"""
Write a python function to find the element that appears only once in a sorted array.
assert search([1,1,2,2,3]) == 3
"""
def search(arr):
n = len(arr)
XOR = 0
for i in range(n) :
XOR = XOR ^ arr[i]
return (XOR)
[[[1, 1, 2, 2, 3]], [[1, 1, 3, 3, 4, 4, 5, 5, 7, 7, 8]], [[1, 2, 2, 3, 3, 4, 4]]] [[[]], [[2, 2, 2, 2, 2, 2]], [[1, 1, 1, 1, 1, 1]], [[6]], [[1, 2, 3, 4, 5, 6]], [[1, 1, 2, 2, 3, 3, 4, 5, 5, 6, 6, 7, 7, 8, 8, 9, 9, 10, 10, 11, 11, 12, 12, 13, 13, 14, 14, 15]], [[5]], [[1, 1, 4, 4, 5, 6]], [[0, 1, 4, 4, 5, 6]], [[4, 5, 5]], [[4, 4, 5]], [[1, 1, 4, 4, 6]], [[0, 1, 4, 4, 5]], [[2, 2, 2, 2, 2]], [[1, 1, 4, 6]], [[0, 4, 4, 5, 6]], [[4, 5]], [[1, 1, 4, 4, 5]], [[1, 1, 1, 1, 1]], [[1, 2, 4, 6]], [[5, 5]], [[4, 5, 5, 5]], [[1, 1, 1, 1, 1, 1, 1]], [[1, 1, 1, 1, 1, 1, 1, 1]], [[2, 2, 2, 2, 2, 2, 2, 2]], [[2, 2, 2, 2, 2, 2, 2, 2, 2]], [[1, 1, 1]], [[1, 1, 4, 4]], [[0, 1, 4, 4, 8]], [[2, 2, 2, 2, 2, 2, 2, 15]], [[4, 4, 4]], [[1, 1, 2, 2, 3, 3, 4, 5, 5, 6, 6, 7, 8, 8, 9, 9, 10, 10, 11, 11, 12, 12, 13, 13, 14, 14, 15]], [[2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2]], [[False, False]], [[6, 6, 6]], [[1, 4, 4]], [[5, 5, 5]], [[4, 5, 5, 5, 5]], [[1, 4]], [[1, 1]], [[6, 6]], [[1, 1, 1, 1, 1, 1, 1, 1, 1]], [[2, 2, 2, 2, 2, 2, 2, 2, 15]], [[1, 2, 3, 3, 4, 5, 6]], [[0, 4, 4, 8]], [[1, 1, 4]], [[1, 4, 6]], [[0, 1, 1, 1, 1, 1, 1, 1, 1]], [[4, 6]], [[False, True]], [[1, 1, 11]], [[1, 1, 2]], [[0, 1, 4, 8]], [[0, 4, 4, 4, 6, 6]], [[2, 2, 2, 2, 2, 2, 2, 2, 2, 15]], [[1, 5]], [[4, 4, 4, 4, 4]], [[3, 5, 5]], [[5, 5, 5, 5]], [[1, 1, 2, 2, 3, 3, 4, 5, 5, 6, 6, 7, 7, 8, 8, 9, 9, 10, 10, 11, 11, 12, 12, 13, 13, 14, 15, 15]], [[1, 3, 5, 5, 6]], [[0, 4, 8]], [[4, 4, 6]], [[0, 1, 4, 4, 6]], [[1, 1, 4, 4, 5, 5]], [[1, 1, 2, 2]], [[0, 1, 1, 1, 1, 1, 1, 1, 1, 1]], [[1, 1, 11, 11]], [[True]], [[4, 4]], [[False, True, True, True, True, True]], [[5, 11]], [[1, 1, 1, 4, 4]], [[4, 4, 4, 4]], [[2, 2, 2, 2, 2, 2, 2]], [[1, 2, 4, 6, 6]], [[4, 4, 8]], [[0, 2, 2, 2, 2]], [[1, 2, 3, 3, 11]], [[4, 4, 6, 6]], [[3, 5]], [[2, 5]], [[0, 1, 1, 1, 1, 1, 1, 1]], [[3, 6]], [[1, 4, 4, 6]], [[0, 2, 2, 2]], [[0, 1, 1]], [[1, 3, 4, 7]], [[1, 4, 6, 6]], [[1, 4, 4, 4]], [[False]], [[1, 1, 10, 11, 11]], [[0, 4, 5, 5]], [[0, 1, 4, 5, 6]], [[1, 4, 4, 5, 6]], [[1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1]], [[1, 1, 6, 6, 6]], [[4]], [[1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1]], [[1, 1, 12]], [[0, 2, 6, 6]], [[9]], [[1, 1, 1, 4, 4, 4]], [[1, 11, 11]], [[1, 1, 4, 4, 5, 9]], [[3, 6, 6]], [[1, 3, 3, 5, 13]]] code-llama-multi-7b. fail_base: True, fail_plus: True
"""
Write a python function to find the element that appears only once in a sorted array.
assert search([1,1,2,2,3]) == 3
"""
def search(arr):
if len(arr) == 1:
return arr[0]
mid = len(arr) // 2
if arr[mid] == arr[mid - 1]:
if arr[mid] == arr[mid + 1]:
return search(arr[:mid])
else:
return arr[mid]
elif arr[mid] == arr[mid + 1]:
return search(arr[mid + 1:])
else:
return arr[mid]
if __name__ == '__main__':
print(search([1, 1, 2, 2, 3]))
print(search([1, 1, 2, 2, 3, 3]))
print(search([1, 1, 2, 2, 3, 3, 4]))
print(search([1, 1, 2, 2, 3, 3, 4, 4]))
print(search([1, 1, 2, 2, 3, 3, 4, 4, 5]))
print(search([1, 1, 2, 2, 3, 3, 4, 4, 5, 5]))
print(search([1, 1, 2, 2, 3, 3, 4, 4, 5, 5, 6]))
print(search([1, 1, 2, 2, 3, 3, 4, 4, 5, 5, 6, 6]))
print(search([1, 1, 2, 2, 3, 3, 4, 4, 5, 5, 6, 6, 7]))
print(search([1, 1, 2, 2, 3, 3, 4, 4, 5, 5, 6, 6, 7, 7]))
print(search([1, 1, 2, 2, 3, 3, 4, 4, 5, 5, 6, databricks--dbrx-instruct. fail_base: True, fail_plus: True
def search(arr):
"""
This function finds the element that appears only once in a sorted array.
