Introduction to Matrix or Grid – Data Structure and Algorithms Tutorial

What is a Matrix?

A matrix is a two-dimensional array that consists of rows and columns. It is an arrangement of elements in horizontal or vertical lines of entries.

Introduction to Matrix or Grid – Data Structure and Algorithms Tutorials

Declaration of Matrix or Grid

The syntax of declaring a Matrix or two-dimensional array is very much similar to that of a one-dimensional array, given as follows.

int arr[number_of_rows][number_of_columns];   

However, It produces a data structure that looks like the following:

Representation of matrix

As you can see from the above image, the elements are organized in rows and columns. As shown in the above image the cell x[0][0] is the first element of the first row and first column. The value in the first square bracket represents the row number and the value inside the second square bracket represents the column number. (i.e, x[row][column]).

Initializing Matrix or Grids

There are two methods to initialize two-dimensional arrays.

Method 1

int arr[4][3]={1, 2, 3, 4, 5, 6, 20, 80, 90, 100, 110, 120};

Method 2

int arr[4][3]={{1, 2, 3}, {4, 5, 6}, {20, 80, 90}, {100, 110, 120}};

Here are two methods of initialization of an element during declaration. Here, the second method is preferred because the second method is more readable and understandable so that you can visualize that arr[][] comprises four rows and three columns.

How to access data in Matrix or Grid

Like one-dimensional arrays, matrices can be accessed randomly by using their indices to access the individual elements. A cell has two indices, one for its row number, and the other for its column number. We can use X[i][j] to access the element which is at the ith row and jth column of the matrix.

Matrix

The syntax for access element from the matrix which is at the ith row and jth column:

int value = X[i][j];

How to Print the Elements of a Matrix or Grid:

Printing elements of a matrix or two-dimensional array can be done using two “for” loops.

C++

#include <bits/stdc++.h>
using namespace std;

int main()
{

    int arr[3][4] = { { 1, 2, 3, 4 },
                      { 5, 6, 7, 8 },
                      { 9, 10, 11, 12 } };

    for (int i = 0; i < 3; i++) {
        for (int j = 0; j < 4; j++) {
            cout << arr[i][j] << " ";
        }
        cout << endl;
    }
    return 0;
}

Java

/*package whatever //do not write package name here */
import java.io.*;

class GFG {
    public static void main(String[] args)
    {
        int[][] arr = { { 1, 2, 3, 4 },
                        { 5, 6, 7, 8 },
                        { 9, 10, 11, 12 } };

        for (int i = 0; i < 3; i++) {
            for (int j = 0; j < 4; j++) {
                System.out.print(arr[i][j] + " ");
            }
            System.out.println();
        }
    }
}

// This code is contributed by lokesh

Python3

arr = [ [ 1, 2, 3, 4 ], [ 5, 6, 7, 8 ], [ 9, 10, 11, 12] ]

for i in range(0,3):
  for j in range(0,4):

    print(arr[i][j],end = " ")
  print("")

# This code is contributed by akashish__

C#

using System;

public class GFG {

  static public void Main()
  {
    int[, ] arr = new int[3, 4] { { 1, 2, 3, 4 },
                                 { 5, 6, 7, 8 },
                                 { 9, 10, 11, 12 } };

    for (int i = 0; i < 3; i++) {
      for (int j = 0; j < 4; j++) {
        Console.Write(arr[i, j]);
        Console.Write(" ");
      }
      Console.WriteLine(" ");
    }
  }
}

// This code is contributed by akashish__

Javascript

// JS code for above approach
let arr = [[1, 2, 3, 4],
           [5, 6, 7, 8],
           [9, 10, 11, 12]];
for (let i = 0; i < 3; i++) {
let s="";
    for (let j = 0; j < 4; j++) {
        s+=(arr[i][j]+" ");
    }
    console.log(s);
}

// This code is contributed by ishankhandelwals.

Output

1 2 3 4 
5 6 7 8 
9 10 11 12 

Some basic problems on Matrix/Grid that you must know:

1. Search in a matrix:

Given a matrix mat[][] of size N x M, where every row and column is sorted in increasing order, and a number X is given. The task is to find whether element X is present in the matrix or not.

Examples:

Input : mat[][] = { {1, 5, 9}, 
                   {14, 20, 21}, 
                   {30, 34, 43} }
       x = 14
Output : YES

Input : mat[][] = { {1, 5, 9, 11}, 
                   {14, 20, 21, 26}, 
                   {30, 34, 43, 50} }
       x = 42
Output : NO

Solution:

There are a lot of ways to solve this problem but let’s discuss the idea of a very naive or brute-force approach here.

A Simple Solution is to one by one compare x with every element of the matrix. If matches, then return true. If we reach the end then return false. The time complexity of this solution is O(n x m).

