CareerCup

My approach:

transpose(A,k,N){
if(k==N)
return;

for(i = k+1;i <= N;i++)
{
// swap a[i,k] with a [k,i]
}
transpose(A,k+1,N);

}

I guess your approach is fine, but would you mind explaining the answer a bit with a simple example for others to follow. thank you.

why is the swapping commented out in your code? w/o that, the current code will just return the orig matrix

I don't remember much.. but can transpose be found only for a square matrix!!

Assumption : Matrix is a square matrix.
@Hello - Non-Square matrix do hava a Transpose matrix associated to it.

void Transpose(int input[][] matrix, int iterationCount){
if(iterationCount == 0) return;
for(int i= 0 ; i< iterationCount ;i++){
Swap(input[iterationCount][i],input[i][iterationCount]);
}
Transpose(matrix, iterationCount-1);
}

gggg

for (int d=0;d<n;d++)
for (int j=1;d+j<n;j++)
swap(a[d,d+j],a[d+j][d]);

void Transpose(int** A, int N) // Square size - 1
{
    if(N == 0) // Transpose of a 1 X 1 Matrix
        return;
    else
    {
        Transpose(A,N-1);
        for(int i=0; i<N; i++)
        {
            int temp = A[i][N] ;
            A[i][N] = A[N][i];
            A[N][i] = temp;
        }
   }
}

Each recursion solves a N X N matrix.

void transpose(int **a, int x)
{
if(x==n-1) return;
int i;
for(i=x;i<n;i++) swap(a[x][i],a[i][x]);
transpose(a,x+1);
}