Transpose of a Matrix
You are given a matrix ‘MAT’. Print the transpose of the matrix. The transpose of a matrix is obtained by changing rows to columns and columns to rows. In other words, transpose of a matrix A[][] is obtained by changing A[i][j] to A[j][i].

For example:

Let matrix be : 1 2 3 4 Then transpose of the matrix will be: 1 3 2 4

Input Format :

The first line of input contains an integer ‘T’, denoting the number of test cases. The first line of each test case contains two space-separated integers, ‘M’ and ‘N’, representing the size of the matrices. The next ‘M’ lines of each test case contain ‘N’ space-separated integers, representing the elements of ‘MAT’.

Output Format :

For each test case, print the transpose of the matrix. Print output of each test case in a separate line.

Note :

You do not need to print anything. It has already been taken care of. Just implement the given function.

Constraints :

1

Question

Transpose of a Matrix

You are given a matrix ‘MAT’. Print the transpose of the matrix. The transpose of a matrix is obtained by changing rows to columns and columns to rows. In other words, transpose of a matrix A[][] is obtained by changing A[i][j] to A[j][i].

For example:

Let matrix be : 
1 2
3 4

Then transpose of the matrix will be: 
1 3
2 4

Input Format :

The first line of input contains an integer ‘T’, denoting the number of test cases.

The first line of each test case contains two space-separated integers, ‘M’ and ‘N’, representing the size of the matrices.

The next ‘M’  lines of each test case contain ‘N’ space-separated integers, representing the elements of ‘MAT’.

Output Format :

For each test case, print the transpose of the matrix.

Print output of each test case in a separate line.

Note :

You do not need to print anything. It has already been taken care of. Just implement the given function.

Constraints :

1 <= T <= 10
1 <= M, N <= 3*10^3
0 <= MAT[i][j] <= 10^7

Time Limit: 1 sec

Accepted Answer

Brute-Force Approach

The basic idea is to create a new matrix with the position of elements different than the original matrix. An element at position (‘i’, ‘j’) will lie at position (‘j’, ‘i’) in the transpose of the matrix. So we simply create a new matrix with the position of elements changed.

Here is the algorithm :

Create a 2-D matrix(say, ‘transMAT’) of size ‘N’ and ‘M’ that will store the transpose of the matrix.
Run a loop from 0 to ‘N’ (say, iterator ‘i’) to traverse all the rows.
- Run a loop from 0 to ‘M’ (say, iterator ‘j’) to traverse all the columns
  - Update ‘transMAT[i][j]’ with ‘MAT[j][i]’.
Return ‘transMAT’.

Space Complexity: O(m*n) - For 2d arraysExplanation:

O(M*N), where ‘M’ and ‘N’ are the size of the matrix.

We use a matrix to store the transpose. Therefore, the overall space complexity will be O(M*N).

Time Complexity: O(m*n) - For 2d arraysExplanation:

O(M*N), where ‘M’ and ‘N’ are the size of the matrix.

We traverse all the cells of both the matrices only once. Therefore, the overall time complexity will be O(M*N).

Java (SE 1.8)


/*
	Time complexity: O(M*N)
	Space complexity: O(M*N)
	
	Where 'M' and 'N' are the size of the matrix.
*/
public class Solution {

	public static int[][] transpose(int m, int n, int[][] mat) {
		// New matrix to store transpose.
		int[][] transMat = new int[n][m];

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

				// Assign values.
				transMat[i][j] = mat[j][i];
			}
		}

		return transMat;

	}

}

Python (3.5)

'''
    Time complexity: O(M*N)
    Space complexity: O(M*N)
    
    Where 'M' and 'N' are the size of the matrix.
'''

from typing import List

def transpose(m: int, n: int, mat: List[List[int]]) -> List[List[int]]:
    
    # New matrix to store transpose.
    transMat = [[0 for i in range(m)] for j in range(n)]
    
    for i in range(n):
        for j in range(m):
            
            # Assign values.
            transMat[i][j] = mat[j][i]
        
    return transMat

C++ (g++ 5.4)

/*

    Time complexity: O(M*N)

    Space complexity: O(M*N)

    

    Where 'M' and 'N' are the size of the matrix.

*/



vector> transpose(int m, int n, vector> &mat) {

    

    // New matrix to store transpose.

    vector> transMat(n, vector(m));

    

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

        for(int j = 0; j < m; ++j) {

            

            // Assign values.

            transMat[i][j] = mat[j][i];

        }

    }

    

    return transMat;

}

You are given a matrix ‘MAT’. Print the transpose of the matrix. The transpose of a matrix is obtained by changing rows to columns and columns to rows. In other words, transpose of a matrix A[][] is obtained by changing A[i][j] to A[j][i].

For example:

Input Format :

Output Format :

Note :

Constraints :

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