Below is the implementation of the above idea:

C++

#include <bits/stdc++.h>
using namespace std;

bool searchInMatrix(vector<vector<int> >& arr, int x)
{
    int m = arr.size(), n = arr[0].size();

    for (int i = 0; i < m; i++) {
        for (int j = 0; j < n; j++) {
            if (arr[i][j] == x)
                return true;
        }
    }
    return false;
}

// Driver program to test above
int main()
{
    int x = 8;
    vector<vector<int> > arr
        = { { 0, 6, 8, 9, 11 },
            { 20, 22, 28, 29, 31 },
            { 36, 38, 50, 61, 63 },
            { 64, 66, 100, 122, 128 } };

    if (searchInMatrix(arr, x))
        cout << "YES" << endl;
    else
        cout << "NO" << endl;

    return 0;
}

Java

// Java code for the above approach

import java.io.*;

class GFG {

  static boolean searchInMatrix(int[][] arr, int x)
  {
    int m = arr.length, n = arr[0].length;

    for (int i = 0; i < m; i++) {
      for (int j = 0; j < n; j++) {
        if (arr[i][j] == x)
          return true;
      }
    }
    return false;
  }

  public static void main(String[] args)
  {
    int x = 8;
    int[][] arr = { { 0, 6, 8, 9, 11 },
                   { 20, 22, 28, 29, 31 },
                   { 36, 38, 50, 61, 63 },
                   { 64, 66, 100, 122, 128 } };

    if (searchInMatrix(arr, x)) {
      System.out.println("YES");
    }
    else {
      System.out.println("NO");
    }
  }
}

// This code is contributed by lokeshmvs21.

Python3

# Python code for above approach
def searchInMatrix(arr, x):
    # m=4,n=5
    for i in range(0, 4):
        for j in range(0, 5):
            if(arr[i][j] == x):
                return 1
    return

x = 8
arr = [[0, 6, 8, 9, 11],
       [20, 22, 28, 29, 31],
       [36, 38, 50, 61, 63],
       [64, 66, 100, 122, 128]]
if(searchInMatrix(arr, x)):
    print("YES")
else:
    print("NO")

    # This code is contributed by ishankhandelwals.

C#

// C# code for the above approach

using System;

public class GFG {
    static bool searchInMatrix(int[,] arr, int x) {
        int m = arr.GetLength(0), n = arr.GetLength(1);

        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (arr[i, j] == x)
                    return true;
            }
        }
        return false;
    }

    public static void Main(string[] args) {
        int x = 8;
        int[,] arr = { { 0, 6, 8, 9, 11 },
            { 20, 22, 28, 29, 31 },
            { 36, 38, 50, 61, 63 },
            { 64, 66, 100, 122, 128 }
        };

        if (searchInMatrix(arr, x)) {
            Console.WriteLine("YES");
        } else {
            Console.WriteLine("NO");
        }
    }
}

Javascript

<script>
       // JavaScript code for the above approach



       function searchInMatrix(arr, x) {
           let m = arr.length, n = arr[0].length;

           for (let i = 0; i < m; i++) {
               for (let j = 0; j < n; j++) {
                   if (arr[i][j] == x)
                       return true;
               }
           }
           return false;
       }

       // Driver program to test above

       let x = 8;
       let arr
           = [[0, 6, 8, 9, 11],
           [20, 22, 28, 29, 31],
           [36, 38, 50, 61, 63],
           [64, 66, 100, 122, 128]];

       if (searchInMatrix(arr, x))
           document.write("YES" + "<br>");
       else
           document.write("NO" + "<br>");


// This code is contributed by Potta Lokesh

   </script>

Output

YES

Time Complexity: O(M*N), where M and N are the numbers of rows and columns respectively.
Auxiliary Space: O(1)

2. Program to print the Diagonals of a Matrix

Given a 2D square matrix, print the Principal and Secondary diagonals.

Examples :

Input: 
1 2 3 4
4 3 2 1
7 8 9 6
6 5 4 3
Output:
Principal Diagonal: 1, 3, 9, 3
Secondary Diagonal: 4, 2, 8, 6

Input:
1 1 1
1 1 1
1 1 1
Output:
Principal Diagonal: 1, 1, 1
Secondary Diagonal: 1, 1, 1

Solution:

The primary diagonal is formed by the elements A00, A11, A22, A33.
Condition for Principal Diagonal: The row-column condition is row = column.

The secondary diagonal is formed by the elements A03, A12, A21, A30. 
Condition for Secondary Diagonal: The row-column condition is row = numberOfRows – column -1.

