1], [[[1, 2], [3, 4]], 4], [[[1, 2, 3, 4, 5], [6, 7, 8, 9, 10], [11, 12, 13, 14, 15], [16, 17, 18, 19, 20], [21, 22, 23, 24, 25]], 16], [[[1, 2], [3, 4]], 7], [[[1, 3, 5, 7], [9, 11, 13, 15], [2, 4, 6, 8], [10, 12, 14, 16]], 4], [[[10, 7, 14, 1], [15, 9, 8, 5], [12, 6, 2, 16], [13, 11, 3, 4]], 19], [[[1, 5, 9, 13], [2, 6, 10, 14], [3, 7, 11, 15], [4, 8, 12, 16]], 17], [[[1, 2], [3, 4]], 6], [[[10, 5, 20, 14, 16], [7, 11, 12, 17, 9], [13, 19, 1, 15, 8], [3, 18, 4, 6, 2], [22, 21, 23, 24, 25]], 18], [[[10, 7, 14, 1], [15, 9, 8, 5], [12, 6, 2, 16], [13, 11, 3, 4]], 6], [[[1, 2], [3, 4]], 10], [[[1, 2], [3, 4]], 23], [[[10, 7, 14, 1], [15, 9, 8, 5], [12, 6, 2, 16], [13, 11, 3, 4]], 14], [[[10, 7, 14, 1], [15, 9, 8, 5], [12, 6, 2, 16], [13, 11, 3, 4]], 18], [[[1, 5, 9, 13], [2, 6, 10, 14], [3, 7, 11, 15], [4, 8, 12, 16]], 9], [[[10, 7, 14, 1], [15, 9, 8, 5], [12, 6, 2, 16], [13, 11, 3, 4]], 7], [[[1, 3, 5, 7], [9, 11, 13, 15], [2, 4, 6, 8], [10, 12, 14, 16]], 23], [[[1, 2], [3, 4]], 8], [[[10, 7, 14, 1], [15, 9, 8, 5], [12, 6, 2, 16], [13, 11, 3, 4]], 17], [[[10, 7, 14, 1], [15, 9, 8, 5], [12, 6, 2, 16], [13, 11, 3, 4]], 22], [[[1, 6, 13, 14], [9, 8, 7, 16], [12, 10, 2, 11], [5, 4, 3, 15]], 20], [[[1, 5, 9, 13], [2, 6, 10, 14], [3, 7, 11, 15], [4, 8, 12, 16]], 15], [[[1, 2, 3, 4, 5], [6, 7, 8, 9, 10], [11, 12, 13, 14, 15], [16, 17, 18, 19, 20], [21, 22, 23, 24, 25]], 14], [[[1, 5, 9, 13], [2, 6, 10, 14], [3, 7, 11, 15], [4, 8, 12, 16]], 14], [[[10, 7, 14, 1], [15, 9, 8, 5], [12, 6, 2, 16], [13, 11, 3, 4]], 8], [[[1, 6, 13, 14], [9, 8, 7, 16], [12, 10, 2, 11], [5, 4, 3, 15]], 1], [[[1, 2], [3, 4]], 11], [[[1, 5, 9, 13], [2, 6, 10, 14], [3, 7, 11, 15], [4, 8, 12, 16]], 3], [[[10, 7, 14, 1], [15, 9, 8, 5], [12, 6, 2, 16], [13, 11, 3, 4]], 1], [[[1, 3, 5, 7], [9, 11, 13, 15], [2, 4, 6, 8], [10, 12, 14, 16]], 2], [[[1, 6, 13, 14], [9, 8, 7, 16], [12, 10, 2, 11], [5, 4, 3, 15]], 6], [[[10, 7, 14, 1], [15, 9, 8, 5], [12, 6, 2, 16], [13, 11, 3, 4]], 15], [[[1, 3, 5, 7], [9, 11, 13, 15], [2, 4, 6, 8], [10, 12, 14, 16]], 22], [[[1, 2], [3, 4]], 12], [[[10, 5, 20, 14, 16], [7, 11, 12, 17, 9], [13, 19, 1, 15, 8], [3, 18, 4, 6, 2], [22, 21, 23, 24, 25]], 14], [[[1, 2], [3, 4]], 9], [[[1, 2, 3, 4, 5], [6, 7, 8, 9, 10], [11, 12, 13, 14, 15], [16, 17, 18, 19, 20], [21, 22, 23, 24, 25]], 21], [[[1, 2], [3, 4]], 20], [[[10, 7, 14, 1], [15, 9, 8, 5], [12, 6, 2, 16], [13, 11, 3, 4]], 20], [[[2, 7, 4], [3, 1, 5], [6, 8, 9]], 11], [[[1, 5, 9, 13], [2, 6, 10, 14], [3, 7, 11, 15], [4, 8, 12, 16]], 4], [[[1, 5, 9, 13], [2, 6, 10, 14], [3, 7, 11, 15], [4, 8, 12, 16]], 11], [[[1, 5, 9, 13], [2, 6, 10, 14], [3, 7, 11, 15], [4, 8, 12, 16]], 13], [[[2, 7, 4], [3, 1, 5], [6, 8, 9]], 23], [[[2, 7, 4], [3, 1, 5], [6, 8, 9]], 4], [[[1, 6, 13, 14], [9, 8, 7, 16], [12, 10, 2, 11], [5, 4, 3, 15]], 25], [[[1, 3, 5, 7], [9, 11, 13, 15], [2, 4, 6, 8], [10, 12, 14, 16]], 20], [[[10, 5, 20, 14, 16], [7, 11, 12, 17, 9], [13, 19, 1, 15, 8], [3, 18, 4, 6, 2], [22, 21, 23, 24, 25]], 20], [[[10, 5, 20, 14, 16], [7, 11, 12, 17, 9], [13, 19, 1, 15, 8], [3, 18, 4, 6, 2], [22, 21, 23, 24, 25]], 1], [[[2, 7, 4], [3, 1, 5], [6, 8, 9]], 14], [[[10, 5, 20, 14, 16], [7, 11, 12, 17, 9], [13, 19, 1, 15, 8], [3, 18, 4, 6, 2], [22, 21, 23, 24, 25]], 3], [[[1, 2, 3, 4, 5], [6, 7, 8, 9, 10], [11, 12, 13, 14, 15], [16, 17, 18, 19, 20], [21, 22, 23, 24, 25]], 17]] claude-3-haiku-20240307. fail_base: True, fail_plus: True
Here's a possible solution to the problem:
```python
from collections import deque
def minPath(grid, k):
n = len(grid)
visited = set()
queue = deque([(i, j, [grid[i][j]]) for i in range(n) for j in range(n)])
min_path = None
while queue:
i, j, path = queue.popleft()
if len(path) == k:
if min_path is None or path < min_path:
min_path = path
continue
for di, dj in [(0, 1), (0, -1), (1, 0), (-1, 0)]:
ni, nj = i + di, j + dj
if 0 <= ni < n and 0 <= nj < n and (ni, nj) not in visited:
visited.add((ni, nj))
queue.append((ni, nj, path + [grid[ni][nj]]))
return min_pathcode-millenials-34b. fail_base: True, fail_plus: True
def minPath(grid, k):
n = len(grid)
visited = set()
ans = []
def dfs(x, y, k, path):
if k == 0:
ans.append(path.copy())
return
for dx, dy in (1, 0), (0, 1), (-1, 0), (0, -1):
nx, ny = x + dx, y + dy
if 0 <= nx < n and 0 <= ny < n and grid[nx][ny] not in visited:
visited.add(grid[nx][ny])
path.append(grid[nx][ny])
dfs(nx, ny, k - 1, path)
path.pop()
visited.remove(grid[nx][ny])
for i in range(n):
for j in range(n):
visited.add(grid[i][j])
dfs(i, j, k - 1, [grid[i][j]])
visited.remove(grid[i][j])
ans.sort()
return ans[0]speechless-coding-7b-16k-tora. fail_base: True, fail_plus: True
def minPath(grid, k):
n = len(grid)
dp = [[[float('inf')] * (k+1) for _ in range(n)] for _ in range(n)]
dp[0][0][0] = grid[0][0]
for i in range(1, n):
dp[0][i][0] = dp[0][i-1][0] + grid[0][i]
for i in range(1, n):
dp[i][0][0] = dp[i-1][0][0] + grid[i][0]
for i in range(1, n):
for j in range(1, n):
dp[i][j][0] = min(dp[i-1][j][0] + grid[i][j], dp[i][j-1][0] + grid[i][j])
for i in range(1, n):
for j in range(1, n):
for l in range(1, k+1):
dp[i][j][l] = min(dp[i-1][j][l], dp[i][j-1][l], dp[i-1][j-1][l-1] + grid[i][j])
return dp[n-1][n-1][k]claude-3-opus-20240229. fail_base: True, fail_plus: True
Content filering policy blocked output
mixtral-8x7b-instruct. fail_base: True, fail_plus: True
from collections import deque
def minPath(grid, k):
n = len(grid)
if not k or n * n < k:
return []
dist = [[float('inf')] * n for _ in range(n)]
dist[0][0] = 0
q = deque([(0, 0, 0, [grid[0][0]] + [0] * (k - 1))])
while q:
d, x, y, path = q.popleft()
if d > dist[x][y]:
continue
for dx, dy in ((1, 0), (-1, 0), (0, 1), (0, -1)):
nx, ny = x + dx, y + dy
if 0 <= nx < n and 0 <= ny < n:
new_path = path[:]
new_path[-1] = grid[nx][ny]
if d + 1 < dist[nx][ny]:
dist[nx][ny] = d + 1
q.append((d + 1, nx, ny, new_path))
elif d + 1 == dist[nx][ny]:
if new_path < path:
dist[nx][ny] = d + 1
q.append((d + 1, nx, ny, new_path))
return dist[-1][-1] == k and dist[n - 1][n - 1] or []mixtral-8x22b-instruct-v0.1. fail_base: True, fail_plus: True
from heapq import heappop, heappush
def minPath(grid, k):
N = len(grid)
visited = [[False]*N for _ in range(N)]
directions = [(0, 1), (0, -1), (1, 0), (-1, 0)]
heap = [(grid[0][0], 0, 0, [grid[0][0]])]
while heap:
val, x, y, path = heappop(heap)
if len(path) == k:
return path
for dx, dy in directions:
nx, ny = x + dx, y + dy
if 0 <= nx < N and 0 <= ny < N and not visited[nx][ny]:
visited[nx][ny] = True
heappush(heap, (grid[nx][ny], nx, ny, path + [grid[nx][ny]]))
visited[nx][ny] = False
# Test casescode-llama-multi-34b. fail_base: True, fail_plus: True
def minPath(grid, k):
"""
Given a grid with N rows and N columns (N >= 2) and a positive integer k,
each cell of the grid contains a value. Every integer in the range [1, N * N]
inclusive appears exactly once on the cells of the grid.
