Maximum and Minimum in a square matrix.
Last Updated :
13 Sep, 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.
C++
#include<bits/stdc++.h>
using namespace std;
void maxMin( int arr[3][3], int n)
{
int min = INT_MAX;
int max = INT_MIN;
for ( int i = 0; i<n; i++){
for ( int j = 0; j<n; j++){
if (max < arr[i][j]) max = arr[i][j];
}
}
for ( int i = 0; i<n; i++){
for ( int j = 0; j<n; j++){
if (min > arr[i][j]) min = arr[i][j];
}
}
cout << "Maximum = " << max << ", Minimum = " << min;
}
int main(){
int arr[3][3] = {{5, 9, 11} , {25, 0, 14} , {21, 6, 4}};
maxMin(arr, 3);
return 0;
}
|
Java
import java.util.*;
class Main {
static void maxMin( int [][] arr, int n) {
int min = Integer.MAX_VALUE;
int max = Integer.MIN_VALUE;
for ( int i = 0 ; i < n; i++) {
for ( int j = 0 ; j < n; j++) {
if (max < arr[i][j]) max = arr[i][j];
}
}
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.println( "Maximum = " + max + ", Minimum = " + min);
}
public static void main(String[] args) {
int [][] arr = {{ 5 , 9 , 11 }, { 25 , 0 , 14 }, { 21 , 6 , 4 }};
maxMin(arr, 3 );
}
}
|
Python3
import sys
def maxMin(arr, n):
min = sys.maxsize
max = - sys.maxsize - 1
for i in range (n):
for j in range (n):
if max < arr[i][j]:
max = arr[i][j]
for i in range (n):
for j in range (n):
if min > arr[i][j]:
min = arr[i][j]
print ( "Maximum = " , max , ", Minimum = " , min )
arr = [[ 5 , 9 , 11 ], [ 25 , 0 , 14 ], [ 21 , 6 , 4 ]]
maxMin(arr, 3 )
|
C#
using System;
class Program
{
static void MaxMin( int [,] arr, int n)
{
int min = int .MaxValue;
int max = int .MinValue;
for ( int i = 0; i<n; i++){
for ( int j = 0; j<n; j++){
if (max < arr[i, j]) max = arr[i, j];
}
}
for ( int i = 0; i<n; i++){
for ( int j = 0; j<n; j++){
if (min > arr[i, j]) min = arr[i, j];
}
}
Console.WriteLine( "Maximum = {0}, Minimum = {1}" , max, min);
}
static void Main( string [] args)
{
int [,] arr = {{5, 9, 11} , {25, 0, 14} , {21, 6, 4}};
MaxMin(arr, 3);
}
}
|
Javascript
function maxMin(arr, n){
let min = +2147483647;
let max = -2147483648;
for (let i = 0; i<n; i++){
for (let j = 0; j<n; j++){
if (max < arr[i][j]) max = arr[i][j];
}
}
for (let i = 0; i<n; i++){
for (let j = 0; j<n; j++){
if (min > arr[i][j]) min = arr[i][j];
}
}
console.log( "Maximum = " + max + ", Minimum = " + min);
}
let arr = [[9,9,11], [25,0,14], [21,6,4]];
maxMin(arr, 3)
|
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.
Implementation:
C++
#include<bits/stdc++.h>
using namespace std;
#define MAX 100
void maxMin( int arr[][MAX], int n)
{
int min = INT_MAX;
int max = INT_MIN;
for ( int i = 0; i < n; i++)
{
for ( int j = 0; j <= n/2; j++)
{
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;
}
int main()
{
int arr[MAX][MAX] = {5, 9, 11,
25, 0, 14,
21, 6, 4};
maxMin(arr, 3);
return 0;
}
|
Java
class GFG
{
static final int MAX = 100 ;
static void maxMin( int arr[][], int n)
{
int min = + 2147483647 ;
int max = - 2147483648 ;
for ( int i = 0 ; i < n; i++)
{
for ( int j = 0 ; j <= n/ 2 ; j++)
{
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);
}
public static void main (String[] args)
{
int arr[][] = {{ 5 , 9 , 11 },
{ 25 , 0 , 14 },
{ 21 , 6 , 4 }};
maxMin(arr, 3 );
}
}
|
Python3
MAX = 100
def MAXMIN(arr, n):
MIN = 10 * * 9
MAX = - 10 * * 9
for i in range (n):
for j in range (n / / 2 + 1 ):
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 )
arr = [[ 5 , 9 , 11 ],
[ 25 , 0 , 14 ],
[ 21 , 6 , 4 ]]
MAXMIN(arr, 3 )
|
C#
using System;
public class GFG {
static void maxMin( int [,] arr, int n)
{
int min = +2147483647;
int max = -2147483648;
for ( int i = 0; i < n; i++)
{
for ( int j = 0; j <= n/2; j++)
{
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);
}
static public void Main ()
{
int [,] arr = { {5, 9, 11},
{25, 0, 14},
{21, 6, 4} };
maxMin(arr, 3);
}
}
|
PHP
<?php
$MAX = 100;
function maxMin( $arr , $n )
{
$min = PHP_INT_MAX;
$max = PHP_INT_MIN;
for ( $i = 0; $i < $n ; $i ++)
{
for ( $j = 0; $j <= $n / 2; $j ++)
{
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 ;
}
$arr = array ( array (5, 9, 11),
array (25, 0, 14),
array (21, 6, 4));
maxMin( $arr , 3);
?>
|
Javascript
<script>
let MAX = 100;
function maxMin(arr,n)
{
let min = +2147483647;
let max = -2147483648;
for (let i = 0; i < n; i++)
{
for (let j = 0; j <= n / 2; j++)
{
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);
}
let arr = [ [ 5, 9, 11 ],
[ 25, 0, 14 ],
[ 21, 6, 4 ] ];
maxMin(arr, 3);
</script>
|
Output
Maximum = 11, Minimum = 0
Time complexity: O(n2).
Auxiliary Space: O(1), since no extra space has been taken.
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