153. Find Minimum in Rotated Sorted Array

Problem Description

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.

Examples

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.

def findMin(nums: List[int]) -> int:
    left, right = 0, len(nums) - 1
    
    while left < right:
        mid = left + (right - left) // 2
        
        if nums[mid] > nums[right]:
            left = mid + 1
        else:
            right = mid
    
    return nums[left]

class Solution {
    public int findMin(int[] nums) {
        int left = 0;
        int right = nums.length - 1;
        
        while (left < right) {
            int mid = left + (right - left) / 2;
            
            if (nums[mid] > nums[right]) {
                left = mid + 1;
            } else {
                right = mid;
            }
        }
        
        return nums[left];
    }
}

class Solution {
public:
    int findMin(vector& nums) {
        int left = 0;
        int right = nums.size() - 1;
        
        while (left < right) {
            int mid = left + (right - left) / 2;
            
            if (nums[mid] > nums[right]) {
                left = mid + 1;
            } else {
                right = mid;
            }
        }
        
        return nums[left];
    }
};

/**
 * @param {number[]} nums
 * @return {number}
 */
var findMin = function(nums) {
    let left = 0;
    let right = nums.length - 1;
    
    while (left < right) {
        const mid = left + Math.floor((right - left) / 2);
        
        if (nums[mid] > nums[right]) {
            left = mid + 1;
        } else {
            right = mid;
        }
    }
    
    return nums[left];
};

public class Solution {
    public int FindMin(int[] nums) {
        int left = 0;
        int right = nums.Length - 1;
        
        while (left < right) {
            int mid = left + (right - left) / 2;
            
            if (nums[mid] > nums[right]) {
                left = mid + 1;
            } else {
                right = mid;
            }
        }
        
        return nums[left];
    }
}