Find minimum and maximum element in an array gfg solution

Given an array A of size N of integers. Your task is to find the minimum and maximum element in the array.

This question to find minimum and maximum element in an array is taken from GeeksforGeeks platform .

Example 1:

Input:

N = 6
A[] = {3, 2, 1, 56, 10000, 167}
Output:
min = 1, max =  10000

Example 2:

Input:

N = 5
A[]  = {1, 345, 234, 21, 56789}
Output:
min = 1, max = 56789

Your Task:  

You don’t need to read input or print anything. Your task is to complete the function getMinMax() which takes the array A[] and its size N as inputs and returns the minimum and maximum element of the array.

Expected Time Complexity: O(N)
Expected Auxiliary Space: O(1)

Constraints:
1 <= N <= 10⁵
1 <= A_i <=10¹²

Solution in Java

//{ Driver Code Starts
//Initial Template for Java

import java.util.*;
import java.lang.*;
import java.io.*;

class pair  
{  
    long first, second;  
    public pair(long first, long second)  
    {  
        this.first = first;  
        this.second = second;  
    }  
}

class GFG {
	public static void main(String[] args) throws IOException
	{
	        BufferedReader br =
            new BufferedReader(new InputStreamReader(System.in));
        int t =
            Integer.parseInt(br.readLine().trim()); // Inputting the testcases
        while(t-->0)
        {
            long n = Long.parseLong(br.readLine().trim());
            long a[] = new long[(int)(n)];
            // long getAnswer[] = new long[(int)(n)];
            String inputLine[] = br.readLine().trim().split(" ");
            for (int i = 0; i < n; i++) {
                a[i] = Long.parseLong(inputLine[i]);
            }
            
            Compute obj = new Compute();
            pair product = obj.getMinMax(a, n); 
            System.out.println(product.first+" "+product.second);
            
        }
	}
}


// } Driver Code Ends


//User function Template for Java

/*
 class pair  
{  
    long first, second;  
    public pair(long first, long second)  
    {  
        this.first = first;  
        this.second = second;  
    }  
}*/

class Compute 
{
    static pair getMinMax(long a[], long n)  
    {
        int i; long max=0;
        
        for(i=0;i<n;i++)
        {
            if(a[i]>max)
            {
                max=a[i];
            }
        }

        long min=a[0];
        for(i=0;i<n;i++)
        {
            if(min>a[i])
            {
                min=a[i];
            }
        }
    
return new pair(min,max);
        
    }
}

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