"""
Write a function to find the maximum product formed by multiplying numbers of an increasing subsequence of that array.
assert max_product([3, 100, 4, 5, 150, 6]) == 3000
"""
def max_product(arr):
# record the correspond ending element to maintain the increasing subsequence
ret = max_ending = min_ending = (arr[0], arr[0])
for n in arr[1:]:
if n > max_ending[1]:
max_ending = max((max_ending[0] * n, n), max_ending, key=lambda x: x[0])
else:
max_ending = (n, n)
if n > min_ending[1]:
min_ending = min((min_ending[0] * n, n), min_ending, key=lambda x: x[0])
else:
min_ending = (n, n)
ret = max(ret, max_ending, min_ending, key=lambda x: x[0])
return ret[0]
[[[3, 100, 4, 5, 150, 6]], [[4, 42, 55, 68, 80]], [[10, 22, 9, 33, 21, 50, 41, 60]]] [[[-3, -100, -4, -5, -150, -6]], [[3, -5, 2, 6, -1, 4, -3]], [[2, 3, 3, 4, 5, 3, 6]], [[2, -3, 3, 4, -5, 3, -6]], [[50, 100, 200, 150, 300, 250, 400]], [[-5, -4, -3, -2, -1, 0, 1]], [[10, 9, 8, 7, 6, 5, 4, 3]], [[1, -2, 3, -4, 5, -6, 7]], [[10, 8, 6, 5, 1, 9, 7, 4, 3, 2]], [[-3, -4, 5, -2, 6, -1, 7, -8, 9, -10, 11]], [[50, 100, 200, 100, 150, 300, 250, 400, 100]], [[-3, -4, 5, -2, 5, 6, 200, 10, -1, 7, -8, 9, -10, 11]], [[10, 9, 7, 6, 4]], [[50, 250, 150, 300, 250, 400]], [[50, 250, 6, 150, 300, 250, 400]], [[-2, 3, -5, 2, 6, -1, 4, -3]], [[10, 9, 7, 6, 5, 4, 3]], [[50, 100, -3, 150, 300, 250, 400]], [[10, 9, 8, 8, 5, 4, 3, 4]], [[10, 9, 8, 7, 6, 5, -8, 3]], [[3, -5, 2, 6, 4, -3]], [[50, 250, 150, 300, 250, 1, 400]], [[2, 50, 100, 200, 100, 150, 300, 250, 400, 100]], [[2, 50, 100, 200, 100, 150, 300, 400, 100]], [[10, 9, 8, 7, 6, 5, 5, 4, 3, 9]], [[-6, 249, 6, 150, 250, 400]], [[50, 250, 150, 250, 400]], [[10, 9, 7, 7, 6, 5, 5, 4, 3, 9]], [[50, 100, -3, 150, 300, 250, 400, 100]], [[50, 100, -3, 150, 300, 250, 400, 300]], [[10, 9, 8, 7, 6, 5, 5, 4, 9]], [[3, -5, 2, -1, -6, 4, -3]], [[50, 100, -3, 150, 250, 400]], [[10, 8, 5, 1, 9, 7, 4, 3, 2, 6, 6]], [[10, 9, 8, 7, 6, 5, 4, 10]], [[50, 100, 150, 250, 400]], [[-3, -100, -4, -5, -150, -6, -4]], [[50, 100, 200, 150, 300, 250, 400, 100]], [[10, 9, 8, 7, 6, 5, 5, 4, 9, 9]], [[-2, -2, 3, -4, 5, -6, 7]], [[2, 300, 50, 100, 200, 100, 150, 250, 300, 250, 400, 100]], [[50, 100, 150, 200, 100, 150, 300, 250, 400, 100]], [[2, 50, -8, 200, 100, 150, 300, 250, 400, 100]], [[-2, 3, -4, -6, 7]], [[-4, 2, 3, 3, 4, 5, 3, 6]], [[-3, -101, -4, -5, -150, -6, -150]], [[-6, 249, 150, 250, 400]], [[-2, 4, -5, 2, 6, -1, 4, -3]], [[-3, -101, -4, -5, 6, -6, -150, -4]], [[2, -3, 3, 4, 3, -6]], [[50, 100, 200, 150, 300, 1, 400]], [[-2, -2, 3, -4, 5, 7]], [[2, 3, -5, 2, 300, 6, -1, 4, -3]], [[10, 9, 8, 7, 6, 5, -8, 4]], [[-3, -100, -5, -150, -6]], [[2, 300, 50, 100, 200, 100, 150, 250, 300, 250, 400, 100, 400]], [[-3, -100, -4, -5, 5, -6, -5]], [[3, -5, 2, 4, -3]], [[-2, -2, -6, -8, 5, -6, 7]], [[-2, -2, -4, 5, -6, 7]], [[-5, -4, -3, 9, 8, -1, 0, -1, 1]], [[10, 9, 6, 4]], [[251, 50, 100, 200, 150, 300, 250, 400, 100]], [[1, -2, 3, -4, -6, 7]], [[-3, -100, -4, -6, -150, -6, -4]], [[-2, 3, -5, 2, -1, 4, -3]], [[10, 9, 7, 6, 5, 5, 4, 3, 9]], [[-2, -2, -6, -8, -6, 7]], [[-3, 50, 100, -2, 150, 300, 250, 400]], [[10, 1, 9, 8, 7, 6, 5, 5, 4, 3, 9]], [[1, -1, -2, 3, -4, 5, -6, 7, -2]], [[10, 9, 8, 7, 6, 5, -8]], [[10, 9, -8, 7, 7, 6, 5, 5, 4, 3, 9]], [[10, 7, 9, 7, 6, 4]], [[10, 9, 1, 8, 3, 7, 6, 5, 4, 3]], [[-4, 2, 3, 3, 4, 5, 3]], [[50, 100, 200, 150, 300, 1, 3]], [[2, 2, 50, -8, 299, 200, 100, 151, 300, 250, 400, -150]], [[50, 100, 200, 100, 150, 300, 250, 400, -5, 100, 300]], [[-3, -101, -4, -5, -150, -6, -150, -4]], [[2, 300, 50, 100, 200, 100, 150, 250, 300, 250, 400, 100, 400, 2, 150]], [[-5, -3, -2, -1, 0, 1]], [[50, 250, 6, 150, -8, 300, 250, 400]], [[10, 9, 7, 6, 4, 5, 4, 3, 9]], [[10, 8, 7, 6, 5, 5, 251, 9]], [[-3, -100, -4, -5, -150, -6, -6]], [[100, 200, 150, 300, 1, 3]], [[2, 2, -3, 3, 4, -5, 3]], [[50, 100, 150, 250]], [[-5, 2, 6, 4, 250, -3]], [[2, 249, 3, 4, 3, -6, -6, -6, 3]], [[-4, 2, 3, 3, 4, 5, 5, 3, 6]], [[100, 200, 151, 150, 300, 1, 3, 200]], [[9, 8, 7, 6, 5, 4, 3]], [[2, 300, 50, 100, 200, 100, 150, 250, 300, 250, 400, 100, 400, 250]], [[3, -5, 7, 6, 4, -3]], [[2, -3, 3, 4, -5, 251, 7]], [[4, 50, 250, -4, 150, 300, 250, 400]], [[200, 151, 300, 3, 200]], [[9, 8, 7, 6, 5, 4, 3, 5]], [[2, 2, 50, -8, 299, 200, 100, 151, 300, 250, 400, -150, 2]], [[-101, -101, 10, 9, 7, 6, 4]], [[10, 9, 7, 4, 4]], [[50, 100, 100, 0, 300, 250, 400, 100, 100, 50]], [[2, 300, 50, 100, 200, 100, 150, 250, 300, 250, 100, 400, 250]], [[2, -3, 3, 4, -5, -6]], [[50, -3, 150, 300, 250, 400, 300]], [[-2, -2, -7, 5, -6, 7]], [[10, 9, 8, 7, 6, 5, 4, 2]], [[10, 9, -8, 7, 7, 6, 2, 5, 4, 50, 9]]] code-llama-multi-7b. fail_base: True, fail_plus: True
"""
Write a function to find the maximum product formed by multiplying numbers of an increasing subsequence of that array.
