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[LeetCode 33] Search in Rotated Sorted Array

Problem

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

Thoughts

  • We can use binary search for this
  • We need check which portion of the array is sorted

Typescript

function search(nums: number[], target: number): number {
    let left = 0, right = nums.length - 1
    
    // equal sign to handle [1] single element case
    while(left <= right) {
        let mid = Math.floor((left + right) / 2)
        
        // return the index if we find it
        if(nums[mid] === target) {
            return mid;
        }
        
        // left sorted part
        if(nums[left] <= nums[mid]) {
	    // search left 
            if(target < nums[mid] && target >= nums[left]) {
                right = mid - 1;
            } else {	// search right
                left = mid + 1;
            }
        } else {  // right sorted part
	    // search right
            if(target > nums[mid] && target <= nums[right]) {
                left = mid + 1;
            } else {	// search left
                right = mid - 1;
            }
        }
    }
    
    return -1;
};

Reference