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• Matrix Data Structure

Maximum and Minimum in a square matrix.

• Difficulty Level : Easy
• Last Updated : 19 Aug, 2022

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.

Implementation:

C++

 `// C++ program for finding maximum and minimum in``// a matrix.``#include``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

 ` ``\$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 = 11, Minimum = 0`

Time complexity: O(n2).
Auxiliary Space: O(1), since no extra space has been taken.

This article is contributed by Aarti_Rathi and 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 review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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