Java Program for Maximum and Minimum in a square matrix.

Last Updated : 22 Feb, 2023

Given a square matrix of order n*n, find the maximum and minimum from the matrix given.

Examples:

Input : arr[][] = {5, 4, 9,
                   2, 0, 6,
                   3, 1, 8};
Output : Maximum = 9, Minimum = 0

Input : arr[][] = {-5, 3, 
                   2, 4};
Output : Maximum = 4, Minimum = -5

Naive Method :
We find maximum and minimum of matrix separately using linear search. Number of comparison needed is n² for finding minimum and n² for finding the maximum element. The total comparison is equal to 2n².

Java

// Java program for finding maximum 
// and minimum in a matrix.
class GFG {
    static void maxMin(int arr[][], int n){
        int min = +2147483647;
        int max = -2147483648;
    
        // for finding the maximum element
        for(int i = 0; i<n; i++){
            for(int j = 0; j<n; j++){
                if(max < arr[i][j]){
                    max = arr[i][j];
                } 
            }
        }
        // for finding the minimum element
        for(int i = 0; i<n; i++){
            for(int j = 0; j<n; j++){
                if(min > arr[i][j]){
                    min = arr[i][j];
                } 
            }
        }
        System.out.print("Maximum = " + max + ", Minimum = "+min);
    }
    
    // Driver program 
    public static void main (String[] args) {
        int arr[][] = {{5, 9, 11},
                       {25, 0, 14},
                       {21, 6, 4}};
        maxMin(arr, 3);
    }
}

// This code is contributed by Kirti Agarwal(kirtiagarwal23121999)

Output

Maximum = 25, Minimum = 0

Pair Comparison (Efficient method):
Select two elements from the matrix one from the start of a row of the matrix another from the end of the same row of the matrix, compare them and next compare smaller of them to the minimum of the matrix and larger of them to the maximum of the matrix. We can see that for two elements we need 3 compare so for traversing whole of the matrix we need total of 3/2 n² comparisons.

Note : This is extended form of method 3 of Maximum Minimum of Array.

Java

// Java program for finding maximum 
// and minimum in a matrix.

class GFG 
{
    static final int MAX = 100;
    
    // Finds maximum and minimum 
    // in arr[0..n-1][0..n-1]
    // using pair wise comparisons
    static void maxMin(int arr[][], int n)
    {
        int min = +2147483647;
        int max = -2147483648;
    
        // Traverses rows one by one
        for (int i = 0; i < n; i++)
        {
            for (int j = 0; j <= n/2; j++)
            {
                // Compare elements from beginning
                // and end of current row
                if (arr[i][j] > arr[i][n - j - 1])
                {
                    if (min > arr[i][n - j - 1])
                        min = arr[i][n - j - 1];
                    if (max< arr[i][j])
                        max = arr[i][j];
                }
                else
                {
                    if (min > arr[i][j])
                        min = arr[i][j];
                    if (max< arr[i][n - j - 1])
                        max = arr[i][n - j - 1];
                }
            }
        }
        System.out.print("Maximum = "+max+
                         ", Minimum = "+min);
    }
    
    // Driver program 
    public static void main (String[] args) 
    {
        int arr[][] = {{5, 9, 11},
                       {25, 0, 14},
                       {21, 6, 4}};
        maxMin(arr, 3);
    }
}

// This code is contributed by Anant Agarwal.