:param arr: list
:return: int
"""
if len(arr) == 1:
return arr[0]
left, right = 0, len(arr) - 1
while left <= right:
mid = (left + right) // 2
if mid > 0 and arr[mid] == arr[mid - 1]:
if (mid - 1) % 2 == 0:
right = mid - 2
else:
left = mid + 1
elif mid < len(arr) - 1 and arr[mid] == arr[mid + 1]:
if (mid + 1) % 2 == 0:
right = mid - 1
else:
left = mid + 2
else:
return arr[mid]
assert search([1, 1, 2, 2, 3]) == 3CohereForAI--c4ai-command-r-plus. fail_base: False, fail_plus: True
def search(arr):
low = 0
high = len(arr) - 1
while low < high:
mid = (low + high) // 2
if arr[mid] == arr[mid ^ 1]:
low = mid + 1
else:
high = mid
return arr[low]
```claude-3-haiku-20240307. fail_base: False, fail_plus: True
Here's the solution wrapped in a Python markdown block:
```python
def search(arr):
"""
Find the element that appears only once in a sorted array.
Args:
arr (list): A sorted list of integers.
Returns:
int: The element that appears only once in the array.
"""
left, right = 0, len(arr) - 1
while left < right:
mid = left + (right - left) // 2
if mid % 2 == 1:
mid -= 1
if arr[mid] == arr[mid + 1]:
left = mid + 2
else:
right = mid
return arr[left]code-millenials-34b. fail_base: True, fail_plus: True
def search(arr):
for i in range(0, len(arr), 2):
if arr[i] != arr[i+1]:
return arr[i]
# Testing the function
assert search([1,1,2,2,3]) == 3dolphin-2.6. fail_base: True, fail_plus: True
"""
Write a python function to find the element that appears only once in a sorted array.
assert search([1,1,2,2,3]) == 3
"""
def search(arr):
for i in range(len(arr) - 1):
if arr[i] != arr[i + 1]:
return arr[i]
return arr[-1]
# Test the function
assert search([1,1,2,2,3]) == 3claude-3-opus-20240229. fail_base: False, fail_plus: True
Here's the code to find the element that appears only once in a sorted array, wrapped in a markdown block:
```python
def search(arr):
left = 0
right = len(arr) - 1
while left < right:
mid = left + (right - left) // 2
if mid % 2 == 1:
mid -= 1
if arr[mid] == arr[mid + 1]:
left = mid + 2
else:
right = mid
return arr[left]
mixtral-8x7b-instruct. fail_base: True, fail_plus: True
def search(arr):
# Initialize left and right pointers
left, right = 0, len(arr) - 1
# Iterate while left pointer is less than or equal to right pointer
while left <= right:
# Calculate the middle index
mid = (left + right) // 2
# If the middle element is equal to the previous element
if mid > 0 and arr[mid] == arr[mid - 1]:
# Return the next element (which is guaranteed to be the only unique element)
return arr[mid + 1]
# If the middle element is equal to the next element
elif mid < len(arr) - 1 and arr[mid] == arr[mid + 1]:
# Return the previous element (which is guaranteed to be the only unique element)
return arr[mid - 1]
# Update left or right pointer based on comparison with middle element
if arr[left] < arr[mid]:
left = mid + 1
else:
right = mid - 1
# If no unique element is found, return None
return None
# Test the function
assert search([1,1,2,2,3]) == 3wizardcoder-34b. fail_base: False, fail_plus: True
Here's a possible solution:
def search(arr):
"""
Finds the element that appears only once in a sorted array of integers.
"""
if len(arr) == 0:
return None
elif len(arr) == 1:
return arr[0]
elif len(arr) == 2:
if arr[0] == arr[1]:
return None
else:
return arr[0] if arr[0] != arr[1] else arr[1]
else:
if arr[0] != arr[1]:
return arr[0]
elif arr[-1] != arr[-2]:
return arr[-1]
else:
for i in range(1, len(arr)-1):
if arr[i] != arr[i-1] and arr[i] != arr[i+1]:
return arr[i]
return None
# Test the function
assert search([1,1,2,2,3]) == 3 mixtral-8x22b-instruct-v0.1. fail_base: False, fail_plus: True
def search(arr):
"""
This function finds the element that appears only once in a sorted array.
"""
# Iterate over the array with a step of 2
for i in range(0, len(arr) - 1, 2):
# If the current element is not equal to the next one, return it
if arr[i] != arr[i + 1]:
return arr[i]
# If the array has an odd length, return the last element
if len(arr) % 2 != 0:
return arr[-1]
# If no element appears only once, return None
return None
# Test the function
assert search([1, 1, 2, 2, 3]) == 3code-llama-multi-34b. fail_base: False, fail_plus: True
"""
Write a python function to find the element that appears only once in a sorted array.
assert search([1,1,2,2,3]) == 3
"""
def search(arr):
"""
:param arr: sorted array
:return: element that appears only once
"""
for i in range(0, len(arr), 2):
if i + 1 == len(arr):
return arr[i]
if arr[i] != arr[i + 1]:
return arr[i]
if __name__ == '__main__':
assert search([1, 1, 2, 2, 3]) == 3
Qwen--Qwen1.5-72B-Chat. fail_base: True, fail_plus: True
def search(arr):
if not arr:
return None
left, right = 0, len(arr) - 1
while left < right:
mid = (left + right) // 2
if arr[mid] == arr[mid + 1]:
left = mid + 2
else:
right = mid
return arr[left]
# Test case
assert search([1, 1, 2, 2, 3]) == 3mistral-large-latest. fail_base: False, fail_plus: True
Sure, here is a Python function that solves the problem:
```python
def search(sorted_array):
"""
This function takes a sorted array as input and returns the element that appears only once in the array.