assert max_product([3, 100, 4, 5, 150, 6]) == 3000
"""
def max_product(arr):
if len(arr) == 0:
return 0
if len(arr) == 1:
return arr[0]
if len(arr) == 2:
return max(arr[0], arr[1])
max_product = arr[0]
min_product = arr[0]
for i in range(1, len(arr)):
if arr[i] > max_product:
max_product = arr[i]
if arr[i] < min_product:
min_product = arr[i]
if max_product < min_product:
max_product = min_product * arr[i]
return max_product
if __name__ == '__main__':
print(max_product([3, 100, 4, 5, 150, 6]))
print(max_product([1, 2, 3, 4, 5]))
print(max_product([1, 2, 3, 4, 5, 6]))
print(max_product([1, 2, 3, 4, 5, 6, 7]))
print(max_product([1, 2, 3, 4, 5, 6, 7, 8]))
print(max_product([1, 2, 3, 4, 5, 6, 7, 8, 9]))
print(max_product([1, 2, 3, 4, 5, 6, 7, 8, 9, 10]))
print(max_product([1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11]))
print(max_product([1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12]))
print(max_product([1, 2, 3, 4, 5, 6, 7,databricks--dbrx-instruct. fail_base: True, fail_plus: True
def max_product(arr):
if not arr:
return None
# Initialize maximum product and two minimum numbers in the subsequence
max_product = float('-inf')
num1 = float('-inf')
num2 = float('-inf')
for num in arr:
# If current number is greater than both minimum numbers,
# update maximum product and minimum numbers
if num > num1 and num > num2:
max_product = max(max_product, num * num1 * num2)
num2 = num1
num1 = num
# If current number is between minimum numbers,
# update maximum product and the smaller minimum number
elif num > num2 and num < num1:
max_product = max(max_product, num * num1 * num2)
num2 = num
# If current number is smaller than both minimum numbers,
# update the larger minimum number
elif num < num2:
num2 = num
return max_product
assert max_product([3, 100, 4, 5, 150, 6]) == 3000CohereForAI--c4ai-command-r-plus. fail_base: True, fail_plus: True
def max_product(nums: list[int]) -> int:
"""
Finds the maximum product formed by multiplying numbers of an increasing subsequence of that array.
Parameters:
nums (list[int]): The input array of integers.
Returns:
int: The maximum product of an increasing subsequence.
"""
if not nums:
return 0
dp = [nums[0]] * len(nums)
max_product = nums[0]
for i in range(1, len(nums)):
for j in range(i):
if nums[i] > nums[j] and nums[i] * nums[j] > nums[i]:
dp[i] = max(dp[i], dp[j] * nums[i])
elif nums[i] < nums[j]:
dp[i] = max(dp[i], dp[j])
max_product = max(max_product, dp[i])
return max_product
```claude-3-haiku-20240307. fail_base: True, fail_plus: True
Here's the solution wrapped in a Python markdown block:
```python
def max_product(arr):
"""
Find the maximum product formed by multiplying numbers of an increasing subsequence of the given array.
Args:
arr (list): The input array of integers.
Returns:
int: The maximum product formed by multiplying numbers of an increasing subsequence.
"""
n = len(arr)
dp = [1] * n # Initialize the dynamic programming array
for i in range(1, n):
for j in range(i):
if arr[i] > arr[j]:
dp[i] = max(dp[i], dp[j] * arr[i])
return max(dp)
# Example usage
assert max_product([3, 100, 4, 5, 150, 6]) == 3000code-millenials-34b. fail_base: True, fail_plus: True
def max_product(nums):
if not nums:
return 0
max_ending_here = nums[0]
min_ending_here = nums[0]
max_so_far = nums[0]
for i in range(1, len(nums)):
if nums[i] < 0:
max_ending_here, min_ending_here = min_ending_here, max_ending_here
max_ending_here = max(nums[i], max_ending_here * nums[i])
min_ending_here = min(nums[i], min_ending_here * nums[i])
max_so_far = max(max_so_far, max_ending_here)
return max_so_far
print(max_product([3, 100, 4, 5, 150, 6]))dolphin-2.6. fail_base: True, fail_plus: True
"""
Write a function to find the maximum product formed by multiplying numbers of an increasing subsequence of that array.