C++

// C++ Program to print the Diagonals of a Matrix

#include <bits/stdc++.h>
using namespace std;

const int MAX = 100;

// Function to print the Principal Diagonal
void printPrincipalDiagonal(int mat[][MAX], int n)
{
    cout << "Principal Diagonal: ";

    for (int i = 0; i < n; i++) {
        for (int j = 0; j < n; j++) {

            // Condition for principal diagonal
            if (i == j)
                cout << mat[i][j] << ", ";
        }
    }
    cout << endl;
}

// Function to print the Secondary Diagonal
void printSecondaryDiagonal(int mat[][MAX], int n)
{
    cout << "Secondary Diagonal: ";

    for (int i = 0; i < n; i++) {
        for (int j = 0; j < n; j++) {

            // Condition for secondary diagonal
            if ((i + j) == (n - 1))
                cout << mat[i][j] << ", ";
        }
    }
    cout << endl;
}

// Driver code
int main()
{
    int n = 4;
    int a[][MAX] = { { 1, 2, 3, 4 },
                     { 5, 6, 7, 8 },
                     { 1, 2, 3, 4 },
                     { 5, 6, 7, 8 } };

    printPrincipalDiagonal(a, n);
    printSecondaryDiagonal(a, n);
    return 0;
}

Java

// Java Program to print the Diagonals of a Matrix
class GFG {

  // Function to print the Principal Diagonal
  public static void printPrincipalDiagonal(int mat[][], int n) {
    System.out.print("Principal Diagonal: ");

    for (int i = 0; i < n; i++) {
      for (int j = 0; j < n; j++) {

        // Condition for principal diagonal
        if (i == j)
          System.out.print(mat[i][j] + ", ");
      }
    }
    System.out.print("\n");
  }

  // Function to print the Secondary Diagonal
  public static void printSecondaryDiagonal(int mat[][], int n) {
    System.out.print("Secondary Diagonal: ");

    for (int i = 0; i < n; i++) {
      for (int j = 0; j < n; j++) {

        // Condition for secondary diagonal
        if ((i + j) == (n - 1))
          System.out.print(mat[i][j] + ", ");
      }
    }
    System.out.print("\n");
  }

  // Driver Code
  public static void main(String[] args) {
    int n = 4;
    int a[][] = { { 1, 2, 3, 4 },
                 { 5, 6, 7, 8 },
                 { 1, 2, 3, 4 },
                 { 5, 6, 7, 8 }
                };

    printPrincipalDiagonal(a, n);
    printSecondaryDiagonal(a, n);
  }
}

// This code is contributed by ajaymakvana.

Python3

# Python3 Program to print the Diagonals of a Matrix
MAX = 100

# Function to print the Principal Diagonal
def printPrincipalDiagonal(mat, n):
    print("Principal Diagonal: ",end="")

    for i in range(0,n):
        for j in range(0,n):

            # Condition for principal diagonal
            if (i == j):
                print(mat[i][j] , ", ", end="")
    print("")

# Function to print the Secondary Diagonal
def printSecondaryDiagonal(mat, n):
    print("Secondary Diagonal: ",end = "")

    for i in range(0,n):
        for j in range(0,n):

            # Condition for secondary diagonal
            if ((i + j) == (n - 1)):
                print(mat[i][j] , ", ", end = "")
    print("")

# Driver code
n = 4
a = [ [ 1, 2, 3, 4 ],
                    [ 5, 6, 7, 8 ],
                    [ 1, 2, 3, 4 ],
                    [ 5, 6, 7, 8 ] ]

printPrincipalDiagonal(a, n)
printSecondaryDiagonal(a, n)

# This code is contributed by akashish__

C#

// C# Program to print the Diagonals of a Matrix
using System;

namespace ConsoleApp1
{
  class Program
  {
    const int MAX = 100;

    // Function to print the Principal Diagonal
    static void printPrincipalDiagonal(int[,] mat, int n)
    {
      Console.Write("Principal Diagonal: ");

      for (int i = 0; i < n; i++)
      {
        for (int j = 0; j < n; j++)
        {

          // Condition for principal diagonal
          if (i == j)
            Console.Write(mat[i, j] + ", ");
        }
      }
      Console.WriteLine();
    }

    // Function to print the Secondary Diagonal
    static void printSecondaryDiagonal(int[,] mat, int n)
    {
      Console.Write("Secondary Diagonal: ");

      for (int i = 0; i < n; i++)
      {
        for (int j = 0; j < n; j++)
        {

          // Condition for secondary diagonal
          if ((i + j) == (n - 1))
            Console.Write(mat[i, j] + ", ");
        }
      }
      Console.WriteLine();
    }

    // Driver code
    static void Main(string[] args)
    {
      int n = 4;
      int[,] a = { { 1, 2, 3, 4 },
                  { 5, 6, 7, 8 },
                  { 1, 2, 3, 4 },
                  { 5, 6, 7, 8 } };

      printPrincipalDiagonal(a, n);
      printSecondaryDiagonal(a, n);
    }
  }
}

// This code is contributed by ishankhandelwals.

