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This approach makes a single pass through the array while keeping a count of occurrences where nums[i] > nums[i+1]. If more than one such occurrence is found, it returns false. During the traversal, if such an occurrence is found, we will attempt to resolve it by modifying nums[i] or nums[i+1] optimally based on their neighbors.
Time Complexity: O(n), where n is the length of the array.
Space Complexity: O(1), since no additional space is used dependent on input size.
1def checkPossibility(nums):
2 count = 0
3 for i in range(len(nums) - 1):
4 if nums[i] > nums[i + 1]:
5 count += 1
6 if count > 1:
7 return False
8 if i > 0 and nums[i + 1] < nums[i - 1]:
9 nums[i + 1] = nums[i]
10 return TrueThe Python solution runs a loop, counting ordinance failures in the array. On the first failure, it optimizes by evaluating the relationship of adjacent elements and making minimal necessary changes.
This approach involves using two pointers to traverse from the start and end of the array. We simulate potential changes to resolve any issues at both ends, using the pointers to converge inwards and analyze changes required to achieve a non-decreasing state.
Time Complexity: O(n)
Space Complexity: O(1)
1publicThe Java solution implements a two-pointer method, making directional modifications from both ends, which converges on problem points needing fixes. Validity is checked by counting necessary adjustments.