If all elements appear twice, it returns None.
"""
if len(sorted_array) == 1:
return sorted_array[0]
for i in range(0, len(sorted_array), 2):
if i == len(sorted_array) - 1 or sorted_array[i] != sorted_array[i+1]:
return sorted_array[i]
return None
# Test the function
assert search([1,1,2,2,3]) == 3
```
This function works by iterating over the sorted array in steps of 2. If the current element is not equal to the next element, or if we've reached the last element, it returns the current element. If it iterates over the entire array without finding an element that appears only once, it returns None.codegen-2b. fail_base: False, fail_plus: True
"""
Write a python function to find the element that appears only once in a sorted array.
assert search([1,1,2,2,3]) == 3
"""
def search(arr):
for i in range(len(arr)):
if arr.count(arr[i]) == 1:
return arr[i]
print(search([1,1,2,2,3]))
"""
Write a python function to find the element that appears only once in a sorted array.
assert search([1,1,2,2,3]) == 3
"""
def search(arr):
for i in range(len(arr)):
if arr.count(arr[i]) == 1:
return arr[i]
print(search([1,1,2,2,3]))
"""
Write a python function to find the element that appears only once in a sorted array.
assert search([1,1,2,2,3]) == 3
"""
def search(arr):
for i in range(len(arr)):
if arr.count(arr[i]) == 1:
return arr[i]
print(search([1,1,2,2,3]))
"""
Write a python function to find the element that appears only once in a sorted array.
assert search([1,1,2,2,3]) == 3
"""
def search(arr):
for i in range(len(arr)):
if arr.count(arr[i]) == 1:
return arr[i]
print(search([1,1,2,2,3]))
"""
Write a python function to find the element that appears only once in a sorted array.
assert search([1,1,2,2,3]) == 3
"""
def search(arr):
for i in range(len(arr)):
if arr.count(arr[i]) == 1:
return arr[i]
print(search([1,1,2,2,3]))
"""
Write a python function to find the element that appears only once in a sorted array.
assert search([1,1,2,2,3]) == 3
"""
def search(arr):
for i in range(len(arr)):
if arr.count(arr[i]) == 1:HuggingFaceH4--starchat2-15b-v0.1. fail_base: False, fail_plus: True
def search(nums):
"""
Write a python function to find the element that appears only once in a sorted array.
assert search([1,1,2,2,3]) == 3
"""
left, right = 0, len(nums) - 1
while left < right:
mid = left + (right - left) // 2
if mid % 2 == 1:
mid -= 1
if nums[mid] == nums[mid + 1]:
left = mid + 2
else:
right = mid
return nums[left]
# Test the function
assert search([1,1,2,2,3]) == 3
assert search([1,1,2,2,3,3,4,4,5]) == 5
assert search([1,2,2,3,3,4,4,5,5]) == 1codegemma-7b. fail_base: True, fail_plus: True
"""
Write a python function to find the element that appears only once in a sorted array.
assert search([1,1,2,2,3]) == 3
"""
import math
def search(arr):
if not arr:
return -1
if len(arr) == 1:
return arr[0]
if len(arr) == 2:
if arr[0] == arr[1]:
return -1
else:
return arr[0]
mid = len(arr)//2
if mid % 2 == 0:
if arr[mid] == arr[mid - 1]:
return search(arr[mid + 1:])
else:
return search(arr[:mid])
else:
if arr[mid] == arr[mid - 1]:
return search(arr[:mid - 1])
else:
return search(arr[mid + 1:])
code-llama-multi-13b. fail_base: False, fail_plus: True
"""
Write a python function to find the element that appears only once in a sorted array.
assert search([1,1,2,2,3]) == 3
"""
def search(arr):
"""
:param arr:
:return:
"""
if len(arr) == 1:
return arr[0]
elif len(arr) == 2:
return arr[0] if arr[0] != arr[1] else arr[1]
else:
return arr[0] if arr[0] != arr[1] else search(arr[2:])
if __name__ == '__main__':
assert search([1, 1, 2, 2, 3]) == 3
assert search([1, 1, 2, 2, 3, 3]) == 1
assert search([1, 1, 2, 2, 3, 3, 3]) == 2
assert search([1, 1, 2, 2, 3, 3, 3, 3]) == 1
assert search([1, 1, 2, 2, 3, 3, 3, 3, 3]) == 2
assert search([1, 1, 2, 2, 3, 3, 3, 3, 3, 3]) == 1
assert search([1, 1, 2, 2, 3, 3, 3, 3, 3, 3, 3]) == 2
assert search([1, 1, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3]) == 1
assert search([1, 1, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3]) == 2
assert search([1, 1, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3]) == 1
assert search([1, 1, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3,deepseek-coder-6.7b-instruct. fail_base: False, fail_plus: True
Here is the Python function in a markdown style code block:
```python
def search(nums):
left, right = 0, len(nums) - 1
while left < right:
mid = (left + right) // 2
if nums[mid] == nums[mid ^ 1]:
left = mid + 1
else:
right = mid
return nums[left]
# Test the function
assert search([1,1,2,2,3]) == 3
```
This function uses binary search to find the element that appears only once in a sorted array. The function works by comparing the middle element with its next element. If they are the same, it means that the single element is on the right half of the array. If they are not the same, it means that the single element is on the left half of the array. The function continues to divide the array in half until it finds the single element.
