7. 153. Find Minimum in Rotated Sorted Array

153. Find Minimum in Rotated Sorted ArrayMedium11K496Companies

Suppose an array of length n sorted in ascending order is rotated between 1 and n times. For example, the array nums = [0,1,2,4,5,6,7] might become:

• [4,5,6,7,0,1,2] if it was rotated 4 times.

• [0,1,2,4,5,6,7] 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 of unique elements, return the minimum element of this array.

You must write an algorithm that runs in O(log n) time.

Example 1:

Input: nums = [3,4,5,1,2]
Output: 1
Explanation: The original array was [1,2,3,4,5] rotated 3 times.

Example 2:

Input: nums = [4,5,6,7,0,1,2]
Output: 0
Explanation: The original array was [0,1,2,4,5,6,7] and it was rotated 4 times.

Example 3:

Input: nums = [11,13,15,17]
Output: 11
Explanation: The original array was [11,13,15,17] and it was rotated 4 times. 

Constraints:

• n == nums.length

• 1 <= n <= 5000

• -5000 <= nums[i] <= 5000

• All the integers of nums are unique.

• nums is sorted and rotated between 1 and n times.

Loading...LeetCode

Time: O(n)

Space: O(1)

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

    while (s < e) {
        const m = s + Math.floor((e - s) / 2);

        if (nums[m] <= nums[e]) {
            e = m;
        } else {
            s = m + 1;
        }

    }

    return nums[e];
};