Find Minimum in Rotated Sorted Array

Problem Statement - link #

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. 

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.`

Solution #

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