Program to find transpose of a matrix

Last Updated : 08 May, 2025

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 mat[n][m] is obtained by changing mat[i][j] to mat[j][i].

Example:

Approach using (N^2) space

Run a nested loop using two integer pointers i and j for 0 <= i < n and 0 <= j < m
Set mat[i][j] equal to mat[j][i]

Below is the implementation of the above approach:

C++

#include <bits/stdc++.h> using namespace std;  // Function to store the transpose of mat in res void transpose(vector<vector<int>>& mat, vector<vector<int>>& res) {     int rows = mat.size();           int cols = mat[0].size();         // Resize res to have dimensions swapped     res.resize(cols, vector<int>(rows));      // Fill res with transposed values of mat     for (int i = 0; i < rows; i++) {         for (int j = 0; j < cols; j++) {             res[j][i] = mat[i][j];          }     } }  // Driver code int main() {     vector<vector<int>> mat = {         { 1, 2, 3 },         { 4, 5, 6 }     };      // Create a result matrix for the transpose     vector<vector<int>> res;         // Function call to calculate the transpose     transpose(mat, res);      // Print the result matrix     cout << "Result matrix is:\n";     for (auto& row : res) {         for (auto& elem : row) {             cout << " " << elem;         }         cout << "\n";      }      return 0; }

// Import necessary libraries #include <stdio.h>  // Define macros for matrix dimensions #define M 2  // Number of rows in the original matrix #define N 3  // Number of columns in the original matrix  // Function to store the transpose of mat in res void transpose(int mat[M][N], int res[N][M]) {        // Fill res with transposed values of mat     for (int i = 0; i < M; i++) {         for (int j = 0; j < N; j++) {             res[j][i] = mat[i][j];          }     } }  // Driver code int main() {     int mat[M][N] = {         { 1, 2, 3 },         { 4, 5, 6 }     };      // Create a result matrix for the transpose     int res[N][M];      // Function call to calculate the transpose     transpose(mat, res);      // Print the result matrix     printf("Result matrix is:\n");     for (int i = 0; i < N; i++) {         for (int j = 0; j < M; j++) {             printf("%d ", res[i][j]);         }         printf("\n");      }     return 0; }

Java

// Import necessary classes import java.util.Arrays; import java.util.Scanner;  public class TransposeMatrix {        // Function to store the transpose of mat in res     public static void transpose(int[][] mat, int[][] res) {         int rows = mat.length;         int cols = mat[0].length;          // Fill res with transposed values of mat         for (int i = 0; i < rows; i++) {             for (int j = 0; j < cols; j++) {                 res[j][i] = mat[i][j];             }         }     }      // Driver code     public static void main(String[] args) {         int[][] mat = {             { 1, 2, 3 },             { 4, 5, 6 }         };          // Create a result matrix for the transpose         int[][] res = new int[mat[0].length][mat.length];          // Function call to calculate the transpose         transpose(mat, res);          // Print the result matrix         System.out.println("Result matrix is:");         for (int[] row : res) {             System.out.println(Arrays.toString(row));         }     } }

Python

# Function to store the transpose of mat in res def transpose(mat):        # Fill res with transposed values of mat     return [[mat[j][i] for j in range(len(mat))] for i in range(len(mat[0]))]  # Driver code if __name__ == '__main__':     mat = [         [1, 2, 3],         [4, 5, 6]     ]      # Function call to calculate the transpose     res = transpose(mat)      # Print the result matrix     print("Result matrix is:")     for row in res:         print(" ".join(map(str, row)))

// Function to store the transpose of mat in res void Transpose(int[][] mat, out int[][] res) {     int rows = mat.Length;     int cols = mat[0].Length;      // Resize res to have dimensions swapped     res = new int[cols][];     for (int i = 0; i < cols; i++) {         res[i] = new int[rows];     }      // Fill res with transposed values of mat     for (int i = 0; i < rows; i++) {         for (int j = 0; j < cols; j++) {             res[j][i] = mat[i][j];         }     } }  // Driver code public static void Main() {     int[][] mat = {         new int[] { 1, 2, 3 },         new int[] { 4, 5, 6 }     };      // Create a result matrix for the transpose     int[][] res;      // Function call to calculate the transpose     Transpose(mat, out res);      // Print the result matrix     Console.WriteLine("Result matrix is:");     foreach (var row in res) {         foreach (var elem in row) {             Console.Write(" " + elem);         }         Console.WriteLine();     } }

JavaScript

// Function to store the transpose of mat in res function transpose(mat) {     let rows = mat.length;     let cols = mat[0].length;      // Create a result matrix for the transpose     let res = Array.from({ length: cols }, () => new Array(rows));      // Fill res with transposed values of mat     for (let i = 0; i < rows; i++) {         for (let j = 0; j < cols; j++) {             res[j][i] = mat[i][j];         }     }     return res; }  // Driver code let mat = [     [ 1, 2, 3 ],     [ 4, 5, 6 ] ];  // Function call to calculate the transpose let res = transpose(mat);  // Print the result matrix console.log("Result matrix is:"); for (let row of res) {     console.log(" "+ row.join(' ')); }

