LeetCode 154. Find Minimum in Rotated Sorted Array II

Suppose an array of length n sorted in ascending order is rotated between 1 and n times. For example, the array nums = [0,1,4,4,5,6,7] might become:

• [4,5,6,7,0,1,4] if it was rotated 4 times.
• [0,1,4,4,5,6,7] if it was rotated 7 times.

Notice that rotating an array [a[0], a[1], a[2], ..., a[n-1]] 1 time results in the array [a[n-1], a[0], a[1], a[2], ..., a[n-2]].

Given the sorted rotated array nums that may contain duplicates, return the minimum element of this array.

Example 1:

Input: nums = [1,3,5]
Output: 1

Example 2:

Input: nums = [2,2,2,0,1]
Output: 0

Constraints:

• n == nums.length
• 1 <= n <= 5000
• -5000 <= nums[i] <= 5000
• nums is sorted and rotated between 1 and n times.

Idea:

Divide and Conquer

Solution:

/**
 * @param {number[]} nums
 * @return {number}
 */
var findMin = function(nums) {
    return searchMin(nums, 0, nums.length - 1);
};

function searchMin(nums, l, r) {
    // exit recursion:
    // only one or two elements in the array
    if (l + 1 >= r)
        return Math.min(nums[l], nums[r]);
    // this array is sorted
    if (nums[l] < nums[r])
        return nums[l];
    
    let mid = l + Math.floor((r - l) / 2);
    return Math.min(searchMin(nums, l , mid), searchMin(nums, mid + 1, r));
}