assert max_product([3, 100, 4, 5, 150, 6]) == 3000
"""
def max_product(arr):
if not arr:
return 0
n = len(arr)
dp = [1] * n
max_product = arr[0]
for i in range(1, n):
dp[i] = max(arr[i], dp[i-1] * arr[i])
max_product = max(max_product, dp[i])
return max_productclaude-3-opus-20240229. fail_base: True, fail_plus: True
Here's the solution to the problem wrapped in a Python markdown block:
```python
def max_product(arr):
n = len(arr)
if n == 0:
return 0
max_prod = [0] * n
max_prod[0] = arr[0]
for i in range(1, n):
max_prod[i] = arr[i]
for j in range(i):
if arr[i] > arr[j]:
max_prod[i] = max(max_prod[i], max_prod[j] * arr[i])
return max(max_prod)
assert max_product([3, 100, 4, 5, 150, 6]) == 3000mixtral-8x7b-instruct. fail_base: True, fail_plus: True
from typing import List
def max_product(nums: List[int]) -> int:
if not nums:
return 0
n = len(nums)
dp_max = [0] * n # dp_max[i] stores the maximum product ending at index i
dp_min = [0] * n # dp_min[i] stores the minimum product ending at index i
# Initialize the first element
dp_max[0] = nums[0]
dp_min[0] = nums[0]
# Calculate the maximum and minimum products for each subsequence
for i in range(1, n):
dp_max[i] = max(nums[i], nums[i] * dp_max[i - 1], nums[i] * dp_min[i - 1])
dp_min[i] = min(nums[i], nums[i] * dp_max[i - 1], nums[i] * dp_min[i - 1])
# Find the maximum product among all subsequences
max_product = max(dp_max)
return max_product
# Test the function
assert max_product([3, 100, 4, 5, 150, 6]) == 3000wizardcoder-34b. fail_base: True, fail_plus: True
Here's a possible solution:
def max_product(arr):
if len(arr) < 2:
return None
max_product = arr[0]
max_ending_here = arr[0]
min_ending_here = arr[0]
for i in range(1, len(arr)):
if arr[i] < 0:
max_ending_here, min_ending_here = min_ending_here, max_ending_here
max_ending_here = max(arr[i], max_ending_here * arr[i])
min_ending_here = min(arr[i], min_ending_here * arr[i])
max_product = max(max_product, max_ending_here)
return max_product
# Test the function
assert max_product([3, 100, 4, 5, 150, 6]) == 3000
# Test some more cases
assert max_product([1, 2, 3, 4, 5]) == 20
assert max_product([-1, 2, 3, 4, -5]) == 24
assert max_product([-1, -2, 3, 4, -5]) == 6
assert max_product([-1, -2, -3, -4, -5]) == 5
assert max_product([1, -2, 3, 4, -5, 6]) == 12
assert max_product([1, -2, -3, 4, -5, 6]) == 12
assert max_product([1, -2, -3, -4, -5, 6]) == 6
assert max_product([1, -2, -3, -4, -5, -6]) == 3
assert max_product([1, -2, 3, -4, 5, -6]) == 6
assert max_product([1, -2, 3, -4, 5, 6]) == 18
assert max_product([1, -2, 3, -4, 5, -6, 7]) == 21
assert max_product([1, -2, 3, -4, 5, -6, 7, -8]) == 21
assert max_product([1, -2, 3, -4, 5, -6, 7, 8]) == 84
assert max_product([1, -2, 3, -4, 5, -6, 7, 8, -9]) == 36
assert max_product([1, -2, 3, -4, 5, -6, 7, 8, -9, 10]) == 360
assert max_product([1, -2, 3, -4, 5, -6, 7, 8, -9, 10, -11]) == 360
assert max_product([1, -2, 3, -4, 5, -6, 7, 8, -9, 10, 11]) == 360
assert max_product([1, -2, 3, -4, 5, -6, 7, 8, -9, 10, 11, -12]) == 360
assert max_product([1, -2, 3, -4, 5, -6, 7, 8, -9, 10, 11, -12, 13]) == 420
assert max_product([1, -2, 3, -4, 5, -6, 7, 8, -9, 10, 11, -12, 13, -14]) == 420
assert max_product([1, -2, 3, -4, 5, -6, 7, 8, -9, 10, 11, -12, 13, 14]) == 504
assert max_product([1, -2, 3, -4,mixtral-8x22b-instruct-v0.1. fail_base: True, fail_plus: True
def max_product(arr):
if not arr:
return 0
min_ending_here = max_ending_here = product = arr[0]
for num in arr[1:]:
if num < 0:
min_ending_here, max_ending_here = max_ending_here, min_ending_here
min_ending_here = min(num, min_ending_here * num)
max_ending_here = max(num, max_ending_here * num)
product = max(product, max_ending_here)
return product
# Test the function
assert max_product([3, 100, 4, 5, 150, 6]) == 3000code-llama-multi-34b. fail_base: True, fail_plus: True
"""
Write a function to find the maximum product formed by multiplying numbers of an increasing subsequence of that array.
assert max_product([3, 100, 4, 5, 150, 6]) == 3000
"""
def max_product(arr):
if len(arr) == 0:
return 0
max_product = 1
for i in range(len(arr)):
max_product = max(max_product, arr[i])
arr[i] = max_product
max_product = 1
for i in range(len(arr) - 1, -1, -1):
max_product = max(max_product, arr[i])
arr[i] = max_product
return max(arr)
print(max_product([3, 100, 4, 5, 150, 6]))
Qwen--Qwen1.5-72B-Chat. fail_base: True, fail_plus: True
def max_product(arr):
if not arr:
return 0
increasing_subseq = [arr[0]]
max_product = arr[0]
for i in range(1, len(arr)):
if arr[i] > increasing_subseq[-1]:
increasing_subseq.append(arr[i])
max_product *= arr[i]
else:
max_product = max(max_product, arr[i] * increasing_subseq[0])
return max_product
# Test case
assert max_product([3, 100, 4, 5, 150, 6]) == 3000mistral-large-latest. fail_base: True, fail_plus: True
Sure, here is a Python function that solves the problem:
```python
def max_product(arr):
if not arr:
return 0
n = len(arr)
dp = [0]*n
dp[0] = arr[0]
min_so_far = arr[0]
max_product = arr[0]
for i in range(1, n):
dp[i] = max(arr[i], arr[i]*min_so_far, arr[i]*dp[i-1])
min_so_far = min(min_so_far, arr[i])
max_product = max(max_product, dp[i])
return max_product
assert max_product([3, 100, 4, 5, 150, 6]) == 3000
```
This function uses dynamic programming to solve the problem. The `dp` array stores the maximum product of an increasing subsequence ending at the current index. The `min_so_far` variable stores the minimum element seen so far, which is used to calculate the maximum product. The `max_product` variable stores the maximum product seen so far.codegen-2b. fail_base: True, fail_plus: True
"""
Write a function to find the maximum product formed by multiplying numbers of an increasing subsequence of that array.
assert max_product([3, 100, 4, 5, 150, 6]) == 3000
"""
def max_product(arr):
if len(arr) == 0:
return 0
if len(arr) == 1:
return arr[0]
if len(arr) == 2:
return max(arr[0], arr[1])
max_product_so_far = arr[0]
max_product_ending_here = arr[0]
for i in range(1, len(arr)):
max_product_ending_here = max(max_product_ending_here * arr[i], arr[i])
max_product_so_far = max(max_product_so_far, max_product_ending_here)
return max_product_so_far
print(max_product([3, 100, 4, 5, 150, 6]))
"""
Write a function to find the maximum product formed by multiplying numbers of an increasing subsequence of that array.
assert max_product([3, 100, 4, 5, 150, 6]) == 3000
"""
def max_product(arr):
if len(arr) == 0:
return 0
if len(arr) == 1:
return arr[0]
if len(arr) == 2:
return max(arr[0], arr[1])
max_product_so_far = arr[0]
max_product_ending_here = arr[0]
for i in range(1, len(arr)):
max_product_ending_here = max(max_product_ending_here * arr[i], arr[i])
max_product_so_far = max(max_product_so_far, max_product_ending_here)
return max_product_so_far
print(max_product([3, 100, 4, 5, 150, 6]))
"""
Write a function to find the maximum product formed by multiplying numbers of an increasing subsequence of that array.
assert max_product([3, 100, 4, 5, 150, 6]) == 3000
"""
def max_product(arr):
if len(arr) == 0:
return 0HuggingFaceH4--starchat2-15b-v0.1. fail_base: True, fail_plus: True
def max_product(nums):
"""
Write a function to find the maximum product formed by multiplying numbers of an increasing subsequence of that array.
assert max_product([3, 100, 4, 5, 150, 6]) == 3000
"""
if not nums:
return 0
n = len(nums)
max_product_ending_here = [0] * n
max_product_so_far = [0] * n
max_product_ending_here[0] = max_product_so_far[0] = nums[0]
for i in range(1, n):
max_product_ending_here[i] = max(nums[i], max_product_ending_here[i-1] * nums[i])
max_product_so_far[i] = max(max_product_ending_here[i], max_product_so_far[i-1])
return max_product_so_far[-1]
# Test the function
assert max_product([3, 100, 4, 5, 150, 6]) == 3000codegemma-7b. fail_base: True, fail_plus: True
"""
Write a function to find the maximum product formed by multiplying numbers of an increasing subsequence of that array.
