3.8 Search in Rotated Sorted Array

3.8.1 Problem Metadata

3.8.2 Description

Given an ascending-sorted array that has been rotated at an unknown pivot, search for a target value. If found, return its index; otherwise return -1. The array contains no duplicates, and the algorithm must run in O(log n).

3.8.3 Examples

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

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

3.8.4 Constraints

  • 1 <= nums.length <= 10^4
  • -10^4 <= nums[i], target <= 10^4
  • All elements are distinct

3.8.5 Solution - Modified Binary Search

3.8.5.1 Walkthrough

At each step, a sorted half still exists. Compare nums[mid] with nums[left] to detect which side is sorted. If the target lies within that half, keep it; otherwise discard it and continue searching the other half.

3.8.5.2 Analysis

  • Time Complexity: O(log n)
  • Space Complexity: O(1)

3.8.5.3 Implementation Steps

  1. Set left = 0, right = n - 1.
  2. While left <= right:
    1. mid = left + (right - left) / 2.
    2. If nums[mid] == target, return mid.
    3. If left half is sorted (nums[left] <= nums[mid]):
      • If target in [nums[left], nums[mid]), set right = mid - 1 else left = mid + 1.
    4. Else the right half is sorted:
      • If target in (nums[mid], nums[right]], set left = mid + 1 else right = mid - 1.
  3. Return -1.

3.8.5.4 Code - Java

public int search(int[] nums, int target) {
    int left = 0;
    int right = nums.length - 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;
}