LeetCode //C - 33. Search in Rotated Sorted Array

33. Search in Rotated Sorted Array

There is an integer array nums sorted in ascending order (with distinct values).

Prior to being passed to your function, nums is possibly rotated at an unknown pivot index k (1 <= k < nums.length) such that the resulting array is [nums[k], nums[k+1], …, nums[n-1], nums[0], nums[1], …, nums[k-1]] (0-indexed). For example, [0,1,2,4,5,6,7] might be rotated at pivot index 3 and become [4,5,6,7,0,1,2].

Given the array nums after the possible rotation and an integer target, return the index of target if it is in nums, or -1 if it is not in nums.

You must write an algorithm with O(log n) runtime complexity.
 

Example 1:

Input: nums = [4,5,6,7,0,1,2], target = 0
Output: 4

Example 2:

Input: nums = [4,5,6,7,0,1,2], target = 3
Output: -1

Example 3:

Input: nums = [1], target = 0
Output: -1

Constraints:

• 1 <= nums.length <= 5000
• $-10^4<=nums[i]<=10^4$
• All values of nums are unique.
• nums is an ascending array that is possibly rotated.
• $-10^4<=target<=10^4$

From: LeetCode
Link: 33. Search in Rotated Sorted Array

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Solution:

Ideas：

1. Binary Search Approach: While the array is rotated, portions of it remain in sorted order. This observation is key to applying a binary search.

2. Midpoint Calculation: Start with the entire array as the search space. Calculate the midpoint of the current search space.

3. Comparison:

• Check if the element at the midpoint is the target. If it is, return the midpoint.
• If nums[left] <= nums[mid], it means the left half of the array is sorted. Then:
  • If the target lies between nums[left] and nums[mid], then narrow the search to the left half.
  • Otherwise, search in the right half.
• If the left half isn’t sorted, then the right half must be sorted. Then:
  • If the target lies between nums[mid] and nums[right], then narrow the search to the right half.
  • Otherwise, search in the left half.

4. Convergence: Continue the binary search process until the target is found or the search space becomes empty.

5. Result: Return the index if the target is found, or -1 if not found.

Code:

int search(int* nums, int numsSize, int target) {
    int left = 0, right = numsSize - 1;
    
    while (left <= right) {
        int mid = left + (right - left) / 2;
        
        if (nums[mid] == target) {
            return mid;
        }
        
        if (nums[left] <= nums[mid]) {
            if (target >= nums[left] && target < nums[mid]) {
                right = mid - 1;
            } else {
                left = mid + 1;
            }
        } else {
            if (target > nums[mid] && target <= nums[right]) {
                left = mid + 1;
            } else {
                right = mid - 1;
            }
        }
    }
    
    return -1;
}
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