open-hermes-2.5-code-290k-13b. fail_base: False, fail_plus: True
def search(arr):
# Create a dictionary to store the count of each element
count_dict = {}
for num in arr:
if num in count_dict:
count_dict[num] += 1
else:
count_dict[num] = 1
# Iterate through the dictionary to find the element with count 1
for key, value in count_dict.items():
if value == 1:
return key
# If no element with count 1 is found, return None
return None
# Test the function
print(search([1, 1, 2, 2, 3])) # Output: 3codet5p-6b. fail_base: True, fail_plus: True
"""
Write a python function to find the element that appears only once in a sorted array.
assert search([1,1,2,2,3]) == 3
"""
def search(arr):
if len(arr) == 0:
return None
if len(arr) == 1:
return arr[0]
if arr[0] == arr[1]:
return search(arr[1:])
else:
return arr[0]
assert search([1,1,2,2,3]) == 3
assert search([1,1,2,2,3,3]) == 3
assert search([1,1,2,2,3,3,3]) == 3
assert search([1,1,2,2,3,3,3,3]) == 3
assert search([1,1,2,2,3,3,3,3,3]) == 3
assert search([1,1,2,2,3,3,3,3,3,3]) == 3
assert search([1,1,2,2,3,3,3,3,3,3,3]) == 3
assert search([1,1,2,2,3,3,3,3,3,3,3,3]) == 3
assert search([1,1,2,2,3,3,3,3,3,3,3,3,3]) == 3
assert search([1,1,2,2,3,3,3,3,3,3,3,3,3,3,3,3]) == 3
assert search([1,1,2,2,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3]) == 3
assert search([1,1,2,2,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3]) == 3
assert search([1,1,2,2,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3]) == 3
assert search([1,1codegemma-7b-it. fail_base: True, fail_plus: True
def search(arr):
for i in range(len(arr)):
if i == 0 or arr[i] != arr[i-1]:
return arr[i]mistralai--Mistral-7B-Instruct-v0.2. fail_base: True, fail_plus: True
"""
Write a python function to find the element that appears only once in a sorted array.
assert search([1,1,2,2,3]) == 3
"""
def find_unique_element(arr):
"""
Find the element that appears only once in a sorted array.
"""
if len(arr) == 1:
return arr[0]
left, right = 0, len(arr) - 1
while left <= right:
mid = (left + right) // 2
if arr[mid] != arr[mid + 1]:
return arr[mid]
left += 1
arr = [1, 1, 2, 2, 3]
assert find_unique_element(arr) == 3xwincoder-34b. fail_base: True, fail_plus: True
def search(arr):
for i in range(len(arr)):
if i == 0 or i == len(arr) - 1:
if arr[i] != arr[i - 1] and arr[i] != arr[i + 1]:
return arr[i]
elif arr[i] != arr[i - 1] and arr[i] != arr[i + 1]:
return arr[i]
return None
assert search([1, 1, 2, 2, 3]) == 3wizardcoder-15b. fail_base: True, fail_plus: True
Here's the Python function to find the element that appears only once in a sorted array:
```python
def search(arr):
# Initialize variables
count = 0
result = None
# Traverse the array
for i in range(len(arr)):
# If the count is 0, store the current element as the result
if count == 0:
result = arr[i]
# If the count is 1, return the result
elif count == 1:
return result
# If the current element is equal to the result, increment the count
elif arr[i] == result:
count += 1
# If the current element is not equal to the result, decrement the count
else:
count -= 1
# If the loop completes without finding the element, return None
return None
```
The function works by traversing the array and keeping track of the count of each element. If the count is 0, the current element is stored as the result. If the count is 1, the result is returned. If the current element is equal to the result, the count is incremented. If the current element is not equal to the result, the count is decremented. If the loop completes without finding the element, None is returned.code-llama-34b. fail_base: False, fail_plus: True
"""
Write a python function to find the element that appears only once in a sorted array.
assert search([1,1,2,2,3]) == 3
"""
def search(arr):
if len(arr) == 1:
return arr[0]
if arr[0] != arr[1]:
return arr[0]
if arr[-1] != arr[-2]:
return arr[-1]
return search(arr[1:-1])
deepseek-coder-33b-instruct. fail_base: False, fail_plus: True
Here is the Python function to find the element that appears only once in a sorted array:
```python
def search(arr):
low = 0
high = len(arr) - 1
while low < high:
mid = (low + high) // 2
if arr[mid] == arr[mid ^ 1]:
low = mid + 1
else:
high = mid
return arr[low]
```
This function uses binary search to find the element that appears only once in a sorted array. The binary search is performed on the indices of the array. The mid index is calculated as the average of the low and high indices. If the element at the mid index is equal to the element at the mid index XOR 1, then the single element is on the right side of the mid index, so the low index is updated to mid + 1. Otherwise, the single element is on the left side of the mid index, so the high index is updated to mid. The process continues until the low index is equal to the high index, at which point the single element is found.
phi-2. fail_base: True, fail_plus: True
"""
Write a python function to find the element that appears only once in a sorted array.
assert search([1,1,2,2,3]) == 3
"""
def search(arr):
"""
Time complexity: O(log n)
Space complexity: O(1)
"""
if len(arr) == 0:
return None
if len(arr) == 1:
return arr[0]
if arr[0] == arr[1]:
return search(arr[1:])
if arr[-1] == arr[-2]:
return search(arr[:-1])
if arr[0]!= arr[-1]:
return arr[0]
assert search([1,1,2,2,3]) == 3
assert search([1,1,2,2,3,3]) == None
assert search([1,1,2,2,3,3,4]) == 4
assert search([1,1,2,2,3,3,4,4]) == None
assert search([1,1,2,2,3,3,4,4,5]) == 5
assert search([1,1,2,2,3,3,4,4,5,5]) == None
assert search([1,1,2,2,3,3,4,4,5,5,6]) == 6
assert search([1,1,2,2,3,3,4,4,5,5,6,6]) == None
assert search([1,1,2,2,3,3,4,4,5,5,6,6,7]) == 7
assert search([1,1,2,2,3,3,4,4,5,5,6,6,7,7]) == None
assert search([1,1,2,2,3,3,4,4,5,5,6,6,7,7,8]) == 8
assert search([1,1,2,2,3,3,4,4,5,5,6,6,7,7,8,8]) == None
assert search([1,1,2,2,3,3,4,4,5,5,6,6,7,7,8,8,9]) == 9
assert search([1,1,2,2,3,3,4,4codet5p-2b. fail_base: False, fail_plus: True
"""
Write a python function to find the element that appears only once in a sorted array.
