Java Program to Find Transpose of Matrix
Transpose of a matrix is obtained by changing rows to columns and columns to rows. In other words, the transpose of A[ ][ ] is obtained by changing A[i][j] to A[j][i].
Example of First Transpose of Matrix
Input: [ [ 1 , 2 , 3 ] ,
[ 4 , 5 , 6 ] ,
[ 7 , 8 , 9 ] ]
Output: [ [ 1 , 4 , 7 ] ,
[ 2 , 5 , 8 ] ,
[ 3 , 6 , 9 ] ]
There are Two cases which we will face while transposing the Matrix as mentioned below:
- Rectangular Matrix ( Row != Column )
- Square Matrix ( Row == Column )
Rectangular Matrix
The below program finds the transpose of A[][] and stores the result in B[][].
// Java Program to find
// transpose of a matrix
class GFG {
static final int M = 3;
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 < M; 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 } };
int B[][] = new int[N][M], i, j;
transpose(A, B);
System.out.print("Result matrix is \n");
for (i = 0; i < N; i++) {
for (j = 0; j < M; j++)
System.out.print(B[i][j] + " ");
System.out.print("\n");
}
}
}
Output
Result matrix is 1 2 3 1 2 3 1 2 3 1 2 3
The complexity of the above method:
Time Complexity: O(n*m)
Auxiliary Space: O(n*m)
Square Matrix
1. Using Extra Space
The below program finds the transpose of A[][] and stores the result in B[][], we can change N for different dimensions.
Below is the implementation of the above method:
// 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;
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");
}
}
}
Output
Result matrix is 1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4
The complexity of the above method:
Time Complexity: O(n2)
Auxiliary Space: O(n2)
2. Transpose of In-Place for Square Matrix
Here in this Method we will swap the elements rather than using some sort of extra space.
Below is the implementation of the above topic:
// Java Program to find
// transpose of a matrix
// Driver Class
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;
}
}
// main function
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");
}
}
}
Output
Modified matrix is 1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4
The complexity of the above method:
Time Complexity: O(n2)
Auxiliary Space: O(1)
Please refer complete article on Program to find transpose of a matrix for more details!
