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Search in a Rotated Sorted Array

Finding a target in a rotated sorted array by identifying which half is sorted at each step.

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Find target in a rotated sorted array (distinct values), returning its index or -1. [4,5,6,7,0,1,2], target 0 → 4. (LeetCode 33)

Approach: at each step, one half of the array (split at mid) is guaranteed sorted. Figure out which half is sorted by comparing nums[low] and nums[mid], then check whether target falls inside that sorted half’s range — if it does, search there, otherwise search the other half.

class Solution {
public:
    int search(vector<int>& nums, int target) {
        int low = 0, high = nums.size() - 1;


        while (low <= high) {
            int mid = low + (high - low) / 2;


            if (nums[mid] == target) return mid;


            if (nums[low] <= nums[mid]) { // left half is sorted
                if (nums[low] <= target && target < nums[mid]) {
                    high = mid - 1;
                } else {
                    low = mid + 1;
                }
            } else { // right half is sorted
                if (nums[mid] < target && target <= nums[high]) {
                    low = mid + 1;
                } else {
                    high = mid - 1;
                }
            }
        }


        return -1;
    }
};

Time: O(log n). Space: O(1).