You have to find the minimum path of length k in the grid. You can start
from any cell, and in each step you can move to any of the neighbor cells,
in other words, you can go to cells which share an edge with you current
cell.
Please note that a path of length k means visiting exactly k cells (not
necessarily distinct).
You CANNOT go off the grid.
A path A (of length k) is considered less than a path B (of length k) if
after making the ordered lists of the values on the cells that A and B go
through (let's call them lst_A and lst_B), lst_A is lexicographically less
than lst_B, in other words, there exist an integer index i (1 <= i <= k)
such that lst_A[i] < lst_B[i] and for any j (1 <= j < i) we have
lst_A[j] = lst_B[j].
It is guaranteed that the answer is unique.
Return an ordered list of the values on the cells that the minimum path go through.
Examples:
Input: grid = [ [1,2,3], [4,5,6], [7,8,9]], k = 3
Output: [1, 2, 1]
Input: grid = [ [5,9,3], [4,1,6], [7,8,2]], k = 1
Output: [1]
"""
pass
if __name__ == "__main__":
print(minPath([[1,2,3], [4,5,6], [7,8,9]], 3))
print(minPath([[5,9,3], [4,1,6], [7,8,2]], 1))
Qwen--Qwen1.5-72B-Chat. fail_base: True, fail_plus: True
def minPath(grid, k):
N = len(grid)
visited = [[False] * N for _ in range(N)]
directions = [(0, 1), (0, -1), (1, 0), (-1, 0)]
def dfs(x, y, path, k):
nonlocal min_path
if len(path) == k:
min_path = path
return
visited[x][y] = True
for dx, dy in directions:
nx, ny = x + dx, y + dy
if 0 <= nx < N and 0 <= ny < N and not visited[nx][ny]:
dfs(nx, ny, path + [grid[nx][ny]], k)
visited[x][y] = False
min_path = None
for i in range(N):
for j in range(N):
dfs(i, j, [grid[i][j]], k)
return min_path
# Test cases
grid1 = [[1, 2, 3], [4, 5, 6], [7, 8, 9]]
k1 = 3mistral-large-latest. fail_base: True, fail_plus: True
Sure, here's a Python function that uses Depth-First Search (DFS) to solve the problem. The function `minPath` takes a grid and an integer `k` as input and returns the minimum path of length `k` in the grid.
```python
from collections import deque
def minPath(grid, k):
n = len(grid)
visited = [[0] * n for _ in range(n)]
directions = [(0, 1), (0, -1), (1, 0), (-1, 0)]
min_path = deque([float('inf')] * k)
def dfs(i, j, path):
if len(path) == k:
if path < min_path:
min_path[:] = path
return
visited[i][j] = 1
for dx, dy in directions:
x, y = i + dx, j + dy
if 0 <= x < n and 0 <= y < n and not visited[x][y]:
dfs(x, y, path + deque([grid[x][y]]))
visited[i][j] = 0
for i in range(n):
for j in range(n):
dfs(i, j, deque([grid[i][j]]))
return list(min_path)
```
This function works by starting a DFS from each cell in the grid. It keeps track of the current path and the minimum path found so far. When the path length reaches `k`, it checks if the current path is lexicographically smaller than the minimum path. If it is, it updates the minimum path. The function returns the minimum path as a list.HuggingFaceH4--starchat2-15b-v0.1. fail_base: True, fail_plus: True
from typing import List
import heapq
def minPath(grid: List[List[int]], k: int) -> List[int]:
n = len(grid)
directions = [(0, 1), (0, -1), (1, 0), (-1, 0)]
heap = [(grid[0][0], 0, 0, [grid[0][0]])] # (value, x, y, path)
visited = set((0, 0))
while heap:
value, x, y, path = heapq.heappop(heap)
if len(path) == k:
return path
for dx, dy in directions:
nx, ny = x + dx, y + dy
if 0 <= nx < n and 0 <= ny < n and (nx, ny) not in visited:
visited.add((nx, ny))
heapq.heappush(heap, (grid[nx][ny], nx, ny, path + [grid[nx][ny]]))
# Test cases
assert minPath([[1,2,3], [4,5,6], [7,8,9]], 3) == [1, 2, 1]
assert minPath([[5,9,3], [4,1,6], [7,8,2]], 1) == [1]
assert minPath([[1,2,3], [4,5,6], [7,8,9]], 6) == [1, 2, 1, 4, 2, 3]
assert minPath([[5,9,3], [4,1,6], [7,8,2]], 3) == [1, 4, 1]codegemma-7b. fail_base: True, fail_plus: True
This problem is a twist on the classic "shortest-path" problem with a couple of additional conditions. Not only must the path be of minimal length, but all the values in the path must be unique (no repeats).
To solve the problem, we can use a depth-first search (DFS) algorithm to find all possible paths of length k, and then return the one with the smallest sum of values.
First, we can flatten the grid into a list and sort it in ascending order. This will help us to easily identify values that are in the same position in two different paths.
Next, we can use a set to keep track of all visited values as we search the grid. This will help us to avoid visiting the same cell twice.
Finally, we can use a recursive function to find all possible paths of length k starting from each cell. We can keep track of the current path and its sum of values, and if it is currently the smallest, we can update our answer.
Here is the Python code for the above approach:
code-llama-multi-13b. fail_base: True, fail_plus: True
def minPath(grid, k):
"""
Given a grid with N rows and N columns (N >= 2) and a positive integer k,
each cell of the grid contains a value. Every integer in the range [1, N * N]
inclusive appears exactly once on the cells of the grid.
You have to find the minimum path of length k in the grid. You can start
from any cell, and in each step you can move to any of the neighbor cells,
in other words, you can go to cells which share an edge with you current
cell.
Please note that a path of length k means visiting exactly k cells (not
necessarily distinct).
You CANNOT go off the grid.
A path A (of length k) is considered less than a path B (of length k) if
after making the ordered lists of the values on the cells that A and B go
through (let's call them lst_A and lst_B), lst_A is lexicographically less
than lst_B, in other words, there exist an integer index i (1 <= i <= k)
such that lst_A[i] < lst_B[i] and for any j (1 <= j < i) we have
lst_A[j] = lst_B[j].
It is guaranteed that the answer is unique.
Return an ordered list of the values on the cells that the minimum path go through.
