Find a Minimum in Rotated Sorted Array (#53 Leetcode)

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.

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.

class Solution {
public:
    int findMin(vector<int>& nums) {       
        if (nums.size() < 1 || nums.size() > 5000) {
            return 0;
        }
        
        int end = nums.size()-1;
        for (int i = 0; i < end; i++) {
            if (nums[i] < -5000 || nums[i] > 5000) {
                return 0;
            }
        }
        
        int start = 0;
        int mid = 0;
        while (start < end) {
           if (nums[start] < nums[end]) {
               //cout << "return start " << start  << " end " << end<< endl;
               return nums[start];
           }
           mid = (end + start)/2;
           int midEle = nums[mid];
           //cout << "midEle " << midEle << endl;
           if (nums[mid] < nums[end]) {
               end = mid;
           } else {
               start = mid+1;
           }
             
           //cout << "loop start " << start  << " end " << end<< endl;
        }
        return nums[start];
    }
};

Since always the nums[end] is smaller than nums[start], we can compare the element with nums[end to find minimum.

Takeaway:

Binary Search will work only in the sorted array/list. 

The Time complexity is O(log n)