Javascript

// javascript Program to print the Diagonals of a Matrix

// Function to print the Principal Diagonal
function printPrincipalDiagonal(mat, n)
{
    let s = "";
    for (let i = 0; i < n; i++) {
        for (let j = 0; j < n; j++) {

            // Condition for principal diagonal
            if (i == j)
               {
                   s += mat[i][j];
                   s += " ";
               }
        }
    }
    console.log("Principal Diagonal: "+s);

}

// Function to print the Secondary Diagonal
function printSecondaryDiagonal(mat,n)
{
    let s="";
    for (let i = 0; i < n; i++) {
        for (let j = 0; j < n; j++) {

            // Condition for secondary diagonal
            if ((i + j) == (n - 1))
               {
                   s += mat[i][j];
                   s +=" ";
               }
        }
    }
    console.log("Secondary Diagonal: "+s);

}

// Driver code

    let n = 4;
    let a = [[ 1, 2, 3, 4 ],
             [ 5, 6, 7, 8 ],
             [ 1, 2, 3, 4 ],
             [ 5, 6, 7, 8 ] ];

    printPrincipalDiagonal(a, n);
    printSecondaryDiagonal(a, n);

   // This code is contributed by garg28harsh.

Output

Principal Diagonal: 1, 6, 3, 8, 
Secondary Diagonal: 4, 7, 2, 5, 

Time Complexity: O(n2), As there is a nested loop involved so the time complexity is squared.
Auxiliary Space: O(1).  

3. Sort the given matrix:

Given a n x n matrix. The problem is to sort the given matrix in strict order. Here strict order means that the matrix is sorted in a way such that all elements in a row are sorted in increasing order and for row ‘i’, where 1 <= i <= n-1, the first element of row ‘i’ is greater than or equal to the last element of row ‘i-1’.

Examples:

Input : mat[][] = { {5, 4, 7}, 
                         {1, 3, 8}, 
                        {2, 9, 6} }
Output : 1 2 3
             4 5 6
             7 8 9

Solution: 

The idea to solve this proble is Create a temp[] array of size n^2. Starting with the first row one by one copy the elements of the given matrix into temp[]. Sort temp[]. Now one by one copy the elements of temp[] back to the given matrix.

Below is the implementation:

C++

// C++ implementation to sort the given matrix
#include <bits/stdc++.h>
using namespace std;

#define SIZE 10

// function to sort the given matrix
void sortMat(int mat[SIZE][SIZE], int n)
{
    // temporary matrix of size n^2
    int temp[n * n];
    int k = 0;

    // copy the elements of matrix one by one
    // into temp[]
    for (int i = 0; i < n; i++)
        for (int j = 0; j < n; j++)
            temp[k++] = mat[i][j];

    // sort temp[]
    sort(temp, temp + k);

    // copy the elements of temp[] one by one
    // in mat[][]
    k = 0;
    for (int i = 0; i < n; i++)
        for (int j = 0; j < n; j++)
            mat[i][j] = temp[k++];
}

// function to print the given matrix
void printMat(int mat[SIZE][SIZE], int n)
{
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < n; j++)
            cout << mat[i][j] << " ";
        cout << endl;
    }
}

// Driver program to test above
int main()
{
    int mat[SIZE][SIZE]
        = { { 5, 4, 7 }, { 1, 3, 8 }, { 2, 9, 6 } };
    int n = 3;

    cout << "Original Matrix:\n";
    printMat(mat, n);

    sortMat(mat, n);

    cout << "\nMatrix After Sorting:\n";
    printMat(mat, n);

    return 0;
}

Java

// Java implementation to sort the given matrix

import java.io.*;
import java.util.*;

class GFG {

    static final int SIZE = 10;

    // function to sort the given matrix
    static void sortMat(int[][] mat, int n)
    {
        // temporary matrix of size n^2
        int[] temp = new int[n * n];
        int k = 0;

        // copy the elements of matrix one by one
        // into temp[]
        for (int i = 0; i < n; i++)
            for (int j = 0; j < n; j++)
                temp[k++] = mat[i][j];

        // sort temp[]
        Arrays.sort(temp);

        // copy the elements of temp[] one by one
        // in mat[][]
        k = 0;
        for (int i = 0; i < n; i++)
            for (int j = 0; j < n; j++)
                mat[i][j] = temp[k++];
    }

    // function to print the given matrix
    static void printMat(int[][] mat, int n)
    {
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++)
                System.out.print(mat[i][j] + " ");
            System.out.println();
        }
    }

    public static void main(String[] args)
    {
        int[][] mat
            = { { 5, 4, 7 }, { 1, 3, 8 }, { 2, 9, 6 } };
        int n = 3;

        System.out.println("Original Matrix:");
        printMat(mat, n);

        sortMat(mat, n);

        System.out.println("\nMatrix After Sorting:");
        printMat(mat, n);
    }
}

// This code is contributed by lokeshmvs21.

