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++
// C++ program for finding maximum and minimum in
// a matrix.
#include<bits/stdc++.h>
using namespace std;
// Finds maximum and minimum in arr[0..n-1][0..n-1]
// using pair wise comparisons
void maxMin(int arr[3][3], int n)
{
int min = INT_MAX;
int max = INT_MIN;
// for finding the max element in given array
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 min element in given array
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;
}
// Driver Program to test above function
int main(){
int arr[3][3] = {{5, 9, 11} , {25, 0, 14} , {21, 6, 4}};
maxMin(arr, 3);
return 0;
}
// THIS CODE IS CONTRIBUTED BY Yash Agarwal(yashagarwal23121999)
import java.util.*;
class Main {
// 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 = Integer.MAX_VALUE;
int max = Integer.MIN_VALUE;
// for finding the max element in given array
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 min element in given array
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);
}
}
import sys
# Finds maximum and minimum in arr[0..n-1][0..n-1]
# using pair wise comparisons
def maxMin(arr, n):
min = sys.maxsize
max = -sys.maxsize - 1
# for finding the max element in given array
for i in range(n):
for j in range(n):
if max < arr[i][j]:
max = arr[i][j]
# for finding the min element in given array
for i in range(n):
for j in range(n):
if min > arr[i][j]:
min = arr[i][j]
print("Maximum = ", max, ", Minimum = ", min)
# Driver Program to test above function
arr = [[5, 9, 11], [25, 0, 14], [21, 6, 4]]
maxMin(arr, 3)
// C# program for finding maximum and minimum in
// a matrix.
using System;
class Program
{
// 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 = int.MaxValue;
int max = int.MinValue;
// for finding the max element in given array
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 min element in given array
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);
}
// Driver Program to test above function
static void Main(string[] args)
{
int[,] arr = {{5, 9, 11} , {25, 0, 14} , {21, 6, 4}};
MaxMin(arr, 3);
}
}
// JavaScript program for finding maximum and minimum in
// a matrix
// 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;
// for finding the max element in givne array
for(let i = 0; i<n; i++){
for(let j = 0; j<n; j++){
if(max < arr[i][j]) max = arr[i][j];
}
}
// for finding the min element in givne array
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);
}
// driver program to test above function
let arr = [[9,9,11], [25,0,14], [21,6,4]];
maxMin(arr, 3)
OutputMaximum = 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++
// 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 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 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# 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 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.
?>
<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>
OutputMaximum = 11, Minimum = 0
Time complexity: O(n2).
Auxiliary Space: O(1), since no extra space has been taken.
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