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.#includeusing namespace std; #define MAX 100 // Finds maximum and minimum in arr[0..n-1][0..n-1]// using pair wise comparisonsvoid 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 comparisonsdef 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 Codearr = [[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

 \$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



Output:

Maximum = 25, Minimum = 0