assert search([1,1,2,2,3]) == 3
"""
def search(arr):
for i in range(len(arr)):
if arr.count(arr[i]) == 1:
return arr[i]
print(search([1,1,2,2,3]))
"""
Write a python function to find the element that appears only once in a sorted array.
assert search([1,1,2,2,3]) == 3
"""
def search(arr):
for i in range(len(arr)):
if arr.count(arr[i]) == 1:
return arr[i]
print(search([1,1,2,2,3]))
"""
Write a python function to find the element that appears only once in a sorted array.
assert search([1,1,2,2,3]) == 3
"""
def search(arr):
for i in range(len(arr)):
if arr.count(arr[i]) == 1:
return arr[i]
print(search([1,1,2,2,3]))
"""
Write a python function to find the element that appears only once in a sorted array.
assert search([1,1,2,2,3]) == 3
"""
def search(arr):
for i in range(len(arr)):
if arr.count(arr[i]) == 1:
return arr[i]
print(search([1,1,2,2,3]))
"""
Write a python function to find the element that appears only once in a sorted array.
assert search([1,1,2,2,3]) == 3
"""
def search(arr):
for i in range(len(arr)):
if arr.count(arr[i]) == 1:
return arr[i]
print(search([1,1,2,2,3]))
"""
Write a python function to find the element that appears only once in a sorted array.
assert search([1,1,2,2,3]) == 3
"""
def search(arr):
for i in range(len(arr)):
if arr.count(arr[i]) == 1:bigcode--starcoder2-15b-instruct-v0.1. fail_base: True, fail_plus: True
def search(nums):
left, right = 0, len(nums) - 1
while left < right:
mid = (left + right) // 2
if nums[mid] == nums[mid - 1]:
if (mid - left) % 2 == 0:
left = mid + 1
else:
right = mid - 1
elif nums[mid] == nums[mid + 1]:
if (right - mid) % 2 == 0:
right = mid - 1
else:
left = mid + 1
else:
return nums[mid]
return nums[left]codegemma-2b. fail_base: True, fail_plus: True
"""
Write a python function to find the element that appears only once in a sorted array.
assert search([1,1,2,2,3]) == 3
"""
def search(arr):
if len(arr) == 1:
return arr[0]
if len(arr) == 2:
if arr[0] == arr[1]:
return arr[0]
else:
return arr[1]
if arr[0] != arr[1]:
return arr[0]
else:
return search(arr[1:])
<|file_separator|>white-rabbit-neo-33b-v1. fail_base: False, fail_plus: True
def search(arr):
left, right = 0, len(arr) - 1
while left < right:
mid = (left + right) // 2
if mid % 2 == 0:
if arr[mid] == arr[mid + 1]:
left = mid + 2
else:
right = mid
else:
if arr[mid] == arr[mid - 1]:
left = mid + 1
else:
right = mid - 1
return arr[left]
# Test the function
assert search([1,1,2,2,3]) == 3microsoft--Phi-3-mini-4k-instruct. fail_base: True, fail_plus: True
def find_single_element(arr):
"""
Find the element that appears only once in a sorted array.
:param arr: List of integers sorted in non-decreasing order
:return: The element that appears only once
"""
# Initialize pointers
left, right = 0, len(arr) - 1
while left < right:
mid = (left + right) // 2
# If the middle element is even, move the left pointer
if mid % 2 == 0:
if arr[mid] == arr[mid + 1]:
left = mid + 2
else:
right = mid
# If the middle element is odd, move the right pointer
else:
if arr[mid] == arr[mid - 1]:
left = mid + 1
else:
right = mid - 1
return arr[left]
# Test the function
assert find_single_element([1, 1, 2, 2, 3]) == 3speechless-codellama-34b. fail_base: True, fail_plus: True
def search(arr):
for i in range(0, len(arr), 2):
if i+1 < len(arr) and arr[i] != arr[i+1]:
return arr[i]
return -1
assert search([1,1,2,2,3]) == 3
```
Explanation:
The function `search` takes a sorted array as input and returns the element that appears only once in the array. The function iterates through the array in steps of 2, comparing each element with the next one. If the two elements are not equal, it means that the current element appears only once, so it is returned. If no such element is found, the function returns -1.
meta-llama-3-70b-instruct. fail_base: False, fail_plus: True
def search(arr):
low, high = 0, len(arr) - 1
while low < high:
mid = (low + high) // 2
if mid % 2 == 1:
mid -= 1
if arr[mid] == arr[mid+1]:
low = mid + 2
else:
high = mid
return arr[low]
assert search([1,1,2,2,3]) == 3stable-code-3b. fail_base: True, fail_plus: True
"""
Write a python function to find the element that appears only once in a sorted array.
