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Java Program for Find Minimum Element in Rotated Sorted Array

Search Minimum number in Rotated and Sorted Array in Java using Binary Search

Search Minimum number in Rotated and Sorted Array in Java using Binary Search

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:

  • [5, 6, 7, 0, 1, 2, 4] if it was rotated 3 times.
  • [2, 4, 5, 6, 7, 0, 1] if it was rotated 5 times.

Example 1 :

Input :
[5, 6, 7, 0, 1, 2, 4]

Output :
0

Example 2 :

Input :
[5, 8, 10, 15, 20, 50, 100, 200]

Output :
5

Example 3 :

Input :
[100, 50, 25, 5, 1]

Output :
1

We will use Binary Search for finding solution.

Solution 1 : Find Minimum Element in Rotated Sorted Array in Java


 import java.util.Scanner;

 public class FindMinimumInRotatedSortedArray {

    public static void main(String[] args) {
        
        Scanner sc = new Scanner(System.in);

        System.out.println("Enter array size : ");
        int size = sc.nextInt();

        int[] array = new int[size];

        // Taking user inputs
        System.out.println("Enter array elements :");
        for (int i = 0; i < array.length; i++) {
            array[i] = sc.nextInt();
        }

        System.out.println("Output : "+findMinimum(array));

    }
    
    public static int findMinimum(int[] nums) { 
        
        int start = 0;
        int end = nums.length-1;
        
        // If array is not rotated
        if (nums[start] <= nums[end]) {
            return nums[0];
        }
        
        while (start <= end) {
            int mid = (start+end)/2;
            
            if (nums[mid] > nums[mid+1]) {
                return nums[mid+1];
            } else if (nums[mid] < nums[mid-1]) {
                return nums[mid];
            } else if (nums[start] <= nums[mid]) {
                start = mid+1;
            } else  if (nums[mid] <= nums[end]) {
                end = mid-1;
            }
        }
        return -1;
    }    
 }

Solution Explanation :

  • Take two variables, start as 0 ad end as array length.
  • In first condition check for array does not rotated. If it does not rotated just return 0th element of array no need to do anything else.   
  • Otherwise go through start <= end
    • Take mid index.
    • Here compare mid with mid+1 and mid-1 and return minimum element accordingly.
    • Else skip left part if start <= mid or skip right part if mid <= end.

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