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)
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.
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.
Output:
Maximum = 25, Minimum = 0
Time complexity: O(n^2) for given n*n matrix
Auxiliary space: O(1) because constant space is being used
Please refer complete article on Maximum and Minimum in a square matrix. for more details!
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