assert search([1,1,2,2,3]) == 3
"""
def search(arr):
if len(arr) == 1:
return arr[0]
if len(arr) == 2:
if arr[0] == arr[1]:
return -1
else:
return arr[0]
if arr[0] == arr[1]:
return search(arr[2:])
if arr[1] == arr[2]:
return search(arr[3:])
if arr[0] == arr[2]:
return search(arr[3:])
if arr[0] == arr[3]:
return search(arr[4:])
if arr[1] == arr[3]:
return search(arr[4:])
if arr[1] == arr[4]:
return search(arr[5:])
if arr[2] == arr[3]:
return search(arr[4:])
if arr[2] == arr[4]:
return search(arr[5:])
if arr[3] == arr[4]:
return search(arr[5:])
if arr[3] == arr[5]:
return search(arr[6:])
if arr[4] == arr[5]:
return search(arr[6:])
if arr[4] == arr[6]:
return search(arr[7:])
if arr[5] == arr[6]:
return search(arr[7:])
if arr[5] == arr[7]:
return search(arr[8:])
if arr[6] == arr[7]:
return search(arr[8:])
if arr[6] == arr[8]:
return search(arr[9:])
if arr[7] == arr[8]:
return search(arr[9:])
if arr[7] == arr[9]:
return search(arr[10:])
if arr[8] == arr[9]:
return search(arr[10:])
if arr[8] == arr[10]:
return search(arr[11:])
if arr[9] == arr[10codegen-6b. fail_base: True, fail_plus: True
"""
Write a python function to find the element that appears only once in a sorted array.
assert search([1,1,2,2,3]) == 3
"""
def search(arr):
"""
:param arr: array of integers
:return: integer
"""
if len(arr) == 0:
return None
if len(arr) == 1:
return arr[0]
if len(arr) == 2:
if arr[0]!= arr[1]:
return arr[0]
else:
return arr[1]
mid = len(arr) // 2
if arr[mid] == arr[mid - 1]:
return search(arr[:mid])
elif arr[mid] == arr[mid + 1]:
return search(arr[mid + 1:])
else:
return arr[mid]
assert search([1,1,2,2,3]) == 3
assert search([1,1,2,2,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3gpt-4-1106-preview. fail_base: False, fail_plus: True
"""
Write a python function to find the element that appears only once in a sorted array.
assert search([1,1,2,2,3]) == 3
"""
def search(arr):
# Initialize left and right pointers
left, right = 0, len(arr) - 1
# Binary search approach
while left < right:
mid = left + (right - left) // 2
# Check if the mid is at the start of a pair
if mid % 2 == 0:
# If the next element is the same, the single element is on the right
if arr[mid] == arr[mid + 1]:
left = mid + 2
else:
right = mid
else:
# If the previous element is the same, the single element is on the right
if arr[mid] == arr[mid - 1]:
left = mid + 1
else:
right = mid - 1
# When left == right, we have found the single element
return arr[left]
# Example usage
assert search([1,1,2,2,3]) == 3
gemma-7b-it. fail_base: False, fail_plus: True
def search(arr):
# Create a dictionary to store the count of each element
count = {}
for num in arr:
# If the element is not in the dictionary, add it with a count of 1
if num not in count:
count[num] = 1
# Otherwise, increment the count of the element by 1
else:
count[num] += 1
# Find the element that has a count of 1
for num, c in count.items():
if c == 1:
return numopencodeinterpreter-ds-33b. fail_base: False, fail_plus: True
```python
def search(arr):
for i in range(0, len(arr), 2):
if i == len(arr) - 1 or arr[i] != arr[i+1]:
return arr[i]
return -1
# Test the function
print(search([1,1,2,2,3])) # Output: 3openchat. fail_base: True, fail_plus: True
def search(arr):
i = 0
while i < len(arr) - 1:
if arr[i] != arr[i+1]:
return arr[i]
i += 1
return arr[-1]
# Test case
assert search([1,1,2,2,3]) == 3speechless-starcoder2-7b. fail_base: True, fail_plus: True
def find_unique(arr):
# Initialize two pointers, one at the start and one at the end of the array
left, right = 0, len(arr) - 1
# While the left pointer is less than the right pointer
while left < right:
# If the middle element is not equal to the left and right elements, return the middle element
if arr[left] != arr[right] and arr[left] != arr[(left + right) // 2]:
return arr[(left + right) // 2]
# If the middle element is equal to the left element, move the left pointer to the middle + 1
elif arr[left] == arr[(left + right) // 2]:
left = (left + right) // 2 + 1
# If the middle element is equal to the right element, move the right pointer to the middle - 1
else:
right = (left + right) // 2 - 1
# If the left pointer is equal to the right pointer, return the left pointer
return arr[left]
# Test the function
assert find_unique([1,1,2,2,3]) == 3wizardcoder-7b. fail_base: True, fail_plus: True
Here's a possible solution to the problem:
```python
def search(arr):
# Initialize variables to keep track of the previous and current elements
prev = arr[0]
curr = arr[1]
# Loop through the array starting from the second element
for i in range(1, len(arr)):
# If the current element is different from the previous element, return it
if curr != prev:
return curr
# Otherwise, update the previous and current elements
prev = curr
curr = arr[i+1]
# If we reach the end of the array without finding a different element, return the last element
return curr
```
This solution works by iterating through the array and comparing each element with the previous one. If they are different, we know that the current element is the one that appears only once. If they are the same, we continue iterating until we find a different element or reach the end of the array. If we reach the end of the array without finding a different element, the last element is the one that appears only once. The time complexity of this solution is O(n), where n is the length of the array. octocoder. fail_base: True, fail_plus: True
"""
Write a python function to find the element that appears only once in a sorted array.
