Search in Rotated Sorted Array II

Problem

There is an integer array nums sorted in non-decreasing order (not necessarily with distinct values).

Before being passed to your function, nums is rotated at an unknown pivot index k (0 <= 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, [0,1,2,4,4,4,5,6,6,7] might be rotated at pivot index 5 and become [4,5,6,6,7,0,1,2,4,4].

Given the array nums after the rotation and an integer target, return true if target is in nums, or false if it is not in nums.

You must decrease the overall operation steps as much as possible.

 

Example 1:

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

Example 2:

Input: nums = [2,5,6,0,0,1,2], target = 3
Output: false

 

Constraints:

• 1 <= nums.length <= 5000
• -104 <= nums[i] <= 104
• nums is guaranteed to be rotated at some pivot.
• -104 <= target <= 104

Solution

/**
 * @param {number[]} nums
 * @param {number} target
 * @return {boolean}
 */
var search = function(nums, target) {
    let lo = 0;
    let hi = nums.length - 1;

    while (lo <= hi) {
        // filter out duplicate values
        while (lo < hi && nums[lo] === nums[lo + 1]) {
            lo++;
        }
        while (lo < hi && nums[hi] === nums[hi - 1]) {
            hi--;
        }

        // binary search
        const mid = Math.floor((lo + hi) / 2);
        if (nums[mid] === target) {
            return true;
        } else if (nums[lo] <= nums[mid]) { // [lo, mid] sorted
            if (nums[lo] <= target && target < nums[mid]) {
                hi = mid - 1;
            } else {
                lo = mid + 1;
            }
        } else { // [lo, mid] not sorted => [mid, hi] sorted
            if (nums[mid] < target && target <= nums[hi]) {
                lo = mid + 1;
            } else {
                hi = mid - 1;
            }
        }
    }
    return false;
};

The solution to this problem is extremely similar to Search in Rotated Sorted Array. The only difference is there are duplicate values in nums, so we need to add some additional logic to filter out duplicate values during each step of our binary search.