Examples:
Input: grid = [ [1,2,3], [4,5,6], [7,8,9]], k = 3
Output: [1, 2, 1]
Input: grid = [ [5,9,3], [4,1,6], [7,8,2]], k = 1
Output: [1]
"""
# TODO: Write your code here
pass
def test_minPath():
assert minPath([[1, 2, 3], [4, 5, 6], [7, 8, 9]], 3) == [1, 2, 1]
assert minPath([[5, 9, 3], [4, 1, 6], [7, 8, 2]], 1) == [1]
assert minPath([[1, 2, 3], [4, 5, 6], [7, 8, 9]], 2) == [1, 2]
assert minPath([[1, 2, 3], [4, 5, 6], [7, 8, 9]], 4) == [1, 2, 3, 4]
assert minPath([[1, 2, 3], [4, 5, 6], [7, 8, 9]], 5) == [1, 2, 3, 4, 5]
assert minPath([[1, 2, 3], [4, 5, 6], [7, 8, 9]], 6) == [1, 2, 3, 4, 5, 6]
assert minPath([[1, 2, 3], [4, 5, 6], [7, 8, 9]], 7) == [1, 2, 3, 4, 5, 6, 7]
assert minPath([[1, 2, 3], [4, 5, 6], [7, 8, 9]], 8) == [1, 2, 3, 4, 5, 6, 7, 8]
assert minPath([[1, 2, 3], [4, 5, 6], [7, 8, 9]], 9) == [1, 2, 3, 4, 5, 6, 7, 8, 9]
assert minPath([[1, 2, 3], [4, 5, 6],deepseek-coder-6.7b-instruct. fail_base: True, fail_plus: True
Here is the Python function `minPath` implemented in a markdown style code block:
```python
def minPath(grid, k):
"""
Given a grid with N rows and N columns (N >= 2) and a positive integer k,
each cell of the grid contains a value. Every integer in the range [1, N * N]
inclusive appears exactly once on the cells of the grid.
You have to find the minimum path of length k in the grid. You can start
from any cell, and in each step you can move to any of the neighbor cells,
in other words, you can go to cells which share an edge with you current
cell.
Please note that a path of length k means visiting exactly k cells (not
necessarily distinct).
You CANNOT go off the grid.
A path A (of length k) is considered less than a path B (of length k) if
after making the ordered lists of the values on the cells that A and B go
through (let's call them lst_A and lst_B), lst_A is lexicographically less
than lst_B, in other words, there exist an integer index i (1 <= i <= k)
such that lst_A[i] < lst_B[i] and for any j (1 <= j < i) we have
lst_A[j] = lst_B[j].
It is guaranteed that the answer is unique.
Return an ordered list of the values on the cells that the minimum path go through.
Examples:
Input: grid = [ [1,2,3], [4,5,6], [7,8,9]], k = 3
Output: [1, 2, 1]
Input: grid = [ [5,9,3], [4,1,6], [7,8,2]], k = 1
Output: [1]
"""
# Your code goes here
```
This function takes a grid and a length k as input and returns the minimum path of length k in the grid. The path is represented as a list of values on the cells that the path goes through. The function uses aopen-hermes-2.5-code-290k-13b. fail_base: True, fail_plus: True
def minPath(grid, k):
n = len(grid)
visited = [[False] * n for _ in range(n)]
min_path = []
def dfs(x, y, k, path):
if k == 0:
min_path = path[:]
return
if k == 1:
min_path.append(grid[x][y])
return
for dx, dy in [(0, 1), (0, -1), (1, 0), (-1, 0)]:
nx, ny = x + dx, y + dy
if 0 <= nx < n and 0 <= ny < n and not visited[nx][ny] and grid[nx][ny] not in path:
visited[nx][ny] = True
dfs(nx, ny, k - 1, path + [grid[nx][ny]])
visited[nx][ny] = False
for i in range(n):
for j in range(n):
visited[i][j] = True
dfs(i, j, k - 1, [grid[i][j]])
visited[i][j] = False
return min_pathspeechless-coder-ds-6.7b. fail_base: True, fail_plus: True
from heapq import heappop, heappush
def minPath(grid, k):
n = len(grid)
directions = [(0, 1), (0, -1), (1, 0), (-1, 0)]
heap = [(grid[0][0], 0, 0, [grid[0][0]])]
visited = set()
while heap:
_, x, y, path = heappop(heap)
if len(path) == k:
return path
for dx, dy in directions:
nx, ny = x + dx, y + dy
if 0 <= nx < n and 0 <= ny < n and (nx, ny) not in visited:
visited.add((nx, ny))
heappush(heap, (grid[nx][ny], nx, ny, path + [grid[nx][ny]]))
return []codegemma-7b-it. fail_base: True, fail_plus: True
def minPath(grid, k):
"""
Given a grid with N rows and N columns (N >= 2) and a positive integer k,
each cell of the grid contains a value. Every integer in the range [1, N * N]
inclusive appears exactly once on the cells of the grid.
You have to find the minimum path of length k in the grid. You can start
from any cell, and in each step you can move to any of the neighbor cells,
in other words, you can go to cells which share an edge with you current
cell.
Please note that a path of length k means visiting exactly k cells (not
necessarily distinct).
You CANNOT go off the grid.
A path A (of length k) is considered less than a path B (of length k) if
after making the ordered lists of the values on the cells that A and B go
through (let's call them lst_A and lst_B), lst_A is lexicographically less
than lst_B, in other words, there exist an integer index i (1 <= i <= k)
such that lst_A[i] < lst_B[i] and for any j (1 <= j < i) we have
lst_A[j] = lst_B[j].
It is guaranteed that the answer is unique.
Return an ordered list of the values on the cells that the minimum path go through.
Examples:
Input: grid = [ [1,2,3], [4,5,6], [7,8,9]], k = 3
Output: [1, 2, 1]
Input: grid = [ [5,9,3], [4,1,6], [7,8,2]], k = 1
Output: [1]
"""
n = len(grid)
def getNeighbors(row, col):
neighbors = []
if row > 0:
neighbors.append((row - 1, col))
if row < n - 1:
neighbors.append((row + 1, col))
if col > 0:
neighbors.append((row, col - 1))
if coldeepseek-coder-33b. fail_base: True, fail_plus: True
def minPath(grid, k):
"""
Given a grid with N rows and N columns (N >= 2) and a positive integer k,
each cell of the grid contains a value. Every integer in the range [1, N * N]
inclusive appears exactly once on the cells of the grid.
You have to find the minimum path of length k in the grid. You can start
from any cell, and in each step you can move to any of the neighbor cells,
in other words, you can go to cells which share an edge with you current
cell.
Please note that a path of length k means visiting exactly k cells (not
necessarily distinct).
You CANNOT go off the grid.
A path A (of length k) is considered less than a path B (of length k) if
after making the ordered lists of the values on the cells that A and B go
through (let's call them lst_A and lst_B), lst_A is lexicographically less
than lst_B, in other words, there exist an integer index i (1 <= i <= k)
such that lst_A[i] < lst_B[i] and for any j (1 <= j < i) we have
lst_A[j] = lst_B[j].
It is guaranteed that the answer is unique.
Return an ordered list of the values on the cells that the minimum path go through.
Examples:
Input: grid = [ [1,2,3], [4,5,6], [7,8,9]], k = 3
Output: [1, 2, 1]
Input: grid = [ [5,9,3], [4,1,6], [7,8,2]], k = 1
Output: [1]
"""
pass
mistralai--Mistral-7B-Instruct-v0.2. fail_base: True, fail_plus: True
def minPath(grid, k):
"""
Given a grid with N rows and N columns (N >= 2) and a positive integer k,
each cell of the grid contains a value. Every integer in the range [1, N * N]
inclusive appears exactly once on the cells of the grid.
You have to find the minimum path of length k in the grid. You can start
from any cell, and in each step you can move to any of the neighbor cells,
in other words, you can go to cells which share an edge with you current
cell.
Please note that a path of length k means visiting exactly k cells (not
necessarily distinct).
You CANNOT go off the grid.
A path A (of length k) is considered less than a path B (of length k) if
after making the ordered lists of the values on the cells that A and B go
through (let's call them lst_A and lst_B), lst_A is lexicographically less
than lst_B, in other words, there exist an integer index i (1 <= i <= k)
such that lst_A[i] < lst_B[i] and for any j (1 <= j < i) we have
lst_A[j] = lst_B[j].
It is guaranteed that the answer is unique.
Return an ordered list of the values on the cells that the minimum path go through.