Python3

# Python Implementation to sort the given matrix
def sortMat(mat, n):

    # temporary matrix of size n^2
    temp = [0]*n*n
    k = 0

    # copy the elements of matrix one by one
    # leto temp[]
    for i in range(0, n):
        for j in range(0, n):
            temp[k] = mat[i][j]
            k += 1

    # sort temp[]
    temp.sort(reverse = True)

    # copy the elements of temp[] one by one
    # in mat[][]
    k = 0
    for i in range(0, n):
        for j in range(0, n):
            mat[i][j] = temp[k]
            k += 1

# function to print the given matrix
def printMat(mat, n):
    for i in range(0, n):
        for j in range(0, n):
            print(mat[i][j], end = ' ')
        print("")

# Driver program to test above
mat = [[5, 4, 7],
       [1, 3, 8],
       [2, 9, 6]]
n = 3

print("Original Matrix:")
printMat(mat, n)
sortMat(mat, n)
print("\nMatrix After Sorting:")
printMat(mat, n)

# This code is contributed by ishankhandelwals.

C#

using System;
using System.Linq;

public class GFG{

  static readonly int SIZE = 10;

  // function to sort the given matrix
  static void sortMat(int[,] mat, int n)
  {

    // temporary matrix of size n^2
    int[] temp = new int[n * n];
    int k = 0;

    // copy the elements of matrix one by one
    // into temp[]
    for (int i = 0; i < n; i++)
      for (int j = 0; j < n; j++)
        temp[k++] = mat[i,j];

    // sort temp[]
    Array.Sort(temp);

    // copy the elements of temp[] one by one
    // in mat[][]
    k = 0;
    for (int i = 0; i < n; i++)
      for (int j = 0; j < n; j++)
        mat[i,j] = temp[k++];
  }

  // function to print the given matrix
  static void printMat(int[,] mat, int n)
  {
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < n; j++)
        Console.Write(mat[i,j] + " ");
      Console.WriteLine();
    }
  }

  static public void Main (){
    int[,] mat
      = { { 5, 4, 7 }, { 1, 3, 8 }, { 2, 9, 6 } };
    int n = 3;

    Console.WriteLine("Original Matrix:");
    printMat(mat, n);

    sortMat(mat, n);

    Console.WriteLine("\nMatrix After Sorting:");
    printMat(mat, n);
  }
}

// This code is contributed by akashish__

Javascript

// JS implementation to sort the given matrix
// function to sort the given matrix
function sortMat(mat, n)
{

    // temporary matrix of size n^2
    let temp = [];
    let k = 0;

    // copy the elements of matrix one by one
    // leto temp[]
    for (let i = 0; i < n; i++)
        for (let j = 0; j < n; j++)
            temp[k++] = mat[i][j];

    // sort temp[]
    temp.sort(function (a, b) { return b - a });

    // copy the elements of temp[] one by one
    // in mat[][]
    k = 0;
    for (let i = 0; i < n; i++)
        for (let j = 0; j < n; j++)
            mat[i][j] = temp[k++];
}

// function to print the given matrix
function printMat(mat, n) {
    for (let i = 0; i < n; i++) {
        for (let j = 0; j < n; j++)
            console.log(mat[i][j]);
    }
}

// Driver program to test above
let mat = [[5, 4, 7], [1, 3, 8], [2, 9, 6]];
let n = 3;
console.log("Original Matrix:\n");
printMat(mat, n);
sortMat(mat, n);
console.log("\nMatrix After Sorting:\n");
printMat(mat, n);

// This code is contributed by ishankhandelwals.

Output

Original Matrix:
5 4 7 
1 3 8 
2 9 6 

Matrix After Sorting:
1 2 3 
4 5 6 
7 8 9 

Time Complexity: O(n2log2n). 
Auxiliary Space: O(n2), since n * n extra space has been taken.

4. Rotate a Matrix by 180 degree

Given a square matrix, the task is that turn it by 180 degrees in an anti-clockwise direction without using any extra space.

Examples :

Input :  1  2  3
        4  5  6
        7  8  9
Output : 9 8 7 
        6 5 4 
        3 2 1

Input :  1 2 3 4 
        5 6 7 8 
        9 0 1 2 
        3 4 5 6 
Output : 6 5 4 3 
        2 1 0 9 
        8 7 6 5 
        4 3 2 1

Solution:

There are four steps that are required to solve this problem:
1- Find the transpose of a matrix. 
2- Reverse columns of the transpose. 
3- Find the transpose of a matrix. 
4- Reverse columns of the transpose

Illustration:

Let the given matrix be
1  2  3  4
5  6  7  8
9  10 11 12
13 14 15 16

First we find transpose.
1 5 9 13
2 6 10 14
3 7 11 15
4 8 12 16

Then we reverse elements of every column.
4 8 12 16
3 7 11 15
2 6 10 14
1 5  9 13

then transpose again 
4 3 2 1 
8 7 6 5 
12 11 10 9
16 15 14 13

Then we reverse elements of every column again
16 15 14 13 
12 11 10 9 
8 7 6 5 
4 3 2 1