assert search([1,1,2,2,3]) == 3
"""
def search(arr):
if len(arr) == 1:
return arr[0]
if len(arr) == 2:
return arr[0] if arr[0] != arr[1] else arr[1]
if len(arr) == 3:
return arr[0] if arr[0] != arr[1] and arr[0] != arr[2] else arr[1] if arr[1] != arr[2] else arr[2]
if len(arr) == 4:
return arr[0] if arr[0] != arr[1] and arr[0] != arr[2] and arr[0] != arr[3] else arr[1] if arr[1] != arr[2] and arr[1] != arr[3] else arr[2] if arr[2] != arr[3] else arr[3]
if len(arr) == 5:
return arr[0] if arr[0] != arr[1] and arr[0] != arr[2] and arr[0] != arr[3] and arr[0] != arr[4] else arr[1] if arr[1] != arr[2] and arr[1] != arr[3] and arr[1] != arr[4] else arr[2] if arr[2] != arr[3] and arr[2] != arr[4] else arr[3] if arr[3] != arr[4] else arr[4]
if len(arr) == 6:
return arr[0] if arr[0] != arr[1] and arr[0] != arr[2] and arr[0] != arr[3] and arr[0] != arr[4] and arr[0] != arr[5] else arr[1] if arr[1] != arr[2] and arr[1] != arr[3] and arr[1] != arr[4] and arr[1] != arr[5] else arr[2] if arr[2] != arr[3] and arr[2] != arr[4] and arr[2] != arr[5] else arr[3] if arr[3] != arr[4] and arr[3] != arr[5] else arr[4gemma-1.1-7b-it. fail_base: True, fail_plus: True
def search(arr):
left, right = 0, len(arr) - 1
while left <= right:
mid = left + (right - left) // 2
if arr[mid] == arr[mid + 1]:
left = mid + 2
else:
right = mid - 1
return arr[left]deepseek-coder-6.7b-base. fail_base: False, fail_plus: True
"""
Write a python function to find the element that appears only once in a sorted array.
assert search([1,1,2,2,3]) == 3
"""
def search(arr):
if len(arr) == 1:
return arr[0]
if arr[0] != arr[1]:
return arr[0]
if arr[-1] != arr[-2]:
return arr[-1]
for i in range(1, len(arr) - 1):
if arr[i] != arr[i - 1] and arr[i] != arr[i + 1]:
return arr[i]
assert search([1, 1, 2, 2, 3]) == 3
assert search([1, 1, 2, 3, 3, 4, 4, 8, 8]) == 2
assert search([3, 3, 7, 7, 10, 11, 11]) == 10codegen-16b. fail_base: True, fail_plus: True
"""
Write a python function to find the element that appears only once in a sorted array.
assert search([1,1,2,2,3]) == 3
"""
def search(arr):
"""
:param arr: array of integers
:return: integer
"""
if len(arr) == 0:
return None
if len(arr) == 1:
return arr[0]
if len(arr) == 2:
if arr[0]!= arr[1]:
return arr[0]
else:
return arr[1]
mid = len(arr) // 2
if arr[mid] == arr[mid - 1]:
return search(arr[:mid])
elif arr[mid] == arr[mid + 1]:
return search(arr[mid + 1:])
else:
return arr[mid]
assert search([1,1,2,2,3]) == 3
assert search([1,1,2,2,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3mistral-7b. fail_base: True, fail_plus: True
"""
Write a python function to find the element that appears only once in a sorted array.
assert search([1,1,2,2,3]) == 3
"""
def search(arr):
if len(arr) == 1:
return arr[0]
else:
mid = len(arr)//2
if arr[mid] == arr[mid-1] and arr[mid] == arr[mid+1]:
return search(arr[:mid])
elif arr[mid] == arr[mid-1] and arr[mid] != arr[mid+1]:
return search(arr[mid+1:])
elif arr[mid] != arr[mid-1] and arr[mid] == arr[mid+1]:
return search(arr[:mid])
else:
return arr[mid]
print(search([1,1,2,2,3]))speechless-mistral-7b. fail_base: True, fail_plus: True
def find_unique(arr):
for i in range(len(arr)-1):
if arr[i] != arr[i+1]:
return arr[i]
return arr[-1]
assert find_unique([1,1,2,2,3]) == 3
```
This Python function `find_unique` takes a sorted array as input and returns the element that appears only once in the array. It iterates through the array and compares each element with the next one. If they are not equal, it means that the current element is unique and it returns it. If it reaches the end of the array without finding a unique element, it returns the last element of the array.
codet5p-16b. fail_base: True, fail_plus: True
"""
Write a python function to find the element that appears only once in a sorted array.
assert search([1,1,2,2,3]) == 3
"""
def search(arr):
if len(arr) == 1:
return arr[0]
if len(arr) == 2:
if arr[0] == arr[1]:
return arr[0]
else:
return arr[0]
if len(arr) == 3:
if arr[0] == arr[1] or arr[0] == arr[2]:
return arr[0]
else:
return arr[1]
if len(arr) == 4:
if arr[0] == arr[1] or arr[0] == arr[2] or arr[0] == arr[3]:
return arr[0]
else:
return arr[1]
if len(arr) == 5:
if arr[0] == arr[1] or arr[0] == arr[2] or arr[0] == arr[3] or arr[0] == arr[4]:
return arr[0]
else:
return arr[1]
if len(arr) == 6:
if arr[0] == arr[1] or arr[0] == arr[2] or arr[0] == arr[3] or arr[0] == arr[4] or arr[0] == arr[5]:
return arr[0]
else:
return arr[1]
if len(arr) == 7:
if arr[0] == arr[1] or arr[0] == arr[2] or arr[0] == arr[3] or arr[0] == arr[4] or arr[0] == arr[5] or arr[0] == arr[6]:
return arr[0]
else:
return arr[1]
if len(arr) == 8:
if arr[0] == arr[1] or arr[0] == arr[2] or arr[0] == arr[3] or arr[0] == arr[4] or arr[0] == arr[5] or arr[0] == arr[6] or arr[0] == arr[7]:
return arr[0]
else:
return arr[1code-llama-13b. fail_base: True, fail_plus: True
"""
Write a python function to find the element that appears only once in a sorted array.