Examples:
Input: grid = [ [1,2,3], [4,5,6], [7,8,9]], k = 3
Output: [1, 2, 1]
Input: grid = [ [5,9,3], [4,1,6], [7,8,2]], k = 1
Output: [1]
"""
import sys
from collections import defaultdict, deque
def minPath(grid, k):
N = len(grid)
visited = set()
graph = defaultdict(list)
def add_edges(i, j):
if i > 0:
graph[i-1].append(i)
graph[i].append(i-1)
if j > 0:
graph[i].append(i+graph[0].index(j))
graph[i+graph[0].index(j)].append(i)
if i < N-1:
graph[i+1].append(i)
graph[i].append(i+1)
if j < N-1:
graph[i].append(i+1+graph[0].index(j))
graph[i+1+graph[0].index(j)].append(i)
for i in range(N):
for j in range(N):
add_edges(i, j)
visited.add((i, j))
queue = deque([(0, 0, [grid[0][0]])])
while queue:
i, j, path = queue.popleft()
if len(path) == k:
return path
for neighbor in graph[i*N+j]:
if (neighbor // N, neighbor % N) not in visited:
visited.add((neighbor // N, neighbor % N))
queue.append((neighbor // N, neighbor % N, path + [grid[neighbor // N][neighbor % N]]))
return []
xwincoder-34b. fail_base: True, fail_plus: True
from heapq import heappush, heappop
def minPath(grid, k):
def neighbors(i, j):
for ni, nj in ((i-1, j), (i+1, j), (i, j-1), (i, j+1)):
if 0 <= ni < len(grid) and 0 <= nj < len(grid[0]):
yield ni, nj
def search(i, j, k):
if k == 0:
return [grid[i][j]]
heap = [(grid[i][j], i, j, 0)]
visited = set()
while heap:
cost, x, y, step = heappop(heap)
if (x, y, step) in visited:
continue
visited.add((x, y, step))
if step == k:
return [cost]
for ni, nj in neighbors(x, y):
if (ni, nj, step + 1) not in visited:
heappush(heap, (grid[ni][nj], ni, nj, step + 1))
result = []
for i in range(len(grid)):
for j in range(len(grid[0])):
result.extend(search(i, j, k))
result.sort()
return result[:k]
# Test cases
print(minPath([[1, 2, 3], [4, 5, 6], [7, 8, 9]], 3)) # Output: [1, 2, 1]
print(minPath([[5, 9, 3], [4, 1, 6], [7, 8, 2]], 1)) # Output: [1]code-13b. fail_base: True, fail_plus: True
def minPath(grid, k):
n = len(grid)
directions = [(0, 1), (0, -1), (1, 0), (-1, 0)]
visited = set()
min_path = []
def dfs(x, y, k, path):
if k == 0:
if len(path) < len(min_path):
min_path = path[:]
return
if (x, y) in visited:
return
visited.add((x, y))
for dx, dy in directions:
nx, ny = x + dx, y + dy
if 0 <= nx < n and 0 <= ny < n and (nx, ny) not in visited:
dfs(nx, ny, k - 1, path + [grid[nx][ny]])
for i in range(n):
for j in range(n):
dfs(i, j, k, [grid[i][j]])
return min_pathwizardcoder-15b. fail_base: True, fail_plus: True
Here's the Python script for the problem:
```python
def minPath(grid, k):
"""
Given a grid with N rows and N columns (N >= 2) and a positive integer k,
each cell of the grid contains a value. Every integer in the range [1, N * N]
inclusive appears exactly once on the cells of the grid.
You have to find the minimum path of length k in the grid. You can start
from any cell, and in each step you can move to any of the neighbor cells,
in other words, you can go to cells which share an edge with you current
cell.
Please note that a path of length k means visiting exactly k cells (not
necessarily distinct).
You CANNOT go off the grid.
A path A (of length k) is considered less than a path B (of length k) if
after making the ordered lists of the values on the cells that A and B go
through (let's call them lst_A and lst_B), lst_A is lexicographically less
than lst_B, in other words, there exist an integer index i (1 <= i <= k)
such that lst_A[i] < lst_B[i] and for any j (1 <= j < i) we have
lst_A[j] = lst_B[j].
It is guaranteed that the answer is unique.
Return an ordered list of the values on the cells that the minimum path go through.
"""
n = len(grid)
dp = [[[float('inf')] * n for _ in range(n)] for _ in range(k+1)]
dp[0][0][0] = 0
for i in range(n):
for j in range(n):
for l in range(k+1):
if i > 0:
dp[l][i][j] = min(dp[l][i][j], dp[l][i-1][j] + grid[i][j])
if j > 0:
dp[l][i][j] = min(dp[l][i][j], dp[l][i][j-1] + grid[i][j])
if i > 0 and j > 0:
dp[l][i][j] = min(dp[l][i][j], dp[l-1][i-1][j-1] + grid[i][j])
path = []
i, j = n-1, n-1
for l in range(k, -1, -1):
if dp[l][i][j] == float('inf'):
continue
path.append(grid[i][j])
if i > 0 and dp[l][i-1][j] + grid[i][j] == dp[l][i][j]:
i -= 1
elif j > 0 and dp[l][i][j-1] + grid[i][j] == dp[l][i][j]:
j -= 1
elif i > 0 and j > 0 and dp[l-1][i-1][j-1] + grid[i][j] == dp[l][i][j]:
i -= 1
j -= 1
return path[::-1]
```
The script uses dynamic programming to find the minimum path of length k in the grid. It uses a 3D array dp to store the minimum cost of reaching the cell (i, j) with a path of length l. The base case is dp[0][0][0] = 0, since we can start from any cell and have a path of length 0.
For each cell (i, j), we can either move to its left cell (i, j-1) or its upper cell (i-1, j) or its diagonal cell (i-1, j-1). We choose the move that results in the minimum cost. We also need to consider the case where we don't move at all, in which case dp[l][i][j] = dp[l-1][i-1][j-1].
After filling the dp array, we can backtrack to find the minimum path. We start from the bottom right cell (n-1, n-1) and move to the cell with the minimum cost, taking into account the constraints of not going off the grid and not visiting the same cell twice. We keep track of the values on the cells that the path goes through and return them in reverse order.mistral-7b-codealpaca. fail_base: True, fail_plus: True
def minPath(grid, k):
"""
Given a grid with N rows and N columns (N >= 2) and a positive integer k,
each cell of the grid contains a value. Every integer in the range [1, N * N]
inclusive appears exactly once on the cells of the grid.
You have to find the minimum path of length k in the grid. You can start
from any cell, and in each step you can move to any of the neighbor cells,
in other words, you can go to cells which share an edge with you current
cell.
Please note that a path of length k means visiting exactly k cells (not
necessarily distinct).
You CANNOT go off the grid.
A path A (of length k) is considered less than a path B (of length k) if
after making the ordered lists of the values on the cells that A and B go
through (let's call them lst_A and lst_B), lst_A is lexicographically less
than lst_B, in other words, there exist an integer index i (1 <= i <= k)
such that lst_A[i] < lst_B[i] and for any j (1 <= j < i) we have
lst_A[j] = lst_B[j].
It is guaranteed that the answer is unique.
Return an ordered list of the values on the cells that the minimum path go through.
Examples:
Input: grid = [ [1,2,3], [4,5,6], [7,8,9]], k = 3
Output: [1, 2, 1]
Input: grid = [ [5,9,3], [4,1,6], [7,8,2]], k = 1
Output: [1]
"""
def dfs(grid, x, y, path, k):
if len(path) == k:
return path
paths = []
for dx, dy in [(-1, 0), (1, 0), (0, -1), (0, 1)]:
nx, ny = x + dx, y + dy
if 0 <= nx < len(grid) and 0 <= ny < len(grid[0]) and grid[nx][ny] not in path:
path = path + [grid[nx][ny]]
paths.append(dfs(grid, nx, ny, path, k))
path = path[:-1]
return paths if not paths else []
res = []
for i in range(len(grid)):
for j in range(len(grid[0])):
paths = dfs(grid, i, j, [], k)
if paths:
res = min(res, paths, key=lambda x: (min(x), max(x)))[::-1]
return res
deepseek-coder-33b-instruct. fail_base: True, fail_plus: True
Here is the Python function that implements the problem:
```python
def minPath(grid, k):
N = len(grid)
dp = [[[float('inf')] * (k + 1) for _ in range(N)] for _ in range(N)]
path = [[[[]] * (k + 1) for _ in range(N)] for _ in range(N)]
for i in range(N):
for j in range(N):
dp[i][j][1] = grid[i][j]
path[i][j][1] = [grid[i][j]]
for l in range(2, k + 1):
for i in range(N):
for j in range(N):
for x in range(N):
for y in range(N):
if abs(i - x) + abs(j - y) == 1:
if dp[i][j][l] > dp[x][y][l - 1] + grid[i][j]:
dp[i][j][l] = dp[x][y][l - 1] + grid[i][j]
path[i][j][l] = path[x][y][l - 1] + [grid[i][j]]
min_path = min(path[i][j][k] for i in range(N) for j in range(N))
return min_path
```
This function uses dynamic programming to find the minimum path of length k in the grid. It first initializes the `dp` and `path` arrays. Then, for each length from 2 to k, it iterates over all cells in the grid and for each cell, it iterates over all its neighbors. If the current cell's value plus the minimum path to the neighbor cell is less than the current minimum path to the current cell, it updates the minimum path and the path. Finally, it returns the minimum path of length k.
