Program to find transpose of a matrix

Last Updated : 12 Oct, 2023

Given a matrix of size N X M, find the transpose of the matrix
Transpose of a matrix is obtained by changing rows to columns and columns to rows. In other words, transpose of A[N][M] is obtained by changing A[i][j] to A[j][i].

Example:

Approach: Follow the given steps to solve the problem:

Run a nested loop using two integer pointers i and j for 0 <= i < N and 0 <= j < M
Set B[i][j] equal to A[j][i]

Below is the implementation of the above approach:

C++

// C++ program to find 
// transpose of a matrix 
#include <bits/stdc++.h> 
using namespace std; 
#define N 4 
  
// This function stores transpose 
// of A[][] in B[][] 
void transpose(int A[][N], int B[][N]) 
{ 
    int i, j; 
    for (i = 0; i < N; i++) 
        for (j = 0; j < N; j++) 
            B[i][j] = A[j][i]; 
} 
  
// Driver code 
int main() 
{ 
    int A[N][N] = { { 1, 1, 1, 1 }, 
                    { 2, 2, 2, 2 }, 
                    { 3, 3, 3, 3 }, 
                    { 4, 4, 4, 4 } }; 
  
    // Note dimensions of B[][] 
    int B[N][N], i, j; 
  
      // Function call 
    transpose(A, B); 
  
    cout << "Result matrix is \n"; 
    for (i = 0; i < N; i++) { 
        for (j = 0; j < N; j++) 
            cout << " " << B[i][j]; 
  
        cout << "\n"; 
    } 
    return 0; 
} 
  
// This code is contributed by shivanisinghss2110

C

// C program to find 
// transpose of a matrix 
  
#include <stdio.h> 
#define N 4 
  
// This function stores transpose of A[][] in B[][] 
void transpose(int A[][N], int B[][N]) 
{ 
    int i, j; 
    for (i = 0; i < N; i++) 
        for (j = 0; j < N; j++) 
            B[i][j] = A[j][i]; 
} 
  
// Driver code 
int main() 
{ 
    int A[N][N] = { { 1, 1, 1, 1 }, 
                    { 2, 2, 2, 2 }, 
                    { 3, 3, 3, 3 }, 
                    { 4, 4, 4, 4 } }; 
  
    // Note dimensions of B[][] 
    int B[N][N], i, j; 
  
      // Function call 
    transpose(A, B); 
  
    printf("Result matrix is \n"); 
    for (i = 0; i < N; i++) { 
        for (j = 0; j < N; j++) 
            printf("%d ", B[i][j]); 
        printf("\n"); 
    } 
  
    return 0; 
}

Java

// Java Program to find 
// transpose of a matrix 
  
class GFG { 
    static final int N = 4; 
  
    // This function stores transpose 
    // of A[][] in B[][] 
    static void transpose(int A[][], int B[][]) 
    { 
        int i, j; 
        for (i = 0; i < N; i++) 
            for (j = 0; j < N; j++) 
                B[i][j] = A[j][i]; 
    } 
  
    // Driver code 
    public static void main(String[] args) 
    { 
        int A[][] = { { 1, 1, 1, 1 }, 
                      { 2, 2, 2, 2 }, 
                      { 3, 3, 3, 3 }, 
                      { 4, 4, 4, 4 } }; 
  
        int B[][] = new int[N][N], i, j; 
  
          // Function call 
        transpose(A, B); 
  
        System.out.print("Result matrix is \n"); 
        for (i = 0; i < N; i++) { 
            for (j = 0; j < N; j++) 
                System.out.print(B[i][j] + " "); 
            System.out.print("\n"); 
        } 
    } 
} 
  
// This code is contributed by Anant Agarwal.

Python3

# Python3 Program to find 
# transpose of a matrix 
  
N = 4
  
# This function stores 
# transpose of A[][] in B[][] 
  
  
def transpose(A, B): 
  
    for i in range(N): 
        for j in range(N): 
            B[i][j] = A[j][i] 
  
  
# Driver code 
if __name__ == '__main__': 
  A = [[1, 1, 1, 1], 
       [2, 2, 2, 2], 
       [3, 3, 3, 3], 
       [4, 4, 4, 4]] 
  
  
  # To store result 
  B = [[0 for x in range(N)] for y in range(N)] 
  
  # Function call 
  transpose(A, B) 
  
  print("Result matrix is") 
  for i in range(N): 
      for j in range(N): 
          print(B[i][j], " ", end='') 
      print() 

C#

// C# Program to find 
// transpose of a matrix 
using System; 
class GFG { 
  
    static int N = 4; 
  
