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Maximum and Minimum in a square matrix.

  • Difficulty Level : Easy
  • Last Updated : 30 Apr, 2021

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.

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.



C++




// C++ program for finding maximum and minimum in
// a matrix.
#include<bits/stdc++.h>
using namespace std;
 
#define MAX 100
 
// Finds maximum and minimum in arr[0..n-1][0..n-1]
// using pair wise comparisons
void maxMin(int arr[][MAX], int n)
{
    int min = INT_MAX;
    int max = INT_MIN;
 
    // 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];
            }
        }
    }
    cout << "Maximum = " << max;
         << ", Minimum = " << min;
}
 
/* Driver program to test above function */
int main()
{
    int arr[MAX][MAX] = {5, 9, 11,
                        25, 0, 14,
                        21, 6, 4};
    maxMin(arr, 3);
    return 0;
}

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.

Python3




# Python3 program for finding
# MAXimum and MINimum in a matrix.
MAX = 100
 
# Finds MAXimum and MINimum in arr[0..n-1][0..n-1]
# using pair wise comparisons
def MAXMIN(arr, n):
 
    MIN = 10**9
    MAX = -10**9
 
    # Traverses rows one by one
    for i in range(n):
        for j in range(n // 2 + 1):
             
        # 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]
 
    print("MAXimum =", MAX, ", MINimum =", MIN)
 
# Driver Code
arr = [[5, 9, 11],
       [25, 0, 14],
       [21, 6, 4]]
 
MAXMIN(arr, 3)
 
# This code is contributed by Mohit Kumar

C#




// C# program for finding maximum
// and minimum in a matrix.
using System;
 
public class GFG {
     
    // 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];
                }
            }
        }
        Console.Write("Maximum = " + max +
                        ", Minimum = " + min);
    }
     
    // Driver code
    static public void Main ()
    {
        int[,] arr = { {5, 9, 11},
                       {25, 0, 14},
                       {21, 6, 4} };
                        
        maxMin(arr, 3);
    }
}
 
// This code is contributed by Shrikant13.

PHP




<?php
// PHP program for finding
// maximum and minimum in
// a matrix.
 
$MAX = 100;
 
// Finds maximum and minimum
// in arr[0..n-1][0..n-1]
// using pair wise comparisons
function maxMin($arr, $n)
{
    $min = PHP_INT_MAX;
    $max = PHP_INT_MIN;
 
    // Traverses rows one by one
    for ($i = 0; $i < $n; $i++)
    {
        for ($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];
            }
        }
    }
    echo "Maximum = " , $max
        ,", Minimum = " , $min;
}
 
    // Driver Code
    $arr = array(array(5, 9, 11),
                array(25, 0, 14),
                array(21, 6, 4));
    maxMin($arr, 3);
     
// This code is contributed by anuj_67.
?>

Javascript




<script>
 
// Javascript program for finding maximum
// and minimum in a matrix.
let MAX = 100;
     
// Finds maximum and minimum
// in arr[0..n-1][0..n-1]
// using pair wise comparisons
function maxMin(arr,n)
{
    let min = +2147483647;
    let max = -2147483648;
 
    // Traverses rows one by one
    for(let i = 0; i < n; i++)
    {
        for(let 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];
            }
        }
    }
    document.write("Maximum = " + max +
                   ", Minimum = " + min);
}
 
// Driver Code
let arr = [ [ 5, 9, 11 ],
            [ 25, 0, 14 ],
            [ 21, 6, 4 ] ];
             
maxMin(arr, 3);
 
// This code is contributed by sravan kumar
 
</script>

Output: 

Maximum = 25, Minimum = 0

This article is contributed by Shivam Pradhan (anuj_charm). If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
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