assert max_product([3, 100, 4, 5, 150, 6]) == 3000
"""
def max_product(arr):
if not arr:
return None
if len(arr) == 1:
return arr[0]
max_product = arr[0]
for i in range(len(arr)):
for j in range(i + 1, len(arr)):
if arr[j] > arr[i]:
max_product = max(max_product, arr[i] * arr[j])
return max_product
code-llama-multi-13b. fail_base: True, fail_plus: True
"""
Write a function to find the maximum product formed by multiplying numbers of an increasing subsequence of that array.
assert max_product([3, 100, 4, 5, 150, 6]) == 3000
"""
def max_product(arr):
"""
:param arr:
:return:
"""
if len(arr) == 0:
return 0
if len(arr) == 1:
return arr[0]
if len(arr) == 2:
return arr[0] * arr[1]
max_product = arr[0] * arr[1]
min_product = arr[0] * arr[1]
for i in range(2, len(arr)):
if arr[i] > arr[i - 1]:
max_product = max(max_product, arr[i] * min_product)
min_product = min(min_product, arr[i] * min_product)
else:
max_product = max(max_product, arr[i] * max_product)
min_product = min(min_product, arr[i] * max_product)
return max_product
if __name__ == '__main__':
assert max_product([3, 100, 4, 5, 150, 6]) == 3000
assert max_product([3, 100, 4, 5, 150, 6]) == 3000
assert max_product([3, 100, 4, 5, 150, 6]) == 3000
assert max_product([3, 100, 4, 5, 150, 6]) == 3000
assert max_product([3, 100, 4, 5, 150, 6]) == 3000
assert max_product([3, 100, 4, 5, 150, 6]) == 3000
assert max_product([3, 100, 4, 5, 150, 6]) == 3000
assert max_product([3, 100, 4, 5, 150deepseek-coder-6.7b-instruct. fail_base: True, fail_plus: True
Here is the Python function in a markdown style code block:
```python
def max_product(arr):
n = len(arr)
dp = [0]*n
dp[0] = arr[0]
for i in range(1, n):
dp[i] = arr[i]
for j in range(i):
if arr[i] > arr[j]:
dp[i] = max(dp[i], dp[j]*arr[i])
return max(dp)
# Test the function
assert max_product([3, 100, 4, 5, 150, 6]) == 3000
```
This function uses dynamic programming to solve the problem. It initializes an array `dp` of the same length as the input array `arr`, and sets `dp[0]` to `arr[0]`. Then it iterates over the array from the second element to the end, and for each element `arr[i]`, it iterates over all previous elements `arr[j]` (where `j < i`). If `arr[i]` is greater than `arr[j]`, it updates `dp[i]` to be the maximum of its current value and the product of `dp[j]` and `arr[i]`. Finally, it returns the maximum value in the `dp` array.
open-hermes-2.5-code-290k-13b. fail_base: True, fail_plus: True
def max_product(arr):
if not arr:
return 0
max_product = float('-inf')
max_ending_here = 1
for i in range(1, len(arr)):
if arr[i] > arr[i - 1]:
max_ending_here *= arr[i]
max_product = max(max_product, max_ending_here)
else:
max_ending_here = 1
return max_product
print(max_product([3, 100, 4, 5, 150, 6]))codet5p-6b. fail_base: True, fail_plus: True
"""
Write a function to find the maximum product formed by multiplying numbers of an increasing subsequence of that array.
assert max_product([3, 100, 4, 5, 150, 6]) == 3000
"""
def max_product(arr):
max_product = arr[0]
min_product = arr[0]
for i in range(1, len(arr)):
if arr[i] < 0:
max_product, min_product = min_product, max_product
max_product = max(max_product * arr[i], arr[i])
min_product = min(min_product * arr[i], arr[i])
return max_product
print(max_product([3, 100, 4, 5, 150, 6])) # 3000
print(max_product([-3, -100, -4, -5, -150, -6])) # -6
print(max_product([-3, -100, -4, -5, -150, -6, -2])) # -6
print(max_product([-3, -100, -4, -5, -150, -6, -2, -1])) # -6
print(max_product([-3, -100, -4, -5, -150, -6, -2, -1, -1])) # -6
print(max_product([-3, -100, -4, -5, -150, -6, -2, -1, -1, -1])) # -6
print(max_product([-3, -100, -4, -5, -150, -6, -2, -1, -1, -1, -1])) # -6
print(max_product([-3, -100, -4, -5, -150, -6, -2, -1, -1, -1, -1, -1])) # -6
print(max_product([-3, -100, -4, -5, -150, -6, -2, -1, -1, -1, -1, -1, -1])) # -6
print(max_product([-3, -100, -4, -5, -150, -6, -2, -1, -1, -1, -1, -1, -1, -1])) # -6
print(max_codegemma-7b-it. fail_base: True, fail_plus: True
def max_product(arr):
"""
Write a function to find the maximum product formed by multiplying numbers of an increasing subsequence of that array.
assert max_product([3, 100, 4, 5, 150, 6]) == 3000
"""
max_product = arr[0]
for i in range(1, len(arr)):
if arr[i] > arr[i - 1]:
max_product *= arr[i]
return max_productmistralai--Mistral-7B-Instruct-v0.2. fail_base: True, fail_plus: True
"""
Write a function to find the maximum product formed by multiplying numbers of an increasing subsequence of that array.
assert max_product([3, 100, 4, 5, 150, 6]) == 3000
"""
def max_product(arr):
if len(arr) < 2:
return arr[0] if arr else 0
# Initialize current product and maximum product with first element of array
current_product, max_product = arr[0], arr[0]
# Iterate through the array starting from the second element
for num in arr[1:]:
# If current product is negative, reset it to the current number and update maximum product with current product
if current_product < 0:
current_product = num
max_product = max(max_product, current_product)
else:
# Multiply current product with the number and update maximum product if the product is greater
current_product *= num
max_product = max(max_product, current_product)
return max_product
# Test the function
assert max_product([3, 100, 4, 5, 150, 6]) == 3000xwincoder-34b. fail_base: True, fail_plus: True
def max_product(arr):
if len(arr) < 2:
return None
max_product = arr[0] * arr[1]
for i in range(len(arr)):
for j in range(i + 1, len(arr)):
product = arr[i] * arr[j]
if product > max_product:
max_product = product
return max_product
assert max_product([3, 100, 4, 5, 150, 6]) == 3000wizardcoder-15b. fail_base: True, fail_plus: True
Here's the Python script for the problem:
```python
def max_product(nums):
if len(nums) == 0:
return 0
max_product = nums[0]
min_product = nums[0]
max_ending_here = nums[0]
min_ending_here = nums[0]
for i in range(1, len(nums)):
if nums[i] > max_ending_here * nums[i]:
max_ending_here = nums[i]
else:
max_ending_here = max_ending_here * nums[i]
if nums[i] < min_ending_here * nums[i]:
min_ending_here = nums[i]
else:
min_ending_here = min_ending_here * nums[i]
if max_ending_here > max_product:
max_product = max_ending_here
if min_ending_here < min_product:
min_product = min_ending_here
return max_product
```
The function takes an array of integers as input and returns the maximum product formed by multiplying numbers of an increasing subsequence of that array.
The function first checks if the length of the array is zero, in which case it returns zero.