    // This function stores transpose 
    // of A[][] in B[][] 
    static void transpose(int[, ] A, int[, ] B) 
    { 
        int i, j; 
        for (i = 0; i < N; i++) 
            for (j = 0; j < N; j++) 
                B[i, j] = A[j, i]; 
    } 
  
    // Driver code 
    public static void Main() 
    { 
        int[, ] A = { { 1, 1, 1, 1 }, 
                      { 2, 2, 2, 2 }, 
                      { 3, 3, 3, 3 }, 
                      { 4, 4, 4, 4 } }; 
  
        int[, ] B = new int[N, N]; 
  
          // Function call 
        transpose(A, B); 
  
        Console.WriteLine("Result matrix is \n"); 
        for (int i = 0; i < N; i++) { 
            for (int j = 0; j < N; j++) 
                Console.Write(B[i, j] + " "); 
            Console.Write("\n"); 
        } 
    } 
} 
  
// This code is contributed by nitin mittal

Javascript

<script> 
// javascript Program to find  
// transpose of a matrix     
  
     var N = 4; 
  
    // This function stores transpose 
    // of A in B 
    function transpose(A , B) { 
        var i, j; 
        for (i = 0; i < N; i++) 
            for (j = 0; j < N; j++) 
                B[i][j] = A[j][i]; 
    } 
  
    // Driver code 
      
        var A = [ [ 1, 1, 1, 1 ], [ 2, 2, 2, 2 ], [ 3, 3, 3, 3 ], [4, 4, 4, 4]]; 
  
        var B = Array(N); 
        for(i=0;i<N;i++) 
        B[i] =Array(N).fill(0); 
  
        transpose(A, B); 
  
        document.write("Result matrix is <br/>"); 
        for (i = 0; i < N; i++) { 
            for (j = 0; j < N; j++) 
                document.write(B[i][j] + " "); 
            document.write("<br/>"); 
        } 
// This code contributed by Rajput-Ji 
</script>

PHP

<?php 
  
// This function stores transpose 
// of A[][] in B[][] 
function transpose(&$A, &$B) 
{ 
    $N = 4; 
    for ($i = 0; $i < $N; $i++) 
        for ($j = 0; $j < $N; $j++) 
            $B[$i][$j] = $A[$j][$i]; 
} 
  
// Driver code 
  
$A = array(array(1, 1, 1, 1), 
           array(2, 2, 2, 2), 
           array(3, 3, 3, 3), 
           array(4, 4, 4, 4)); 
  
$N = 4; 
  
// Function call 
transpose($A, $B); 
  
echo "Result matrix is \n"; 
for ($i = 0; $i < $N; $i++) 
{ 
    for ($j = 0; $j < $N; $j++) 
    { 
    echo $B[$i][$j]; 
    echo " "; 
    } 
    echo "\n"; 
} 
  
// This code is contributed 
// by Shivi_Aggarwal  
?>

Output

Result matrix is 
 1 2 3 4
 1 2 3 4
 1 2 3 4
 1 2 3 4

Time complexity: O(M x N).
Auxiliary Space: O(M x N), since M x N extra space has been used.

Program to find the transpose of a matrix using constant space:

Note – This approach works only for square matrices (i.e., – where no. of rows are equal to the number of columns). This algorithm is also known as an “in-place” algorithm as it uses no extra space to solve the problem.

Follow the given steps to solve the problem:

Run a nested loop using two integer pointers i and j for 0 <= i < N and i+1 <= j < N
Swap A[i][j] with A[j][i]

Below is the implementation of the above approach:

C++

#include <bits/stdc++.h> 
using namespace std; 
  
#define N 4 
  
// Converts A[][] to its transpose 
void transpose(int A[][N]) 
{ 
    for (int i = 0; i < N; i++) 
        for (int j = i + 1; j < N; j++) 
            swap(A[i][j], A[j][i]); 
} 
  
int main() 
{ 
    int A[N][N] = { { 1, 1, 1, 1 }, 
                    { 2, 2, 2, 2 }, 
                    { 3, 3, 3, 3 }, 
                    { 4, 4, 4, 4 } }; 
  
    transpose(A); 
  
    printf("Modified matrix is \n"); 
    for (int i = 0; i < N; i++) { 
        for (int j = 0; j < N; j++) 
            printf("%d ", A[i][j]); 
        printf("\n"); 
    } 
  
    return 0; 
}

Java

// Java Program to find 
// transpose of a matrix 
  
class GFG { 
    static final int N = 4; 
  
    // Finds transpose of A[][] in-place 
    static void transpose(int A[][]) 
    { 
        for (int i = 0; i < N; i++) 
            for (int j = i + 1; j < N; j++) { 
                int temp = A[i][j]; 
                A[i][j] = A[j][i]; 
                A[j][i] = temp; 
            } 
    } 
  