<|EOT|>phi-2. fail_base: True, fail_plus: True
def minPath(grid, k):
"""
Given a grid with N rows and N columns (N >= 2) and a positive integer k,
each cell of the grid contains a value. Every integer in the range [1, N * N]
inclusive appears exactly once on the cells of the grid.
You have to find the minimum path of length k in the grid. You can start
from any cell, and in each step you can move to any of the neighbor cells,
in other words, you can go to cells which share an edge with you current
cell.
Please note that a path of length k means visiting exactly k cells (not
necessarily distinct).
You CANNOT go off the grid.
A path A (of length k) is considered less than a path B (of length k) if
after making the ordered lists of the values on the cells that A and B go
through (let's call them lst_A and lst_B), lst_A is lexicographically less
than lst_B, in other words, there exist an integer index i (1 <= i <= k)
such that lst_A[i] < lst_B[i] and for any j (1 <= j < i) we have
lst_A[j] = lst_B[j].
It is guaranteed that the answer is unique.
Return an ordered list of the values on the cells that the minimum path go through.
Examples:
Input: grid = [ [1,2,3], [4,5,6], [7,8,9]], k = 3
Output: [1, 2, 1]
Input: grid = [ [5,9,3], [4,1,6], [7,8,2]], k = 1
Output: [1]
"""
# TODO: Your code here
# TODO: Test your code with the following examples
# grid = [ [1,2,3], [4,5,6], [7,8,9] ]
# k = 3
# grid = [ [5,9,3], [4,1,6], [7,8,2] ]
# k = 1
# grid = [ [1,2,3], [4,5,6], [7,8,9] ]
# k = 2
# grid = [ [1,2,3], [4,5,6], [7,8,9] ]
# k = 3
# grid = [ [1,2,3], [4,5,6], [7,8,9] ]
# k = 4
# grid = [ [1,2,3], [4,5,6], [7,8,9] ]
# k = 5
# grid = [ [1,2,3], [4,5,6], [7,8,9] ]
# k = 6
# grid = [ [1,2,3], [4,5,6], [7,8,9] ]
# k = 7
# grid = [ [1,2,3], [4,5,6], [7,8,9] ]
# k = 8
# grid = [ [1,2,3], [4,5,6], [7,8,9] ]
# k = 9
# grid = [ [1,2,3], [4,5,6], [7,8,9] ]
# k = 10
# grid = [ [1,2,3], [4,5,6], [7,8,9] ]
# k = 11
# grid = [ [1,2,3], [4,5,6], [7,8,9] ]
# k = 12
# grid = [ [1,2,3], [4,5,6], [7,8,9] ]
# k = 13
# grid = [ [1,2,3], [4bigcode--starcoder2-15b-instruct-v0.1. fail_base: True, fail_plus: True
def minPath(grid, k):
N = len(grid)
min_path = []
min_path_val = []
for i in range(N):
for j in range(N):
curr_path = [grid[i][j]]
curr_path_val = [grid[i][j]]
dfs(grid, i, j, k-1, curr_path, curr_path_val, min_path, min_path_val)
return min_path
def dfs(grid, i, j, k, curr_path, curr_path_val, min_path, min_path_val):
if k == 0:
if min_path == [] or curr_path_val < min_path_val:
min_path[:] = curr_path[:]
min_path_val[:] = curr_path_val[:]
return
N = len(grid)
for x, y in [(i+1, j), (i-1, j), (i, j+1), (i, j-1)]:
if 0 <= x < N and 0 <= y < N and grid[x][y] not in curr_path:
curr_path.append(grid[x][y])
curr_path_val.append(grid[x][y])
dfs(grid, x, y, k-1, curr_path, curr_path_val, min_path, min_path_val)
curr_path.pop()
curr_path_val.pop()white-rabbit-neo-33b-v1. fail_base: True, fail_plus: True
def minPath(grid, k):
def dfs(i, j, path, visited):
if len(path) == k:
return path
if i < 0 or i >= len(grid) or j < 0 or j >= len(grid[0]) or visited[i][j]:
return None
visited[i][j] = True
path.append(grid[i][j])
for ni, nj in [(i+1, j), (i-1, j), (i, j+1), (i, j-1)]:
result = dfs(ni, nj, path, visited)
if result:
return result
visited[i][j] = False
path.pop()
return None
min_path = None
for i in range(len(grid)):
for j in range(len(grid[0])):
visited = [[False] * len(grid[0]) for _ in range(len(grid))]
path = dfs(i, j, [], visited)
if path and (not min_path or path < min_path):
min_path = path
return min_pathmicrosoft--Phi-3-mini-4k-instruct. fail_base: True, fail_plus: True
def minPath(grid, k):
n = len(grid)
visited = [[False] * n for _ in range(n)]
min_path = []
def dfs(x, y, path):
nonlocal min_path
visited[x][y] = True
path.append(grid[x][y])
if len(path) == k:
if not min_path or path < min_path:
min_path = path[:]
else:
for dx, dy in [(0, 1), (1, 0), (0, -1), (-1, 0)]:
nx, ny = x + dx, y + dy
if 0 <= nx < n and 0 <= ny < n and not visited[nx][ny]:
dfs(nx, ny, path)
visited[x][y] = False
for i in range(n):
for j in range(n):
dfs(i, j, [])
return min_path
# Test cases
assert minPath([[1, 2, 3], [4, 5, 6], [7, 8, 9]], 3) == [1, 2, 1]
assert minPath([[5, 9, 3], [4, 1, 6], [7, 8, 2]], 1) == [1]
assert minPath([[1, 2, 3], [4, 5, 6], [7, 8, 9]], 2) == [1, 2]
assert minPath([[1, 2, 3], [4, 5, 6], [7, 8, 9]], 4) == [1, 2, 3, 1]
assert minPath([[1, 2, 3], [4, 5, 6], [7, 8, 9]], 5) == [1, 2, 3, 4, 1]
assert minPath([[1, 2, 3], [4, 5, 6], [7, 8, 9speechless-codellama-34b. fail_base: True, fail_plus: True
def minPath(grid, k):
n = len(grid)
visited = [[False] * n for _ in range(n)]
min_path = [0] * k
min_path_val = float('inf')
def dfs(row, col, path, path_val):
nonlocal min_path, min_path_val
if len(path) == k:
if path_val < min_path_val:
min_path = path[:]
min_path_val = path_val
return
for dr, dc in ((-1, 0), (1, 0), (0, -1), (0, 1)):
new_row, new_col = row + dr, col + dc
if 0 <= new_row < n and 0 <= new_col < n and not visited[new_row][new_col]:
visited[new_row][new_col] = True
path.append(grid[new_row][new_col])
dfs(new_row, new_col, path, path_val + grid[new_row][new_col])
visited[new_row][new_col] = False
path.pop()
for row in range(n):
for col in range(n):
visited[row][col] = True
dfs(row, col, [grid[row][col]], grid[row][col])
visited[row][col] = False
return min_path
```
Explanation:
The algorithm uses a depth-first search (DFS) approach to find the minimum path of length k in the grid. It starts by marking the current cell as visited and recursively explores all its neighboring cells. If the path length reaches k, it compares the path value with the current minimum path value and updates it if necessary. After exploring all possible paths, it returns the minimum path of length k.
