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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.
1public class Solution {
2 public bool Search(int[] nums, int target) {
3 int left = 0, right = nums.Length - 1;
4 while (left <= right) {
5 int mid = left + (right - left) / 2;
6 if (nums[mid] == target) return true;
7 while (left < mid && nums[left] == nums[mid]) left++;
8 while (mid < right && nums[right] == nums[mid]) right--;
9 if (nums[left] <= nums[mid]) {
10 if (nums[left] <= target && target < nums[mid]) {
11 right = mid - 1;
12 } else {
13 left = mid + 1;
}
} else {
if (nums[mid] < target && target <= nums[right]) {
left = mid + 1;
} else {
right = mid - 1;
}
}
}
return false;
}
}This C# solution modifies the standard binary search to accommodate potential duplicates and rotational impacts, identifying sorted sections effectively.
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.
1In this simplistic Python solution, we loop through the entire array, checking each element for equality with the target, making it a reliable brute-force method.