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This approach uses a modified binary search to handle the rotated sorted array with duplicates. Normally, in a binary search, you would compare the middle element with the target and adjust your search range accordingly. However, because this array is rotated, the sorted order is disrupted, and duplicates can complicate matters.
The key steps are:
The time complexity is generally O(log N), but in the worst case with duplicates, it can degrade to O(N).
The time complexity is O(log N) in the average case, but O(N) in the worst case due to duplicates. The space complexity is O(1) since no extra storage is needed.
1def search(nums, target):
2 left, right = 0, len(nums) - 1
3 while left <= right:
4 mid =This Python implementation of the modified binary search accounts for duplicates by checking for duplicate middle elements. If they exist, the algorithm adjusts the search boundaries while maintaining the sorted distinction in either half of the array.
In cases of heavy element duplication, when the above methods become inefficient, a linear search ensures completion of the task. Though not ideal for large-sized arrays, this method provides a simple, guaranteed solution without involving complicated logic.
The complexity for linear search is both time and space O(N).
Both time and space complexities are O(N), since we traverse the entire array.
1This C implementation uses a straightforward iteration to look for the target element within the array.