meta-llama-3-70b-instruct. fail_base: True, fail_plus: True
from collections import deque
def minPath(grid, k):
N = len(grid)
directions = [(0, 1), (0, -1), (1, 0), (-1, 0)]
queue = deque([(grid[0][0], [(0, 0)])])
visited = {(0, 0)}
min_path = []
while queue:
val, path = queue.popleft()
if len(path) == k:
if not min_path or val < min_path[0]:
min_path = [grid[r][c] for r, c in path]
else:
for dr, dc in directions:
r, c = path[-1][0] + dr, path[-1][1] + dc
if 0 <= r < N and 0 <= c < N and (r, c) not in visited:
visited.add((r, c))
queue.append((val * N * N + grid[r][c], path + [(r, c)]))
return min_path
# Test cases
grid1 = [[1, 2, 3], [4, 5, 6], [7, 8, 9]]
k1 = 3
print(minPath(grid1, k1)) # [1, 2, 1]
grid2 = [[5, 9, 3], [4, 1, 6], [7, 8, 2]]
k2 = 1
print(minPath(grid2, k2)) # [1]code-llama-70b-instruct. fail_base: True, fail_plus: True
def minPath(grid, k):
n = len(grid)
# Create a set to store the visited cells
visited = set()
# Create a list to store the path
path = []
# Initialize the minimum path
min_path = None
# Find the minimum path
find_min_path(grid, k, 0, 0, visited, path, min_path)
return min_path
def find_min_path(grid, k, i, j, visited, path, min_path):
# If the path is complete, check if it is the minimum path
if len(path) == k:
if min_path is None or path < min_path:
min_path = path.copy()
return
# Check if the current cell is valid
if i < 0 or i >= len(grid) or j < 0 or j >= len(grid[0]) or (i, j) in visited:
return
# Mark the current cell as visited
visited.add((i, j))
# Add the current cell to the path
path.append(grid[i][j])
# Move to the adjacent cells
find_min_path(grid, k, i-1, j, visited, path, min_path)
find_min_path(grid, k, i+1, j, visited, path, min_path)
find_min_path(grid, k, i, j-1, visited, path, min_path)
find_min_path(grid, k, i, j+1, visited, path, min_path)
# Remove the current cell from the path
path.pop()
# Mark the current cell as unvisited
visited.remove((i, j))opencodeinterpreter-ds-33b. fail_base: True, fail_plus: True
```python
def minPath(grid, k):
from heapq import heappop, heappush
from collections import defaultdict
N = len(grid)
directions = [(0, 1), (0, -1), (1, 0), (-1, 0)]
heap = [(grid[0][0], 0, 0, [grid[0][0]])]
visited = defaultdict(bool)
visited[(0, 0)] = True
while heap:
val, x, y, path = heappop(heap)
if len(path) == k:
return path
for dx, dy in directions:
nx, ny = x + dx, y + dy
if 0 <= nx < N and 0 <= ny < N and not visited[(nx, ny)]:
visited[(nx, ny)] = True
heappush(heap, (grid[nx][ny], nx, ny, path + [grid[nx][ny]]))
openchat. fail_base: True, fail_plus: True
def minPath(grid, k):
N = len(grid)
visited = set()
path = []
def dfs(x, y, path):
if len(path) == k:
return path
if (x, y) in visited:
return None
visited.add((x, y))
path.append(grid[x][y])
for dx, dy in [(-1, 0), (1, 0), (0, -1), (0, 1)]:
nx, ny = x + dx, y + dy
if 0 <= nx < N and 0 <= ny < N:
result = dfs(nx, ny, path)
if result is not None:
return result
path.pop()
return None
for i in range(N):
for j in range(N):
result = dfs(i, j, [])
if result is not None:
return result
return []
# Example usage:
grid = [[1, 2, 3], [4, 5, 6], [7, 8, 9]]
k = 3
print(minPath(grid, k)) # Output: [1, 2, 1]
grid = [[5, 9, 3], [4, 1, 6], [7, 8, 2]]
k = 1
print(minPath(grid, k)) # Output: [1]deepseek-coder-7b-instruct-v1.5. fail_base: True, fail_plus: True
def minPath(grid, k):
"""
Given a grid with N rows and N columns (N >= 2) and a positive integer k,
each cell of the grid contains a value. Every integer in the range [1, N * N]
inclusive appears exactly once on the cells of the grid.
You have to find the minimum path of length k in the grid. You can start
from any cell, and in each step you can move to any of the neighbor cells,
in other words, you can go to cells which share an edge with you current
cell.
Please note that a path of length k means visiting exactly k cells (not
necessarily distinct).
You CANNOT go off the grid.
A path A (of length k) is considered less than a path B (of length k) if
after making the ordered lists of the values on the cells that A and B go
through (let's call them lst_A and lst_B), lst_A is lexicographically less
than lst_B, in other words, there exist an integer index i (1 <= i <= k)
such that lst_A[i] < lst_B[i] and for any j (1 <= j < i) we have
lst_A[j] = lst_B[j].
It is guaranteed that the answer is unique.
Return an ordered list of the values on the cells that the minimum path go through.
Examples:
Input: grid = [ [1,2,3], [4,5,6], [7,8,9]], k = 3
Output: [1, 2, 1]
Input: grid = [ [5,9,3], [4,1,6], [7,8,2]], k = 1
Output: [1]
"""
# Initialize the direction vectors for the four possible moves
directions = [(0, 1), (1, 0), (0, -1), (-1, 0)]
# Get the size of the grid
n = len(grid)
# Initialize the queue with the starting cell
queue = [(0, 0, 0, [grid[0][0]])]
# Initialize the visited set
visited = {(0, 0)}
# Loop until the queue is empty
while queue:
# Dequeue the current cell
x, y, steps, path = queue.pop(0)
# If we have reached the destination
if steps == k:
# Return the path
return path
# For each direction
for dx, dy in directions:
# Calculate the new cell
nx, ny = x + dx, y + dy
# If the new cell is within the grid and not visited
if 0 <= nx < n and 0 <= ny < n and (nx, ny) not in visited:
# Add the new cell to the visited set
visited.add((nx, ny))
# Calculate the new path
new_path = path + [grid[nx][ny]]
# Sort the new path
new_path.sort()
# Enqueue the new cell
queue.append((nx, ny, steps + 1, new_path))
# If we have not reached the destination
return []speechless-starcoder2-7b. fail_base: True, fail_plus: True
def minPath(grid, k):
n = len(grid)
visited = [[False]*n for _ in range(n)]
directions = [(0,1),(0,-1),(1,0),(-1,0)]
min_path = []
min_val = float('inf')
def dfs(i, j, path, visited, k):
nonlocal min_val, min_path
if len(path) == k:
if min_val > path[-1]:
min_val = path[-1]
min_path = path
return
for dx, dy in directions:
x, y = i + dx, j + dy
if 0 <= x < n and 0 <= y < n and not visited[x][y]:
visited[x][y] = True
dfs(x, y, path + [grid[x][y]], visited, k)
visited[x][y] = False
for i in range(n):
for j in range(n):
visited[i][j] = True
dfs(i, j, [grid[i][j]], visited, k)
visited[i][j] = False
return min_pathwizardcoder-7b. fail_base: True, fail_plus: True
Here's a possible implementation for the `minPath` function:
def minPath(grid, k):
n = len(grid)
# Initialize a set to store the visited cells and a list to store the path
visited = set()
path = []
# Start from the first cell and add it to the path and visited set
visited.add((0, 0))
path.append(grid[0][0])
# Initialize the minimum path as a list of N zeros
min_path = [0] * n
# Initialize the minimum path length as infinity
min_path_length = float('inf')
# Initialize the current path length as 1
current_path_length = 1
# Initialize the current cell as the first cell
current_cell = (0, 0)
# Initialize the current path as the first cell value
current_path = [grid[0][0]]
# Initialize the current minimum path as the first cell value
current_min_path = [grid[0][0]]
# Initialize the minimum path length as 1
min_path_length = 1
# Initialize the direction vectors for the four possible movements
directions = [(0, 1), (1, 0), (0, -1), (-1, 0)]
# Loop until the path length is equal to k
while current_path_length < k:
# Loop through the four possible movements
for dx, dy in directions:
# Calculate the new cell coordinates
new_x, new_y = current_cell[0] + dx, current_cell[1] + dy
# Check if the new cell is within the grid boundaries
if 0 <= new_x < n and 0 <= new_y < n:
# Calculate the new cell value
new_value = grid[new_x][new_y]
# Check if the new cell is not visited and add it to the visited set
if (new_x, new_y) not in visited:
# Add the new cell to the path and visited set
visited.add((new_x, new_y))
path.append(new_value)
# Update the current path length
current_path_length += 1
# Update the current path
current_path.append(new_value)
# Check if the current path length is equal to k
if current_path_length == k:
# Check if the current path is lexicographically less than the minimum path
if current_path < current_min_path:
# Update the minimum path and its length
current_min_path = current_path
min_path_length = current_path_length
# Reset the current path and path length
current_path = [grid[new_x][new_y]]
current_path_length = 1
# Move to the new cell
current_cell = (new_x, new_y)
# Return the minimum path
return current_min_path
# Test the function with the given examples
print(minPath([[1,2,3], [4,5,6], [7,8,9]], 3)) # Output: [1, 2, 1]
print(minPath([[5,9,3], [4,1,6], [7,8,2]], 1)) # Output: [1] octocoder. fail_base: True, fail_plus: True
def minPath(grid, k):
"""
Given a grid with N rows and N columns (N >= 2) and a positive integer k,
each cell of the grid contains a value. Every integer in the range [1, N * N]
inclusive appears exactly once on the cells of the grid.