Output

Result matrix is:  1 4  2 5  3 6

Time complexity: O(m x n).
Auxiliary Space: O(m x n)

Approach using constant space for Square Matrix

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 mat[i][j] with mat[j][i]

Below is the implementation of the above approach:

C++

#include <bits/stdc++.h> using namespace std;  // Function to convert mat to its transpose void transpose(vector<vector<int>>& mat) {     int n = mat.size();      for (int i = 0; i < n; i++) {         for (int j = i + 1; j < n; j++) {             swap(mat[i][j], mat[j][i]);          }     } }  // Driver code int main() {     vector<vector<int>> mat = {         { 1, 1, 1, 1 },         { 2, 2, 2, 2 },         { 3, 3, 3, 3 },         { 4, 4, 4, 4 }     };      transpose(mat);      cout << "Modified matrix is:" << endl;     for (const auto& row : mat) {         for (int elem : row) {             cout << elem << " ";          }         cout << endl;      }      return 0; }

#include <stdio.h> #define N 4  // Function to convert mat to its transpose void transpose(int mat[N][N]) {     for (int i = 0; i < N; i++) {         for (int j = i + 1; j < N; j++) {             int temp = mat[i][j];             mat[i][j] = mat[j][i];             mat[j][i] = temp;         }     } }  // Driver code int main() {     int mat[N][N] = {         { 1, 1, 1, 1 },         { 2, 2, 2, 2 },         { 3, 3, 3, 3 },         { 4, 4, 4, 4 }     };      transpose(mat);      printf("Modified matrix is:\n");     for (int i = 0; i < N; i++) {         for (int j = 0; j < N; j++) {             printf("%d ", mat[i][j]);         }         printf("\n");     }      return 0; }

Java

import java.util.Arrays;  public class GfG {     static void transpose(int[][] mat) {         int n = mat.length;         for (int i = 0; i < n; i++) {             for (int j = i + 1; j < n; j++) {                 int temp = mat[i][j];                 mat[i][j] = mat[j][i];                 mat[j][i] = temp;             }         }     }      // Driver code     public static void main(String[] args) {         int[][] mat = {             { 1, 1, 1, 1 },             { 2, 2, 2, 2 },             { 3, 3, 3, 3 },             { 4, 4, 4, 4 }         };          transpose(mat);          System.out.println("Modified matrix is:");         for (int[] row : mat) {             System.out.println(Arrays.toString(row));         }     } }

Python

# Function to convert mat to its transpose  def transpose(mat):     n = len(mat)     for i in range(n):         for j in range(i + 1, n):             mat[i][j], mat[j][i] = mat[j][i], mat[i][j]  # Driver code if __name__ == '__main__':     mat = [         [1, 1, 1, 1],         [2, 2, 2, 2],         [3, 3, 3, 3],         [4, 4, 4, 4]     ]      transpose(mat)      print("Modified matrix is:")     for row in mat:         print(' '.join(map(str, row)))

using System;  class Program {        // Function to convert mat to its transpose     static void Transpose(int[,] mat) {         int n = mat.GetLength(0);         for (int i = 0; i < n; i++) {             for (int j = i + 1; j < n; j++) {                 int temp = mat[i, j];                 mat[i, j] = mat[j, i];                 mat[j, i] = temp;             }         }     }      // Driver code     static void Main() {         int[,] mat = {             { 1, 1, 1, 1 },             { 2, 2, 2, 2 },             { 3, 3, 3, 3 },             { 4, 4, 4, 4 }         };          Transpose(mat);          Console.WriteLine("Modified matrix is:");         for (int i = 0; i < mat.GetLength(0); i++) {             for (int j = 0; j < mat.GetLength(1); j++) {                 Console.Write(mat[i, j] + " ");             }             Console.WriteLine();         }     } }

JavaScript

// Function to convert mat to its transpose function transpose(mat) {     let n = mat.length;     for (let i = 0; i < n; i++) {         for (let j = i + 1; j < n; j++) {             [mat[i][j], mat[j][i]] = [mat[j][i], mat[i][j]];         }     } }  // Driver code const mat = [     [1, 1, 1, 1],     [2, 2, 2, 2],     [3, 3, 3, 3],     [4, 4, 4, 4] ];  transpose(mat);  console.log("Modified matrix is:"); mat.forEach(row => {     console.log(row.join(' ')); });

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)

Adjoint and Inverse of a Matrix

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Program to find transpose of a matrix

Approach using (N^2) space

Approach using constant space for Square Matrix

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