5. Search in Rotated Sorted Array

Description

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
• -104 <= nums[i] <= 104
• All values of nums are unique.
• nums is an ascending array that is possibly rotated.
• -104 <= target <= 104

Approach

• Compute mid then check if it matches target then return mid
• Check if mid is in left sorted or right sorted
• For left check if it is out -> like target > mid or target < l then go left else right
• For right check if it is out -> target < mid or target > rigth

class Solution {
    public int search(int[] nums, int target) {
        int l = 0, r = nums.length - 1;
 
        while (l <= r) {
            int m = l + (r-l) / 2;
            if (target == nums[m]) return m;
 
            if (nums[l] <= nums[m]) {
                if (target > nums[m] || target < nums[l])
                    l = m + 1;
                else
                    r = m - 1;    
            } else {
                if (target < nums[m] || target > nums[r])
                    r = m - 1;
                else
                    l = m + 1;    
            }
        }
        return -1;
    }
}

Brute Force Approach: Linear Scan

Intuition

Iterate through the array from left to right and compare each element with target. If found, return its index; otherwise, return -1.

class Solution {
    public int search(int[] nums, int target) {
        for (int i = 0; i < nums.length; i++) {
            if (nums[i] == target) {
                return i;
            }
        }
        return -1;
    }
}
 

Complexity

• Time Complexity: $O(N)$ — Scans through all $N$ elements in the worst case.
• Space Complexity: $O(1)$ — Uses constant extra memory.

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Most Optimized Solution: Modified Binary Search

Intuition

When a sorted array is rotated, splitting it at any index mid divides it into two halves where at least one half is guaranteed to be completely sorted.

1. Calculate mid = low + (high - low) / 2. If nums[mid] == target, return mid.
2. Identify which half is sorted:

• **If nums[low] <= nums[mid]**, the left half is sorted.

• Check if target falls within [nums[low], nums[mid]). If yes, search left (high = mid - 1); otherwise, search right (low = mid + 1).

• Otherwise, the right half is sorted.

• Check if target falls within (nums[mid], nums[high]]. If yes, search right (low = mid + 1); otherwise, search left (high = mid - 1).

class Solution {
    public int search(int[] nums, int target) {
        int low = 0, high = nums.length - 1;
 
        while (low <= high) {
            int mid = low + (high - low) / 2;
 
            if (nums[mid] == target) {
                return mid;
            }
 
            // Check if left half is sorted
            if (nums[low] <= nums[mid]) {
                if (nums[low] <= target && target < nums[mid]) {
                    high = mid - 1; // Target is in left half
                } else {
                    low = mid + 1;  // Target is in right half
                }
            } 
            // Otherwise, right half is sorted
            else {
                if (nums[mid] < target && target <= nums[high]) {
                    low = mid + 1;  // Target is in right half
                } else {
                    high = mid - 1; // Target is in left half
                }
            }
        }
 
        return -1;
    }
}
 

Complexity

• Time Complexity: $O(\log N)$ — Halves search space in each iteration.
• Space Complexity: $O(1)$ — Uses constant space.

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Easy Memory Rule

“At least one half is always sorted. Find the sorted half, check if target lies within its range, and adjust low/high.”