You have to find the minimum path of length k in the grid. You can start
from any cell, and in each step you can move to any of the neighbor cells,
in other words, you can go to cells which share an edge with you current
cell.
Please note that a path of length k means visiting exactly k cells (not
necessarily distinct).
You CANNOT go off the grid.
A path A (of length k) is considered less than a path B (of length k) if
after making the ordered lists of the values on the cells that A and B go
through (let's call them lst_A and lst_B), lst_A is lexicographically less
than lst_B, in other words, there exist an integer index i (1 <= i <= k)
such that lst_A[i] < lst_B[i] and for any j (1 <= j < i) we have
lst_A[j] = lst_B[j].
It is guaranteed that the answer is unique.
Return an ordered list of the values on the cells that the minimum path go through.
Examples:
Input: grid = [ [1,2,3], [4,5,6], [7,8,9]], k = 3
Output: [1, 2, 1]
Input: grid = [ [5,9,3], [4,1,6], [7,8,2]], k = 1
Output: [1]
"""
# TODO: Write your code here
pass
gemma-1.1-7b-it. fail_base: True, fail_plus: True
def minPath(grid, k):
rows, cols = len(grid), len(grid[0])
dp = [[float('inf') for _ in range(k + 1)] for _ in range(rows)]
for i in range(rows):
for j in range(cols):
dp[i][0] = grid[i][j]
for i in range(1, rows):
for j in range(cols):
for l in range(1, k + 1):
dp[i][l] = min(dp[i][l], dp[i][l - 1] + grid[i][j])
for j in range(cols):
for l in range(1, k + 1):
dp[0][l] = min(dp[0][l], dp[0][l - 1] + grid[0][j])
res = []
i, j, l = 0, 0, k
while l > 0:
res.append(grid[i][j])
if l - 1 >= 0:
dp[i][l - 1] = float('inf')
if i > 0 and dp[i - 1][l] < dp[i][l]:
i -= 1
else:
j += 1
l -= 1
return res[::-1]gpt-3.5-turbo. fail_base: True, fail_plus: True
Here is the code to solve the problem:
```python
def minPath(grid, k):
N = len(grid)
visited = [[False] * N for _ in range(N)]
path = []
def dfs(i, j, count):
if count == k:
return True
visited[i][j] = True
path.append(grid[i][j])
if i > 0 and not visited[i-1][j]:
if dfs(i-1, j, count+1):
return True
if i < N-1 and not visited[i+1][j]:
if dfs(i+1, j, count+1):
return True
if j > 0 and not visited[i][j-1]:
if dfs(i, j-1, count+1):
return True
if j < N-1 and not visited[i][j+1]:
if dfs(i, j+1, count+1):
return True
visited[i][j] = False
path.pop()
return False
for i in range(N):
for j in range(N):
if dfs(i, j, 1):
return path
```codebooga-34b. fail_base: True, fail_plus: True
def minPath(grid, k):
"""
Given a grid with N rows and N columns (N >= 2) and a positive integer k,
each cell of the grid contains a value. Every integer in the range [1, N * N]
inclusive appears exactly once on the cells of the grid.
You have to find the minimum path of length k in the grid. You can start
from any cell, and in each step you can move to any of the neighbor cells,
in other words, you can go to cells which share an edge with you current
cell.
Please note that a path of length k means visiting exactly k cells (not
necessarily distinct).
You CANNOT go off the grid.
A path A (of length k) is considered less than a path B (of length k) if
after making the ordered lists of the values on the cells that A and B go
through (let's call them lst_A and lst_B), lst_A is lexicographically less
than lst_B, in other words, there exist an integer index i (1 <= i <= k)
such that lst_A[i] < lst_B[i] and for any j (1 <= j < i) we have
lst_A[j] = lst_B[j].
It is guaranteed that the answer is unique.
Return an ordered list of the values on the cells that the minimum path go through.
Examples:
Input: grid = [ [1,2,3], [4,5,6], [7,8,9]], k = 3
Output: [1, 2, 1]
Input: grid = [ [5,9,3], [4,1,6], [7,8,2]], k = 1
Output: [1]
"""
n = len(grid)
visited = set()
min_path = []
def dfs(row, col, k, path):
if k == 0:
if len(path) < len(min_path) or (len(path) == len(min_path) and path < min_path):
min_path[:] = path
return
for r, c in ((row - 1, col), (row + 1, col), (row, col - 1), (row, col + 1)):
if 0 <= r < n and 0 <= c < n and grid[r][c] not in visited:
visited.add(grid[r][c])
path.append(grid[r][c])
dfs(r, c, k - 1, path)
path.pop()
visited.remove(grid[r][c])
for i in range(n):
for j in range(n):
visited.add(grid[i][j])
dfs(i, j, k - 1, [grid[i][j]])
visited.remove(grid[i][j])
return min_path
deepseek-coder-1.3b-instruct. fail_base: True, fail_plus: True
Here is the implementation of the function in Python:
```python
def minPath(grid, k):
N = len(grid)
visited = [[False]*N for _ in range(N)]
path = []
directions = [(0, 1), (0, -1), (1, 0), (-1, 0)]
def dfs(x, y, k):
if k == 0:
path.append(grid[x][y])
return
visited[x][y] = True
for dx, dy in directions:
nx, ny = x + dx, y + dy
if 0 <= nx < N and 0 <= ny < N and not visited[nx][ny]:
dfs(nx, ny, k-1)
visited[x][y] = False
for i in range(N):
for j in range(N):
dfs(i, j, k-1)
path.sort()
return path[:k]
```
This function uses depth-first search (DFS) to find the minimum path of length k in the grid. It starts from any cell and moves to any of the neighbor cells, and it stops when it has visited k cells. The path is then sorted and returned.
speechless-mistral-7b. fail_base: True, fail_plus: True
def minPath(grid, k):
n = len(grid)
visited = [[False for _ in range(n)] for _ in range(n)]
path = []
for i in range(n):
for j in range(n):
if not visited[i][j]:
dfs(grid, visited, path, i, j, k)
if len(path) == k:
return path
path.pop()
return []
def dfs(grid, visited, path, i, j, k):
visited[i][j] = True
path.append(grid[i][j])
if len(path) == k:
return
for dx, dy in [(-1, 0), (1, 0), (0, -1), (0, 1)]:
x, y = i + dx, j + dy
if 0 <= x < len(grid) and 0 <= y < len(grid) and not visited[x][y]:
dfs(grid, visited, path, x, y, k)
visited[i][j] = False
path.pop()
grid = [[1,2,3], [4,5,6], [7,8,9]]
k = 3
print(minPath(grid, k)) # Output: [1, 2, 1]
grid = [[5,9,3], [4,1,6], [7,8,2]]
k = 1
print(minPath(grid, k)) # Output: [1]
```
This Python script defines a function `minPath` that takes a grid and a positive integer `k` as input. The function uses depth-first search (DFS) to find the minimum path of length `k` in the grid. The DFS function `dfs` is defined recursively to explore the grid and backtrack when the path length reaches `k`. The function returns the minimum path as a list of values on the cells that the path goes through.
xdan-l1-chat. fail_base: True, fail_plus: True
from heapq import heappush, heappop
def minPath(grid, k):
n = len(grid)
visited = [[False] * n for _ in range(n)]
pq = []
path = []
for i in range(n):
for j in range(n):
heappush(pq, (grid[i][j], i, j))
while k:
value, i, j = heappop(pq)
if visited[i][j]:
continue
visited[i][j] = True
path.append(value)
k -= 1
if k == 0:
break
directions = [(i - 1, j), (i + 1, j), (i, j - 1), (i, j + 1)]
for di, dj in directions:
if 0 <= di < n and 0 <= dj < n:
heappush(pq, (grid[di][dj], di, dj))
return pathsolar-10.7b-instruct. fail_base: True, fail_plus: True
def minPath(grid, k):
"""
Given a grid with N rows and N columns (N >= 2) and a positive integer k,
each cell of the grid contains a value. Every integer in the range [1, N * N]
inclusive appears exactly once on the cells of the grid.
You have to find the minimum path of length k in the grid. You can start
from any cell, and in each step you can move to any of the neighbor cells,
in other words, you can go to cells which share an edge with you current
cell.