    // Driver code 
    public static void main(String[] args) 
    { 
        int A[][] = { { 1, 1, 1, 1 }, 
                      { 2, 2, 2, 2 }, 
                      { 3, 3, 3, 3 }, 
                      { 4, 4, 4, 4 } }; 
  
        transpose(A); 
  
        System.out.print("Modified matrix is \n"); 
        for (int i = 0; i < N; i++) { 
            for (int j = 0; j < N; j++) 
                System.out.print(A[i][j] + " "); 
            System.out.print("\n"); 
        } 
    } 
}

Python3

# Python3 Program to find 
# transpose of a matrix 
  
N = 4
  
# Finds transpose of A[][] in-place 
  
  
def transpose(A): 
  
    for i in range(N): 
        for j in range(i+1, N): 
            A[i][j], A[j][i] = A[j][i], A[i][j] 
  
  
# driver code 
A = [[1, 1, 1, 1], 
     [2, 2, 2, 2], 
     [3, 3, 3, 3], 
     [4, 4, 4, 4]] 
  
transpose(A) 
  
print("Modified matrix is") 
for i in range(N): 
    for j in range(N): 
        print(A[i][j], " ", end='') 
    print() 
  
# This code is contributed 
# by Anant Agarwal. 

C#

// C# Program to find transpose of 
// a matrix 
using System; 
  
class GFG { 
  
    static int N = 4; 
  
    // Finds transpose of A[][] in-place 
    static void transpose(int[, ] A) 
    { 
        for (int i = 0; i < N; i++) 
            for (int j = i + 1; j < N; j++) { 
                int temp = A[i, j]; 
                A[i, j] = A[j, i]; 
                A[j, i] = temp; 
            } 
    } 
  
    // Driver code 
    public static void Main() 
    { 
        int[, ] A = { { 1, 1, 1, 1 }, 
                      { 2, 2, 2, 2 }, 
                      { 3, 3, 3, 3 }, 
                      { 4, 4, 4, 4 } }; 
  
        transpose(A); 
  
        Console.WriteLine("Modified matrix is "); 
        for (int i = 0; i < N; i++) { 
            for (int j = 0; j < N; j++) 
                Console.Write(A[i, j] + " "); 
  
            Console.WriteLine(); 
        } 
    } 
} 
  
// This code is contributed by anuj_67.

Javascript

<script> 
  
// JavaScript Program to find  
// transpose of a matrix     
var N = 4; 
  
    // Finds transpose of A in-place 
    function transpose(A) { 
        for (i = 0; i < N; i++) 
            for (j = i + 1; j < N; j++) { 
                var temp = A[i][j]; 
                A[i][j] = A[j][i]; 
                A[j][i] = temp; 
            } 
    } 
  
    // Driver code 
      
        var A = [ [ 1, 1, 1, 1 ],  
                  [ 2, 2, 2, 2 ], 
                  [ 3, 3, 3, 3 ],  
                  [ 4, 4, 4, 4 ] ]; 
  
        transpose(A); 
  
        document.write("Modified matrix is <br/>"); 
        for (i = 0; i < N; i++) { 
            for (j = 0; j < N; j++) 
                document.write(A[i][j] + " "); 
            document.write("\<br/>"); 
        } 
  
// This code is contributed by aashish1995 
  
</script>

PHP

<?php 
// Converts A[][] to its transpose 
function transpose(&$A) 
{ 
    $N = 4; 
    for ($i = 0; $i < $N; $i++) 
        for ($j = $i + 1; $j < $N; $j++) 
            { 
                $temp = $A[$i][$j]; 
                $A[$i][$j] = $A[$j][$i]; 
                $A[$j][$i] = $temp; 
            } 
} 
  
// Driver Code 
$N = 4; 
$A = array(array(1, 1, 1, 1), 
            array(2, 2, 2, 2), 
           array(3, 3, 3, 3), 
           array(4, 4, 4, 4)); 
  
transpose($A); 
  
echo "Modified matrix is " . "\n"; 
for ($i = 0; $i < $N; $i++) 
{ 
    for ($j = 0; $j < $N; $j++) 
        echo $A[$i][$j] . " "; 
    echo "\n"; 
} 
  
// This code is contributed 
// by Akanksha Rai(Abby_akku) 
?>

Output

Modified matrix is 
1 2 3 4 
1 2 3 4 
1 2 3 4 
1 2 3 4

Time complexity: O(N²).
Auxiliary Space: O(1)

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Program to find transpose of a matrix

C++

C

Java

Python3

C#

Javascript

PHP

Program to find the transpose of a matrix using constant space:

C++

Java

Python3

C#

Javascript

PHP

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