Below is the implementation:

C++

// C++ program for left rotation of matrix by 180
#include <bits/stdc++.h>
using namespace std;

#define R 4
#define C 4

// Function to rotate the matrix by 180 degree
void reverseColumns(int arr[R][C])
{
    for (int i = 0; i < C; i++)
        for (int j = 0, k = C - 1; j < k; j++, k--)
            swap(arr[j][i], arr[k][i]);
}

// Function for transpose of matrix
void transpose(int arr[R][C])
{
    for (int i = 0; i < R; i++)
        for (int j = i; j < C; j++)
            swap(arr[i][j], arr[j][i]);
}

// Function for display the matrix
void printMatrix(int arr[R][C])
{
    for (int i = 0; i < R; i++) {
        for (int j = 0; j < C; j++)
            cout << arr[i][j] << " ";
        cout << '\n';
    }
}

// Function to anticlockwise rotate matrix
// by 180 degree
void rotate180(int arr[R][C])
{
    transpose(arr);
    reverseColumns(arr);
    transpose(arr);
    reverseColumns(arr);
}

// Driven code
int main()
{
    int arr[R][C] = { { 1, 2, 3, 4 },
                      { 5, 6, 7, 8 },
                      { 9, 10, 11, 12 },
                      { 13, 14, 15, 16 } };
    rotate180(arr);
    printMatrix(arr);
    return 0;
}

Java

// Java program for left rotation of matrix by 180
import java.util.*;

class GFG {

    public static int R = 4;
    public static int C = 4;

    // Function to rotate the matrix by 180 degree
    public static void reverseColumns(int[][] arr) {
        for (int i = 0; i < C; i++)
            for (int j = 0, k = C - 1; j < k; j++, k--) {
                int temp = arr[j][i];
                arr[j][i] = arr[k][i];
                arr[k][i] = temp;
            }
    }

    // Function for transpose of matrix
    public static void transpose(int[][] arr) {
        for (int i = 0; i < R; i++)
            for (int j = i; j < C; j++) {
                int temp = arr[i][j];
                arr[i][j] = arr[j][i];
                arr[j][i] = temp;
            }
    }

    // Function for display the matrix
    public static void printMatrix(int[][] arr) {
        for (int i = 0; i < R; i++) {
            for (int j = 0; j < C; j++)
                System.out.print(arr[i][j] + " ");
            System.out.println();
        }
    }

    // Function to anticlockwise rotate matrix by 180 degree
    public static void rotate180(int[][] arr) {
        transpose(arr);
        reverseColumns(arr);
        transpose(arr);
        reverseColumns(arr);
    }

    // Driver code
    public static void main(String[] args) {
        int[][] arr = { { 1, 2, 3, 4 },
            { 5, 6, 7, 8 },
            { 9, 10, 11, 12 },
            { 13, 14, 15, 16 }
        };
        rotate180(arr);
        printMatrix(arr);
    }
}

Python3

# Python program for left rotation of matrix by 180
R = 4
C = 4

# Function to rotate the matrix by 180 degree
def reverseColumns(arr):
  for i in range(C):
    for j in range(0, int(C / 2)):
      x = arr[j][i]
      arr[j][i] = arr[C - 1 - j][i]
      arr[C - 1 - j][i] = x

# Function for transpose of matrix
def transpose(arr):
  for i in range(R):
    for j in range(i, C):
      x = arr[j][i]
      arr[j][i] = arr[i][j]
      arr[i][j] = x

# Function for display the matrix
def printMatrix(arr):
  for i in range(R):
    for j in range(C):
      print(arr[i][j], end = ' ')
    print()

# Function to anticlockwise rotate matrix
# by 180 degree
def rotate180(arr):
  transpose(arr)
  reverseColumns(arr)
  transpose(arr)
  reverseColumns(arr)

# Driven code
arr = [[1, 2, 3, 4 ],
              [ 5, 6, 7, 8 ],
              [ 9, 10, 11, 12 ],
              [ 13, 14, 15, 16 ]]
rotate180(arr)
printMatrix(arr)

# This code is contributed by akashish__

C#

// C# code
using System;

namespace ConsoleApp1 {
  class Program {
    const int R = 4;
    const int C = 4;
    static void Main(string[] args)
    {

      int[, ] arr = new int[, ] { { 1, 2, 3, 4 },
                                 { 5, 6, 7, 8 },
                                 { 9, 10, 11, 12 },
                                 { 13, 14, 15, 16 } };