Please note that a path of length k means visiting exactly k cells (not
necessarily distinct).
You CANNOT go off the grid.
A path A (of length k) is considered less than a path B (of length k) if
after making the ordered lists of the values on the cells that A and B go
through (let's call them lst_A and lst_B), lst_A is lexicographically less
than lst_B, in other words, there exist an integer index i (1 <= i <= k)
such that lst_A[i] < lst_B[i] and for any j (1 <= j < i) we have
lst_A[j] = lst_B[j].
It is guaranteed that the answer is unique.
Return an ordered list of the values on the cells that the minimum path go through.
Examples:
Input: grid = [ [1,2,3], [4,5,6], [7,8,9]], k = 3
Output: [1, 2, 1]
Input: grid = [ [5,9,3], [4,1,6], [7,8,2]], k = 1
Output: [1]
"""
def minPath(grid, k):
# Initialize the result list and check for invalid inputs
result = []
if len(grid) < 2 or k < 1:
return []
# Create a dictionary to store the visited cells and their values
visited = {}
def dfs(x, y, path, visited):
# Base case: If the path length is equal to k, check if it's the minimum
if len(path) == k:
# Create a list of values in the path
path_values = [grid[i][j] for i, j in path]
# Check if the path is lexicographically smaller than any existing path
if not result or lex_smaller(path_values, result):
result = path_values
return
# Recursively explore neighboring cells
for dx, dy in [(-1, 0), (1, 0), (0, -1), (0, 1)]:
nx, ny = x + dx, y + dy
if 0 <= nx < len(grid) and 0 <= ny < len(grid[0]) and (nx, ny) not in visited:
visited[(nx, ny)] = True
path.append((nx, ny))
dfs(nx, ny, path, visited)
visited[(nx, ny)] = False
path.pop()
# Start the depth-first search from any cell
dfs(0, 0, [], visited)
# Return the result
return result
# Helper function to check if one list is lexicographically smaller than another
def lex_smaller(lst1, lst2):
for i in range(min(len(lst1), len(lst2))):
if lst1[i] < lst2[i]:
return True
if lst1[i] > lst2[i]:
return False
return len(lst1) < len(lst2)claude-3-sonnet-20240229. fail_base: True, fail_plus: True
Here's the solution in Python:
```python
from collections import deque
def minPath(grid, k):
n = len(grid)
visited = set()
queue = deque([(0, 0, [grid[0][0]])])
visited.add((0, 0))
while queue:
row, col, path = queue.popleft()
if len(path) == k:
return path
for dr, dc in [(0, 1), (0, -1), (1, 0), (-1, 0)]:
new_row, new_col = row + dr, col + dc
if (
0 <= new_row < n
and 0 <= new_col < n
and (new_row, new_col) not in visited
):
new_path = path + [grid[new_row][new_col]]
visited.add((new_row, new_col))
queue.append((new_row, new_col, new_path))
return []code-33b. fail_base: True, fail_plus: True
def minPath(grid, k):
N = len(grid)
directions = [(0, 1), (0, -1), (1, 0), (-1, 0)]
visited = [[False]*N for _ in range(N)]
ans = [grid[0][0]]
for _ in range(k-1):
ans.append(grid[0][0])
grid[0][0] = -1
for dx, dy in directions:
x, y = 0, 0
while 0 <= x < N and 0 <= y < N and not visited[x][y] and grid[x][y] != -1:
visited[x][y] = True
ans.append(grid[x][y])
grid[x][y] = -1
x += dx
y += dy
return ansstarcoder2-15b-oci. fail_base: True, fail_plus: True
def minPath(grid, k):
"""
Given a grid with N rows and N columns (N >= 2) and a positive integer k,
each cell of the grid contains a value. Every integer in the range [1, N * N]
inclusive appears exactly once on the cells of the grid.
You have to find the minimum path of length k in the grid. You can start
from any cell, and in each step you can move to any of the neighbor cells,
in other words, you can go to cells which share an edge with you current
cell.
Please note that a path of length k means visiting exactly k cells (not
necessarily distinct).
You CANNOT go off the grid.
A path A (of length k) is considered less than a path B (of length k) if
after making the ordered lists of the values on the cells that A and B go
through (let's call them lst_A and lst_B), lst_A is lexicographically less
than lst_B, in other words, there exist an integer index i (1 <= i <= k)
such that lst_A[i] < lst_B[i] and for any j (1 <= j < i) we have
lst_A[j] = lst_B[j].
It is guaranteed that the answer is unique.
Return an ordered list of the values on the cells that the minimum path go through.
Examples:
Input: grid = [ [1,2,3], [4,5,6], [7,8,9]], k = 3
Output: [1, 2, 1]
Input: grid = [ [5,9,3], [4,1,6], [7,8,2]], k = 1
Output: [1]
"""
# TODO: Implement this function
pass
```
The function `minPath` takes two parameters: a 2D array `grid` representing the grid of values and an integer `k` representing the length of the path. The function should return an ordered list of the values on the cells that the minimum path goes through.
Your task is to implement the function `minPath` to find the minimum path of length `k` in the grid. You can start from any cell and in each step you can move to any of the neighbor cells (cells that share an edge with your current cell). Please note that a path of length `k` means visiting exactly `k` cells (not necessarily distinct). You cannot go off the grid.
A path `A` (of length `k`) is considered less than a path `B` (of length `k`) if after making the ordered lists of the values on the cells that `A` and `B` go through (let's call them `lst_A` and `lst_B`), `lst_A` is lexicographically less than `lst_B`, i.e., there exists an integer index `i` (1 <= `i` <= `k`) such that `lst_A[i] < lst_B[i]` and for any `j` (1 <= `j` < `i`) we have `lst_A[j] = lst_B[j]`.
It is guaranteed that the answer is unique.
You should return an ordered list of the values on the cells that the minimum path goes through.
Note: You can assume that the input grid is always valid and the answer is unique.
Example 1:
Input: grid = [[1,2,3],[4,5,6],[7,8,9]], k = 3
Output: [1,2,1]
Explanation: The minimum path of length 3 is [1,2,1].
Example 2:
Input: grid = [[5,9,3],[4,1,6],[7,8,2]], k = 1
Output: [1]
Explanation: The minimum path of length 1 is [1].
Constraints:
- `N` is the length of the grid.
- `N` is in the range [2, 10speechless-starcoder2-15b. fail_base: True, fail_plus: True
def minPath(grid, k):
n = len(grid)
dp = [[float('inf')] * n for _ in range(n)]
dp[0][0] = grid[0][0]
for _ in range(k):
for i in range(n):
for j in range(n):
if i > 0:
dp[i][j] = min(dp[i][j], dp[i - 1][j] + grid[i][j])
if j > 0:
dp[i][j] = min(dp[i][j], dp[i][j - 1] + grid[i][j])
return dp[n - 1][n - 1]python-code-13b. fail_base: True, fail_plus: True
def minPath(grid, k):
# Helper function to get the neighbors of a cell
def getNeighbors(i, j):
neighbors = []
for dx, dy in [(0, 1), (0, -1), (1, 0), (-1, 0)]:
nx, ny = i + dx, j + dy
if 0 <= nx < len(grid) and 0 <= ny < len(grid[0]) and grid[nx][ny] != 0:
neighbors.append((nx, ny))
return neighbors
# Helper function to get the minimum path of length k starting from a cell
def getMinPath(i, j, path, visited):
if len(path) == k:
return path
for dx, dy in getNeighbors(i, j):
nx, ny = dx, dy
if not visited[nx][ny]:
visited[nx][ny] = True
path.append(grid[nx][ny])
getMinPath(nx, ny, path, visited)
visited[nx][ny] = False
# Initialize the visited array and the path
visited = [[False] * len(grid) for _ in range(len(grid))]
path = []
# Start from the top-left cell
i, j = 0, 0
visited[i][j] = True
path.append(grid[i][j])
# Try to find the minimum path of length k
while len(path) < k:
for dx, dy in getNeighbors(i, j):
nx, ny = dx, dy
if not visited[nx][ny]:
visited[nx][ny] = True
path.append(grid[nx][ny])
getMinPath(nx, ny, path, visited)
visited[nx][ny] = False
return pathopencodeinterpreter-ds-6.7b. fail_base: True, fail_plus: True
```python
def minPath(grid, k):
n = len(grid)
path = [0] * k
for i in range(k):
path[i] = grid[0][0]
if i < n:
for j in range(i+1):
if j < n:
path[i] = min(path[i], grid[j][0])
else:
for j in range(n):
path[i] = min(path[i], grid[j][0])
return path