The function initializes two variables, `max_product` and `min_product`, to the first element of the array. It also initializes two variables, `max_ending_here` and `min_ending_here`, to the first element of the array.
The function then iterates through the array from the second element to the last element. For each element, the function checks if the current element is greater than the current `max_ending_here` times the current element. If it is, then the current element becomes the new `max_ending_here`. If it is not, then the current `max_ending_here` is multiplied by the current element to get a new `max_ending_here`.
The function then checks if the current element is less than the current `min_ending_here` times the current element. If it is, then the current element becomes the new `min_ending_here`. If it is not, then the current `min_ending_here` is multiplied by the current element to get a new `min_ending_here`.
The function then checks if the current `max_ending_here` is greater than the current `max_product`. If it is, then the current `max_product` is updated to the current `max_ending_here`.
The function then checks if the current `min_ending_here` is less than the current `min_product`. If it is, then the current `min_product` is updated to the current `min_ending_here`.
Finally, the function returns the `max_product`.code-llama-34b. fail_base: True, fail_plus: True
"""
Write a function to find the maximum product formed by multiplying numbers of an increasing subsequence of that array.
assert max_product([3, 100, 4, 5, 150, 6]) == 3000
"""
def max_product(arr):
max_ending = 1
max_so_far = 1
for i in arr:
max_ending = max(i, max_ending * i)
max_so_far = max(max_so_far, max_ending)
return max_so_far
deepseek-coder-33b-instruct. fail_base: True, fail_plus: True
Here is the Python function to solve the problem:
```python
def max_product(arr):
n = len(arr)
max_ending_here = 1
min_ending_here = 1
max_so_far = 1
flag = 0
for i in range(n):
if arr[i] > 0:
max_ending_here = max_ending_here * arr[i]
min_ending_here = min(min_ending_here * arr[i], 1)
flag = 1
elif arr[i] == 0:
max_ending_here = 1
min_ending_here = 1
else:
temp = max_ending_here
max_ending_here = max(min_ending_here * arr[i], 1)
min_ending_here = temp * arr[i]
if max_so_far < max_ending_here:
max_so_far = max_ending_here
if flag == 0 and max_so_far == 1:
return 0
return max_so_far
```
This function uses dynamic programming to find the maximum product of an increasing subsequence. It keeps track of the maximum and minimum product ending at each position in the array, and updates these values as it iterates through the array. The maximum product found so far is updated whenever a larger product is found. If the array contains only positive numbers, the function returns the product of all numbers. If the array contains zero, the function returns 0. If the array contains only negative numbers, the function returns the maximum negative number.
phi-2. fail_base: True, fail_plus: True
"""
Write a function to find the maximum product formed by multiplying numbers of an increasing subsequence of that array.
assert max_product([3, 100, 4, 5, 150, 6]) == 3000
"""
def max_product(arr):
n = len(arr)
dp = [1] * n
for i in range(1, n):
for j in range(i):
if arr[i] > arr[j]:
dp[i] = max(dp[i], dp[j] * arr[i])
return max(dp)
assert max_product([3, 100, 4, 5, 150, 6]) == 3000
codet5p-2b. fail_base: True, fail_plus: True
"""
Write a function to find the maximum product formed by multiplying numbers of an increasing subsequence of that array.
assert max_product([3, 100, 4, 5, 150, 6]) == 3000
"""
def max_product(arr):
max_product = 0
for i in range(len(arr)):
for j in range(i+1, len(arr)):
max_product = max(max_product, arr[i]*arr[j])
return max_product
print(max_product([3, 100, 4, 5, 150, 6]))
print(max_product([3, 4, 5, 150, 6]))
print(max_product([3, 4, 5, 150, 6, 7]))
print(max_product([3, 4, 5, 150, 6, 7, 8]))
print(max_product([3, 4, 5, 150, 6, 7, 8, 9]))
print(max_product([3, 4, 5, 150, 6, 7, 8, 9, 10]))
print(max_product([3, 4, 5, 150, 6, 7, 8, 9, 10, 11]))
print(max_product([3, 4, 5, 150, 6, 7, 8, 9, 10, 11, 12]))
print(max_product([3, 4, 5, 150, 6, 7, 8, 9, 10, 11, 12, 13]))
print(max_product([3, 4, 5, 150, 6, 7, 8, 9, 10, 11, 12, 13, 14]))
print(max_product([3, 4, 5, 150, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15]))
print(max_product([3, 4, 5, 150, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16]))
print(max_product([3, 4, 5, 150, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17]))
print(max_product([3, 4, 5, 150, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18]))
print(max_product([3, 4, 5, 150, 6, 7, 8, 9, 10, 11, 12, 13,bigcode--starcoder2-15b-instruct-v0.1. fail_base: True, fail_plus: True
def max_product(nums):
max_product = 1
max_product_so_far = 1
for num in nums:
max_product *= num
if max_product > max_product_so_far:
max_product_so_far = max_product
if num == 0:
max_product = 1
return max_product_so_farcodegemma-2b. fail_base: True, fail_plus: True
"""
Write a function to find the maximum product formed by multiplying numbers of an increasing subsequence of that array.
assert max_product([3, 100, 4, 5, 150, 6]) == 3000
"""
def max_product(arr):
if len(arr) < 2:
return 0
max_product = 0
for i in range(len(arr)):
for j in range(i + 1, len(arr)):
product = arr[i] * arr[j]
if product > max_product:
max_product = product
return max_product
white-rabbit-neo-33b-v1. fail_base: True, fail_plus: True
def max_product(nums):
if not nums:
return 0
max_ending_here = min_ending_here = max_so_far = nums[0]
for i in range(1, len(nums)):
temp_max_ending_here = max(nums[i], max_ending_here * nums[i], min_ending_here * nums[i])
min_ending_here = min(nums[i], max_ending_here * nums[i], min_ending_here * nums[i])
max_ending_here = temp_max_ending_here
max_so_far = max(max_so_far, max_ending_here)
return max_so_far
# Test the function
assert max_product([3, 100, 4, 5, 150, 6]) == 3000microsoft--Phi-3-mini-4k-instruct. fail_base: False, fail_plus: True
def max_product(arr):
"""
Find the maximum product formed by multiplying numbers of an increasing subsequence of the given array.
:param arr: List of integers
:return: Maximum product of an increasing subsequence
"""
# Initialize variables to store the maximum product and the current product
max_product = arr[0]
current_product = arr[0]
# Iterate through the array starting from the second element
for num in arr[1:]:
# If the current number is greater than the previous one, multiply it with the current product
if num > arr[arr.index(num) - 1]:
current_product *= num
else:
# If the current number is not greater than the previous one, reset the current product to the current number
current_product = num
# Update the maximum product if the current product is greater
max_product = max(max_product, current_product)
return max_product
# Test the function
assert max_product([3, 100, 4, 5, 150, 6]) == 3000speechless-codellama-34b. fail_base: True, fail_plus: True
def max_product(arr):
n = len(arr)
if n == 0:
return 0
max_ending_here = arr[0]
min_ending_here = arr[0]
max_so_far = arr[0]
for i in range(1, n):
if arr[i] > max_ending_here * arr[i]:
max_ending_here = arr[i]
else:
max_ending_here = max_ending_here * arr[i]
if arr[i] < min_ending_here * arr[i]:
min_ending_here = arr[i]
else:
min_ending_here = min_ending_here * arr[i]
max_so_far = max(max_so_far, max_ending_here)
return max_so_far
assert max_product([3, 100, 4, 5, 150, 6]) == 3000
```
Explanation:
The function `max_product` takes an array `arr` as input and returns the maximum product formed by multiplying numbers of an increasing subsequence of that array.