      Rotate180(arr);
      PrintMatrix(arr);
      Console.Read();
    }

    // Function to rotate the matrix by 180 degree
    static void ReverseColumns(int[, ] arr)
    {
      for (int i = 0; i < C; i++)
        for (int j = 0, k = C - 1; j < k; j++, k--)
          Swap(ref arr[j, i], ref arr[k, i]);
    }

    // Function for transpose of matrix
    static void Transpose(int[, ] arr)
    {
      for (int i = 0; i < R; i++)
        for (int j = i; j < C; j++)
          Swap(ref arr[i, j], ref arr[j, i]);
    }

    // Function for display the matrix
    static void PrintMatrix(int[, ] arr)
    {
      for (int i = 0; i < R; i++) {
        for (int j = 0; j < C; j++)
          Console.Write(arr[i, j] + " ");
        Console.WriteLine();
      }
    }

    // Function to anticlockwise rotate matrix
    // by 180 degree
    static void Rotate180(int[, ] arr)
    {
      Transpose(arr);
      ReverseColumns(arr);
      Transpose(arr);
      ReverseColumns(arr);
    }

    // Swap two numbers
    static void Swap(ref int a, ref int b)
    {
      int temp = a;
      a = b;
      b = temp;
    }
  }
}

Javascript

//  Javascript program for left rotation of matrix by 180


let R= 4
let C= 4

// Function to rotate the matrix by 180 degree
function reverseColumns(arr)
{
    for (let i = 0; i < C; i++)
        for (let j = 0, k = C - 1; j < k; j++, k--)
            {
            let x= arr[j][i];
               arr[j][i] = arr[k][i];
             arr[k][i]=x;
            }

}

// Function for transpose of matrix
function transpose(arr)
{
    for (let i = 0; i < R; i++)
        for (let j = i; j < C; j++)
            {
            let x= arr[j][i];
               arr[j][i] = arr[i][j];
             arr[i][j]=x;
        }
}

// Function for display the matrix
function printMatrix(arr)
{
    document.write(arr);
}

// Function to anticlockwise rotate matrix
// by 180 degree
function rotate180(arr)
{
    transpose(arr);
    reverseColumns(arr);
    transpose(arr);
    reverseColumns(arr);
}

// Driven code

    let arr = [[1, 2, 3, 4 ],
                      [ 5, 6, 7, 8 ],
                      [ 9, 10, 11, 12 ],
                      [ 13, 14, 15, 16 ]];
    rotate180(arr);
    printMatrix(arr);

Output

16 15 14 13 
12 11 10 9 
8 7 6 5 
4 3 2 1 

Time complexity: O(R*C) 
Auxiliary Space: O(1)

5. Find unique elements in a matrix

Given a matrix mat[][] having n rows and m columns. We need to find unique elements in the matrix i.e, those elements not repeated in the matrix or those whose frequency is 1.

Examples:

Input :  20  15  30  2
        2   3   5   30
        6   7   6   8
Output : 3  20  5  7  8  15

Input :  1  2  3  
        5  6  2
        1  3  5
        6  2  2
Output : No unique element in the matrix

Solution:

The idea is to use hashing and traverse through all the elements of the matrix, If an element is present in the dictionary, then increment its count. Otherwise insert an element with value = 1. 

Below is the implementation:

C++

// C++ program to find unique
// element in matrix
#include <bits/stdc++.h>
using namespace std;
#define R 4
#define C 4

// function that calculate unique element
int unique(int mat[R][C], int n, int m)
{
    int maximum = 0, flag = 0;
    for (int i = 0; i < n; i++)
        for (int j = 0; j < m; j++)
            // Find maximum element in
            // a matrix
            if (maximum < mat[i][j])
                maximum = mat[i][j];

    // Take 1-D array of (maximum + 1)
    // size
    int b[maximum + 1] = { 0 };
    for (int i = 0; i < n; i++)
        for (int j = 0; j < m; j++){
            int y = mat[i][j];
            b[y]++;
        }

    // print unique element
    for (int i = 1; i <= maximum; i++)
        if (b[i] == 1)
            cout << i << " ";
    flag = 1;

    if (!flag) {
        cout << "No unique element in the matrix";
    }
}

// Driver program
int main()
{
    int mat[R][C] = { { 1, 2, 3, 20 },
                      { 5, 6, 20, 25 },
                      { 1, 3, 5, 6 },
                      { 6, 7, 8, 15 } };

    // function that calculate unique element
    unique(mat, R, C);
    return 0;
}

// This code is contributed by Naman_Garg.

