43. Find Minimum in Rotated Sorted Array

Topic :

binary search

Difficulty :

medium

Problem Link :

problem statement

Or Find the Pivot element in a rotated sorted array

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.

solution

import java.io.*;
import java.util.*;
class Find_Minimum_In_a_Sorted_Rotated_Array
{
    public static void main(String args[])
    {
        int nums[]={3,4,5,1,2};
        System.out.println(findMin(nums));
    }
    static int findMin(int[] nums) {
        int low=0;
        int high=nums.length-1;
                
        while(low<high)
        {
            int mid=low+(high-low)/2;
            
            if(nums[mid]>nums[high]) // pivot/min element is the right half
                low=mid+1;
            else // pivot/min element is the left half
                high=mid;
        }
        return nums[low];
     }
}