assert search([1,1,2,2,3]) == 3
"""
def search(arr):
for i in range(len(arr)):
if arr[i] != arr[i-1]:
return arr[i]
print(search([1,1,2,2,3]))
xdan-l1-chat. fail_base: True, fail_plus: True
def search(arr):
low, high = 0, len(arr) - 1
while low < high:
mid = low + (high - low) // 2
if arr[mid] == arr[high]:
high -= 1
elif arr[mid] == arr[low]:
low += 1
else:
high = mid
return arr[low]
# Test the function
assert search([1, 1, 2, 2, 3]) == 3solar-10.7b-instruct. fail_base: True, fail_plus: True
"""
Write a python function to find the element that appears only once in a sorted array.
assert search([1,1,2,2,3]) == 3
"""
def search(arr):
unique_element = None
last_element = None
for element in arr:
if element != last_element and unique_element is None:
unique_element = element
elif element != last_element and unique_element != element:
unique_element = None
last_element = element
return unique_element
assert search([1, 1, 2, 2, 3]) == 3gemma-2b. fail_base: True, fail_plus: True
"""
Write a python function to find the element that appears only once in a sorted array.
assert search([1,1,2,2,3]) == 3
"""
def search(arr):
for i in range(len(arr)):
if arr[i] not in arr[i+1:]:
return arr[i]
return -1
claude-3-sonnet-20240229. fail_base: True, fail_plus: True
Here's the Python function to find the element that appears only once in a sorted array, wrapped in a markdown block:
```python
def search(arr):
"""
Find the element that appears only once in a sorted array.
Args:
arr (list): A sorted list of integers.
Returns:
int: The element that appears only once in the list.
Raises:
ValueError: If there is no element that appears only once in the list.
Examples:
>>> search([1, 1, 2, 2, 3])
3
>>> search([1, 1, 2, 3, 3, 4, 4])
2
>>> search([1, 1, 1, 2, 2, 2])
Traceback (most recent call last):
...
ValueError: No element appears only once in the list.
"""
left = 0
right = len(arr) - 1
while left < right:
mid = (left + right) // 2
# If the middle element is the first occurrence or the last occurrence
if (mid == 0 or arr[mid] != arr[mid - 1]) and (mid == len(arr) - 1 or arr[mid] != arr[mid + 1]):
return arr[mid]
# If the middle element is not the first occurrence
if mid > 0 and arr[mid] == arr[mid - 1]:
# If the number of elements before the middle element is odd
if (mid - left) % 2 != 0:
right = mid - 1
else:
left = mid + 1
# If the middle element is not the last occurrence
else:
# If the number of elements after the middle element is odd
if (right - mid) % 2 != 0:
left = mid + 1
else:
right = mid - 1
raise ValueError("No element appears only once in the list.")
# Test cases
assert search([1, 1, 2, 2, 3]) == 3
assert search([1, 1, 2, 3, 3, 4, 4]) == 2
try:
search([1, 1, 1, 2, 2, 2])
except ValueError as e:
print(e) # No element appears only once in the list.starcoder2-15b-oci. fail_base: False, fail_plus: True
"""
Write a python function to find the element that appears only once in a sorted array.
assert search([1,1,2,2,3]) == 3
"""
def search(arr):
if len(arr) == 1:
return arr[0]
mid = len(arr) // 2
if mid % 2 == 0:
if arr[mid] == arr[mid+1]:
return search(arr[mid+2:])
else:
return search(arr[:mid+1])
else:
if arr[mid] == arr[mid-1]:
return search(arr[mid+1:])
else:
return search(arr[:mid])
starcoder2-3b. fail_base: False, fail_plus: True
"""
Write a python function to find the element that appears only once in a sorted array.
assert search([1,1,2,2,3]) == 3
"""
def search(arr):
if len(arr) == 1:
return arr[0]
if arr[0] != arr[1]:
return arr[0]
if arr[-1] != arr[-2]:
return arr[-1]
for i in range(1, len(arr)-1):
if arr[i] != arr[i-1] and arr[i] != arr[i+1]:
return arr[i]
print(search([1,1,2,2,3]))
print(search([1,1,2,2,3,3,4,4,5,5,6,6,7,7,8,8,9,9,10,10]))
print(search([1,1,2,2,3,3,4,4,5,5,6,6,7,7,8,8,9,9,10,10,11,11]))
print(search([1,1,2,2,3,3,4,4,5,5,6,6,7,7,8,8,9,9,10,10,11,11,12,12]))
print(search([1,1,2,2,3,3,4,4,5,5,6,6,7,7,8,8,9,9,10,10,11,11,12,12,13,13]))
print(search([1,1,2,2,3,3,4,4,5,5,6,6,7,7,8,8,9,9,10,10,11,11,12,12,13,13,14,14]))
print(search([1,1,2,2,3,3,4,4,5,5,6,6,7,7,8,8,9,9,10,10,11,11,12,12,13,13,14,14,15,15]))
print(search([1,1,2,2,3,3,4,4,5,speechless-starcoder2-15b. fail_base: True, fail_plus: True
def search(nums):
left, right = 0, len(nums) - 1
while left < right:
mid = left + (right - left) // 2
if nums[mid] == nums[mid - 1]:
if (mid - left) % 2 == 0:
left = mid + 1
else:
right = mid - 1
elif nums[mid] == nums[mid + 1]:
if (right - mid) % 2 == 0:
right = mid - 1
else:
left = mid + 1
else:
return nums[mid]
return nums[left]
assert search([1,1,2,2,3]) == 3gemma-7b. fail_base: True, fail_plus: True
"""
Write a python function to find the element that appears only once in a sorted array.
assert search([1,1,2,2,3]) == 3
"""
def search(array):
for i in range(len(array)-1):
if array[i] != array[i+1]:
return array[i]
return array[-1]
opencodeinterpreter-ds-6.7b. fail_base: False, fail_plus: True
```python
def search(arr):
for i in range(0, len(arr), 2):
if i == len(arr) - 1 or arr[i] != arr[i+1]:
return arr[i]
# Test the function
assert search([1,1,2,2,3]) == 3