Java

// Java program to find unique
// element in matrix

import java.io.*;

class GFG {

    // function that calculate unique element
    static void unique(int[][] mat, int n, int m)
    {
        int maximum = 0, flag = 0;
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                // Find maximum element in
                // a matrix
                if (maximum < mat[i][j]) {
                    maximum = mat[i][j];
                }
            }
        }

        // Take 1-D array of (maximum + 1)
        // size
        int[] b = new int[maximum + 1];
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                int y = mat[i][j];
                b[y]++;
            }
        }

        // print unique element
        for (int i = 1; i <= maximum; i++) {
            if (b[i] == 1) {
                System.out.print(i + " ");
            }
        }
        flag = 1;

        if (flag == 0) {
            System.out.print(
                "No unique element in the matrix");
        }
    }

    public static void main(String[] args)
    {
        int R = 4, C = 4;
        int[][] mat = { { 1, 2, 3, 20 },
                        { 5, 6, 20, 25 },
                        { 1, 3, 5, 6 },
                        { 6, 7, 8, 15 } };

        // function that calculate unique element
        unique(mat, R, C);
    }
}

// This code is contributed by lokeshmvs21.

Python3

# Python program to find unique
# element in matrix

# Function that calculate unique element
def unique(mat, n, m):
    maximum = 0
    flag = 0
    for i in range(n):
        for j in range(m):
            # Find maximum element in
            # a matrix
            if maximum < mat[i][j]:
                maximum = mat[i][j]

    # Take 1-D array of (maximum + 1)
    # size
    b = [0] * (maximum + 1)
    for i in range(n):
        for j in range(m):
            y = mat[i][j]
            b[y] += 1

    # print unique element
    for i in range(1, maximum+1):
        if b[i] == 1:
            print(i, end=' ')
    flag = 1

    if flag == 0:
        print("No unique element in the matrix")

R = 4
C = 4
mat = [[1, 2, 3, 20],
       [5, 6, 20, 25],
       [1, 3, 5, 6],
       [6, 7, 8, 15]]

# function that calculate unique element
unique(mat, R, C)

# This code is contributed by lokesh.

C#

// C# program to find unique element in matrix
using System;
public class GFG {

    // function that calculate unique element
    static void unique(int[, ] mat, int n, int m)
    {
        int maximum = 0, flag = 0;
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {

                // Find maximum element in
                // a matrix
                if (maximum < mat[i, j]) {
                    maximum = mat[i, j];
                }
            }
        }

        // Take 1-D array of (maximum + 1)
        // size
        int[] b = new int[maximum + 1];
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                int y = mat[i, j];
                b[y]++;
            }
        }

        // print unique element
        for (int i = 1; i <= maximum; i++) {
            if (b[i] == 1) {
                Console.Write(i + " ");
            }
        }
        flag = 1;

        if (flag == 0) {
            Console.Write(
                "No unique element in the matrix");
        }
    }

    public static void Main(string[] args)
    {
        int R = 4, C = 4;
        int[, ] mat = { { 1, 2, 3, 20 },
                        { 5, 6, 20, 25 },
                        { 1, 3, 5, 6 },
                        { 6, 7, 8, 15 } };

        // function that calculate unique element
        unique(mat, R, C);
    }
}

// This code is contributed by ajaymakavana.

Javascript

// JS code

let mat = [[1, 2, 3, 20], [5, 6, 20, 25], [1, 3, 5, 6], [6, 7, 8, 15]];

let max = 0;
let flag = 0;
for (let i = 0; i < mat.length; i++) {
  for (let j = 0; j < mat.length; j++) {
    if (max < mat[i][j]) {
      max = mat[i][j];
    }
  }
}

let b = new Array(max + 1).fill(0);
for (let i = 0; i < mat.length; i++) {
  for (let j = 0; j < mat.length; j++) {
    let y = mat[i][j];
    b[y]++;
  }
}

for (let i = 1; i <= max; i++) {
  if (b[i] == 1) {
    console.log(i + " ");
    flag = 1;
  }
}

if (!flag) {
  console.log("No unique element in the matrix");
}

Output

2 7 8 15 25 

Time Complexity: O(m*n) where m is the number of rows & n is the number of columns.
Auxiliary Space: O(max(matrix)). 

More Practice problems on Matrix/Grid:

Question	Practice
Rotate Matrix Elements	Link
Inplace rotate square matrix by 90 degrees | Set 1	Link
Sort the given matrix	Link
Find the row with the maximum number of 1s	Link
Program to multiply two matrices	Link
Print a given matrix in spiral form	Link
Find unique elements in a matrix	Link
Rotate the matrix right by K times	Link
Find the number of islands	Link
Check given matrix is a magic square or not	Link
Diagonally Dominant Matrix	Link
Submatrix Sum Queries	Link
Counting sets of 1s and 0s in a binary matrix	Link
Possible moves of a knight	Link
Boundary elements of a Matrix	Link
Check horizontal and vertical symmetry in binary matrix	Link
Print matrix in a diagonal pattern	Link
Find if the given matrix is Toeplitz or not	Link
Shortest path in a Binary Maze	Link
Minimum Initial Points to Reach Destination	Link
Find a common element in all rows of a given row-wise sorted matrix	Link