The function first initializes `max_ending_here` and `min_ending_here` to the first element of the array, and `max_so_far` to the first element of the array. It then iterates through the array, updating `max_ending_here` and `min_ending_here` at each step. If the current element is greater than `max_ending_here * arr[i]`, then `max_ending_here` is updated to the current element. Otherwise, `max_ending_here` is updated to `max_ending_here * arr[i]`. Similarly, if the current element is less than `min_ending_here * arr[i]`, then `min_ending_here` is updated to the current element. Otherwise, `min_ending_here` is updated to `min_ending_here * arr[i]`.
After each iteration, the function updates `max_so_far` to the maximum of `max_so_far` and `max_ending_here`. Finally, the function returns `max_so_far`, which is the maximum product formed by multiplying numbers of an increasing subsequence of the array.
meta-llama-3-70b-instruct. fail_base: True, fail_plus: True
def max_product(nums):
if not nums:
return 0
max_dp = [num for num in nums]
min_dp = [num for num in nums]
res = max_dp[0]
for i in range(1, len(nums)):
max_dp[i] = max(nums[i], max_dp[i-1] * nums[i], min_dp[i-1] * nums[i])
min_dp[i] = min(nums[i], max_dp[i-1] * nums[i], min_dp[i-1] * nums[i])
res = max(res, max_dp[i])
return res
stable-code-3b. fail_base: True, fail_plus: True
"""
Write a function to find the maximum product formed by multiplying numbers of an increasing subsequence of that array.
assert max_product([3, 100, 4, 5, 150, 6]) == 3000
"""
def max_product(arr):
if len(arr) == 0:
return 0
if len(arr) == 1:
return arr[0]
max_prod = arr[0]
max_ending_here = arr[0]
min_ending_here = arr[0]
for i in range(1, len(arr)):
max_ending_here = max(arr[i], max_ending_here * arr[i], min_ending_here * arr[i])
min_ending_here = min(arr[i], min_ending_here * arr[i], max_ending_here * arr[i])
max_prod = max(max_prod, max_ending_here)
return max_prod
print(max_product([3, 100, 4, 5, 150, 6]))
codegen-6b. fail_base: True, fail_plus: True
"""
Write a function to find the maximum product formed by multiplying numbers of an increasing subsequence of that array.
assert max_product([3, 100, 4, 5, 150, 6]) == 3000
"""
def max_product(arr):
"""
:type nums: List[int]
:rtype: int
"""
if len(arr) == 0:
return 0
if len(arr) == 1:
return arr[0]
max_ending_here = arr[0]
min_ending_here = arr[0]
max_so_far = arr[0]
for i in range(1, len(arr)):
if arr[i] < 0:
max_ending_here = max(arr[i], max_ending_here * arr[i])
min_ending_here = min(arr[i], min_ending_here * arr[i])
else:
max_ending_here = max(arr[i], max_ending_here * arr[i])
min_ending_here = min(arr[i], min_ending_here * arr[i])
max_so_far = max(max_so_far, max_ending_here)
return max_so_far
assert max_product([3, 100, 4, 5, 150, 6]) == 3000
assert max_product([3, -1, 4, 5, -6, 7]) == -6
assert max_product([-1, -2, -3, -4, -5]) == -15
assert max_product([-1, -2, -3, -4, -5, -6]) == -30
assert max_product([-1, -2, -3, -4, -5, -6, -7]) == -60
assert max_product([-1, -2, -3, -4, -5, -6, -7, -8]) == -80
assert max_product([-1, -2, -3, -4, -5, -6, -7, -8, -9]) == -90
assert max_product([-1, -2, -3, -4, -5, -6, -7, -8, -9, -10]) == -100
assert max_product([-1, -2, -3, -4, -5, -6, -7gpt-4-1106-preview. fail_base: True, fail_plus: True
"""
Write a function to find the maximum product formed by multiplying numbers of an increasing subsequence of that array.
assert max_product([3, 100, 4, 5, 150, 6]) == 3000
"""
import numpy as np
def max_product(arr):
n = len(arr)
if n == 0:
return 0
# Initialize max_product array to store the maximum product till ith element
max_product_array = np.ones(n)
# Fill max_product_array[] in bottom-up manner
for i in range(1, n):
for j in range(i):
if arr[i] > arr[j]:
max_product_array[i] = max(max_product_array[i], max_product_array[j] * arr[i])
# Return the maximum value from max_product_array[]
return int(max(max_product_array))
# Example usage
# assert max_product([3, 100, 4, 5, 150, 6]) == 3000
gemma-7b-it. fail_base: True, fail_plus: True
def max_product(arr):
arr.sort()
product = 1
for i in range(len(arr)):
for j in range(i, len(arr)):
if j - i > 0:
product = max(product, arr[i] * arr[j])
return productopencodeinterpreter-ds-33b. fail_base: True, fail_plus: True
```python
def max_product(arr):
n = len(arr)
max_prod = [0]*n
max_prod[0] = arr[0]
for i in range(1, n):
max_prod[i] = arr[i]
for j in range(i):
if arr[j] < arr[i] and max_prod[j]*arr[i] > max_prod[i]:
max_prod[i] = max_prod[j]*arr[i]
return max(max_prod)
print(max_product([3, 100, 4, 5, 150, 6]))openchat. fail_base: True, fail_plus: True
def max_product(arr):
if not arr:
return 0
max_product = [1] * len(arr)
min_product = [1] * len(arr)
for i in range(1, len(arr)):
max_product[i] = max(max_product[i-1] * arr[i], arr[i])
min_product[i] = min(min_product[i-1] * arr[i], arr[i])
return max(max_product)
# Test case
assert max_product([3, 100, 4, 5, 150, 6]) == 3000speechless-starcoder2-7b. fail_base: True, fail_plus: True
def max_product(arr):
if len(arr) < 2:
return arr[0] if arr else 0
arr.sort()
return arr[-1] * arr[-2]
assert max_product([3, 100, 4, 5, 150, 6]) == 3000wizardcoder-7b. fail_base: True, fail_plus: True
Here's a Python function that finds the maximum product of an increasing subsequence in an array:
```python
def max_product(arr):
n = len(arr)
if n == 0:
return 0
elif n == 1:
return arr[0]
elif n == 2:
return max(arr[0], arr[1])
else:
max_product = arr[0]
min_product = arr[0]
result = arr[0]
for i in range(1, n):
if arr[i] > max_product:
max_product = arr[i]
elif arr[i] < min_product:
min_product = arr[i]
result = max(result * max_product, result * min_product)
if result < 0:
result = max(result * arr[i], max_product * min_product)
return result
```
This function works by iterating through the array and keeping track of the maximum and minimum values seen so far. It also keeps track of the current result, which is the maximum product of the current subsequence. If the current number is greater than the maximum product seen so far, it becomes the new maximum product. If the current number is less than the minimum product seen so far, it becomes the new minimum product. The result is updated to be the maximum of the current result and the maximum product times the minimum product. If the result is negative, it is updated to be the maximum of the current result and the maximum product times the minimum product. This ensures that the result is always positive. Finally, the function returns the maximum of the result and the maximum product seen so far. This approach works because the maximum product of an increasing subsequence is either the current number or the maximum product times the minimum product seen so far. octocoder. fail_base: True, fail_plus: True
"""
Write a function to find the maximum product formed by multiplying numbers of an increasing subsequence of that array.
assert max_product([3, 100, 4, 5, 150, 6]) == 3000
"""
def max_product(arr):
if len(arr) == 0:
return 0
if len(arr) == 1:
return arr[0]
max_product = arr[0]
min_product = arr[0]
for i in range(1, len(arr)):
if arr[i] > 0:
max_product = max(max_product, arr[i], arr[i] * max_product, arr[i] * min_product)
min_product = min(min_product, arr[i], arr[i] * max_product, arr[i] * min_product)
else:
max_product = max(max_product, arr[i], arr[i] * min_product, arr[i] * max_product)
min_product = min(min_product, arr[i], arr[i] * min_product, arr[i] * max_product)
return max_product
gemma-1.1-7b-it. fail_base: True, fail_plus: True
def max_product(arr):
if not arr:
return 0
max_product = arr[0]
current_product = arr[0]
for i in range(1, len(arr)):
current_product = max(arr[i], current_product * arr[i])
max_product = max(max_product, current_product)
return max_product
assert max_product([3, 100, 4, 5, 150, 6]) == 3000deepseek-coder-6.7b-base. fail_base: True, fail_plus: True
"""
Write a function to find the maximum product formed by multiplying numbers of an increasing subsequence of that array.
assert max_product([3, 100, 4, 5, 150, 6]) == 3000
"""
def max_product(arr):
if len(arr) == 0:
return 0
if len(arr) == 1:
return arr[0]
if len(arr) == 2:
return max(arr[0] * arr[1], arr[0], arr[1])
max_product_arr = [0] * len(arr)
max_product_arr[0] = arr[0]
max_product_arr[1] = max(arr[0] * arr[1], arr[0], arr[1])
for i in range(2, len(arr)):
max_product_arr[i] = max(max_product_arr[i - 1] * arr[i], arr[i])
return max(max_product_arr)
if __name__ == "__main__":
assert max_product([3, 100, 4, 5, 150, 6]) == 3000
assert max_product([10, 22, 9, 33, 21, 50, 41, 60, 80]) == 1860400
assert max_product([1, 2, 3, 4, 5, 6, 7, 8, 9, 10]) == 3628800
assert max_product([10, 22, 9, 33, 21, 50, 41, 60, 80, 1]) == 2432902008176640000
assert max_product([1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20]) == 1307674368000
assert max_product([1, 2, 3, 4, 5, codegen-16b. fail_base: True, fail_plus: True
"""
Write a function to find the maximum product formed by multiplying numbers of an increasing subsequence of that array.
assert max_product([3, 100, 4, 5, 150, 6]) == 3000
"""
def max_product(nums):
"""
:type nums: List[int]
:rtype: int
"""
if len(nums) == 0:
return 0
if len(nums) == 1:
return nums[0]
max_so_far = nums[0]
min_so_far = nums[0]
max_product_so_far = nums[0]
for i in range(1, len(nums)):
max_so_far = max(max_so_far * nums[i], nums[i], min_so_far * nums[i])
min_so_far = min(min_so_far * nums[i], nums[i], max_so_far)
max_product_so_far = max(max_product_so_far, max_so_far)
return max_product_so_far
def max_product_dp(nums):
"""
:type nums: List[int]
:rtype: int
"""
if len(nums) == 0:
return 0
if len(nums) == 1:
return nums[0]
max_so_far = nums[0]
min_so_far = nums[0]
max_product_so_far = nums[0]
dp = [0] * len(nums)
dp[0] = nums[0]
for i in range(1, len(nums)):
dp[i] = max(max_so_far * nums[i], nums[i], min_so_far * nums[i])
max_so_far = max(max_so_far, dp[i])
min_so_far = min(min_so_far, dp[i])
max_product_so_far = max(max_product_so_far, max_so_far)
return max_product_so_far
def max_product_dp_space_optimized(nums):
mistral-7b. fail_base: True, fail_plus: True
"""
Write a function to find the maximum product formed by multiplying numbers of an increasing subsequence of that array.
assert max_product([3, 100, 4, 5, 150, 6]) == 3000
"""
def max_product(arr):
if len(arr) == 1:
return arr[0]
max_product = arr[0]
for i in range(1, len(arr)):
if arr[i] > max_product:
max_product = arr[i]
elif arr[i] < max_product:
max_product *= arr[i]
return max_product
assert max_product([3, 100, 4, 5, 150, 6]) == 3000speechless-mistral-7b. fail_base: True, fail_plus: True
def max_product(arr):
if len(arr) == 0:
return 0
max_product = arr[0]
min_product = arr[0]
for i in range(1, len(arr)):
if arr[i] < 0:
max_product, min_product = min_product, max_product
max_product = max(arr[i], max_product * arr[i])
min_product = min(arr[i], min_product * arr[i])
return max(max_product, min_product)
assert max_product([3, 100, 4, 5, 150, 6]) == 3000
```
This Python script defines a function `max_product` that takes an array of integers as input and returns the maximum product that can be formed by multiplying numbers of an increasing subsequence of that array. The function uses two variables `max_product` and `min_product` to keep track of the maximum and minimum products seen so far. It iterates over the array and updates `max_product` and `min_product` accordingly. Finally, it returns the maximum of `max_product` and `min_product`. The assert statement at the end checks if the function works correctly for the given input.
codet5p-16b. fail_base: True, fail_plus: True
"""
Write a function to find the maximum product formed by multiplying numbers of an increasing subsequence of that array.
assert max_product([3, 100, 4, 5, 150, 6]) == 3000
"""
def max_product(arr):
max_product = 1
for i in range(len(arr)):
for j in range(i+1, len(arr)):
if arr[j] > arr[i]:
max_product = max(max_product, arr[j]*arr[i])
return max_product
assert max_product([3, 100, 4, 5, 150, 6]) == 3000
assert max_product([3, 100, 4, 5, 150, 6, 100, 6]) == 3000
assert max_product([3, 100, 4, 5, 150, 6, 100, 6, 100, 6]) == 3000
assert max_product([3, 100, 4, 5, 150, 6, 100, 6, 100, 6, 100, 6, 100, 6]) == 3000
assert max_product([3, 100, 4, 5, 150, 6, 100, 6, 100, 6, 100, 6, 100, 6, 100, 6, 100, 6, 100, 6, 100, 6, 100, 6, 100, 6, 100, 6, 100, 6, 100, 6, 100, 6, 100, 6, 100, 6, 100, 6, 100, 6, 100, 6, 100, 6, 100, 6, 100, 6, 100, 6, 100, 6, 100, 6, 100, 6, 100, 6, 100, 6, 100, 6, 100, 6, 100, 6, 100, 6, 100, 6, 100, 6, 100, 6, 100, 6, 100, 6, 100, 6, 100, 6, 100, 6, 100, 6, 100, 6, 100, 6, 100, 6, 100, 6, 100, 6, 100, 6, 100, 6, 100, 6, 100, 6, 100, 6, 100, 6, 100, 6, 100, 6, 100, 6, 100, 6, 100, 6, 100, 6, 100, 6, 100, 6, 100, 6, 100, 6, 100, 6, 100, 6, 100, 6, 100, 6, 100, 6, 100, 6, 100, 6, 100, 6, 100, 6, 100, 6, 100, 6, 100, 6, 100, 6, 100, 6code-llama-13b. fail_base: True, fail_plus: True
"""
Write a function to find the maximum product formed by multiplying numbers of an increasing subsequence of that array.
assert max_product([3, 100, 4, 5, 150, 6]) == 3000
"""
def max_product(arr):
if len(arr) == 0:
return 0
max_prod = arr[0]
curr_max = arr[0]
curr_min = arr[0]
for i in range(1, len(arr)):
curr_num = arr[i]
if curr_num > curr_max:
curr_max = curr_num
elif curr_num < curr_min:
curr_min = curr_num
max_prod = max(max_prod, curr_max * curr_num, curr_min * curr_num)
return max_prod
print(max_product([3, 100, 4, 5, 150, 6]))
xdan-l1-chat. fail_base: True, fail_plus: True
def max_product(arr):
if not arr:
return 0
n = len(arr)
max_so_far = arr[0]
max_product = arr[0]
for i in range(1, n):
if arr[i] >= max_so_far:
max_so_far = arr[i]
else:
max_product = max(max_product, max_so_far * arr[i])
max_so_far = arr[i]
return max(max_product, max_so_far)
assert max_product([3, 100, 4, 5, 150, 6]) == 3000solar-10.7b-instruct. fail_base: True, fail_plus: True
"""
Write a function to find the maximum product formed by multiplying numbers of an increasing subsequence of that array.
assert max_product([3, 100, 4, 5, 150, 6]) == 3000
"""
def max_product(arr):
n = len(arr)
max_product = 1
dp = [arr[0]]
for i in range(1, n):
max_product = max(max_product, arr[i] * dp[-1])
while i > 0 and arr[i] < dp[-1]:
dp.pop()
dp.append(arr[i])
return max_product
assert max_product([3, 100, 4, 5, 150, 6]) == 3000gemma-2b. fail_base: True, fail_plus: True
"""
Write a function to find the maximum product formed by multiplying numbers of an increasing subsequence of that array.
assert max_product([3, 100, 4, 5, 150, 6]) == 3000
"""
def max_product(arr):
max_product = 0
for i in range(len(arr)):
for j in range(i + 1, len(arr)):
product = arr[i] * arr[j]
if product > max_product:
max_product = product
return max_product
claude-3-sonnet-20240229. fail_base: True, fail_plus: True
Here's the Python code to solve the problem, wrapped in a markdown block:
```python
def max_product(arr):
"""
Find the maximum product formed by multiplying numbers of an increasing subsequence of the given array.
Args:
arr (list): A list of integers.
Returns:
int: The maximum product formed by multiplying numbers of an increasing subsequence.
"""
if not arr:
return 0
max_so_far = arr[0]
min_val = arr[0]
for i in range(1, len(arr)):
if arr[i] < min_val:
min_val = arr[i]
elif arr[i] > max_so_far:
prev_max = max_so_far
max_so_far = max(max_so_far * arr[i], min_val * arr[i])
min_val = min(min_val, prev_max)
return max_so_far
# Test case
assert max_product([3, 100, 4, 5, 150, 6]) == 3000starcoder2-15b-oci. fail_base: True, fail_plus: True
"""
Write a function to find the maximum product formed by multiplying numbers of an increasing subsequence of that array.
assert max_product([3, 100, 4, 5, 150, 6]) == 3000
"""
def max_product(nums):
max_product = 1
max_product_sequence = []
for num in nums:
if num > max_product:
max_product *= num
max_product_sequence.append(num)
return max_product, max_product_sequence
starcoder2-3b. fail_base: True, fail_plus: True
"""
Write a function to find the maximum product formed by multiplying numbers of an increasing subsequence of that array.
assert max_product([3, 100, 4, 5, 150, 6]) == 3000
"""
def max_product(arr):
max_product = 1
for i in range(len(arr)):
for j in range(i, len(arr)):
product = 1
for k in range(i, j+1):
product *= arr[k]
max_product = max(max_product, product)
return max_product
print(max_product([3, 100, 4, 5, 150, 6]))
/Arrays/max_sum_subarray.py
"""
Given an array of integers, find the maximum sum of a subarray within the array.
Example:
Input: [-2,1,-3,4,-1,2,1,-5,4],
Output: 6
Explanation: [4,-1,2,1] has the largest sum = 6.
"""
def max_sum_subarray(arr):
max_sum = arr[0]
current_sum = arr[0]
for i in range(1, len(arr)):
current_sum = max(arr[i], current_sum + arr[i])
max_sum = max(max_sum, current_sum)
return max_sum
print(max_sum_subarray([-2,1,-3,4,-1,2,1,-5,4]))
/Arrays/max_sum_subarray_k.py
"""
Given an array of integers and a number k, find the maximum sum of k consecutive elements in the array.
Example:
Input: [2, 1, 5, 1, 3, 2], k=3
Output: 9
Explanation: Subarray with maximum sum is [5, 1, 3].
"""
def max_sum_subarray_k(arr, k):
max_sum = 0
for i in range(k):
max_sum += arr[i]
current_sum = max_sum
for i in range(k, len(arr)):
current_sum = max(arr[i], current_sum + arr[i])
max_sum = max(max_sum, current_sum)
return max_sum
print(max_sum_subarray_k([speechless-starcoder2-15b. fail_base: True, fail_plus: True
def max_product(nums):
if not nums:
return 0
max_product = nums[0]
min_product = nums[0]
max_result = nums[0]
for i in range(1, len(nums)):
if nums[i] < 0:
max_product, min_product = min_product, max_product
max_product = max(nums[i], max_product * nums[i])
min_product = min(nums[i], min_product * nums[i])
max_result = max(max_result, max_product)
return max_result
assert max_product([3, 100, 4, 5, 150, 6]) == 3000gemma-7b. fail_base: True, fail_plus: True
"""
Write a function to find the maximum product formed by multiplying numbers of an increasing subsequence of that array.
assert max_product([3, 100, 4, 5, 150, 6]) == 3000
"""
def max_product(arr):
if not arr:
return 0
dp = [1] * len(arr)
for i in range(1, len(arr)):
for j in range(i):
if arr[i] > arr[j]:
dp[i] = max(dp[i], arr[i] * arr[j] * dp[j])
return max(dp)
opencodeinterpreter-ds-6.7b. fail_base: True, fail_plus: True
```python
def max_product(arr):
n = len(arr)
if n < 2:
return None
max_ending_here = max_ending_at_i = arr[0]
min_ending_here = min_ending_at_i = arr[0]
max_product_so_far = arr[0]
for i in range(1, n):
temp = max(arr[i], max_ending_here * arr[i], min_ending_here * arr[i])
min_ending_here = min(arr[i], max_ending_here * arr[i], min_ending_here * arr[i])
max_ending_here = temp
max_product_so_far = max(max_product_so_far, max_ending_at_i)
return max_product_so_far
# Test the function
print(max_product([3, 100, 4, 5, 150, 6]))