Given an image, how will you turn it by 90 degrees? A vague question. Minimize the browser and try your solution before going further. An image can be treated as 2D matrix which can be stored in a buffer. We are provided with matrix dimensions and it’s base address. How can we turn it? For example see the below picture,
* * * ^ * * * * * * | * * * * * * | * * * * * * | * * *
After rotating right, it appears (observe arrow direction)
* * * * * * * * * * * * -- - - > * * * * * * * * * * * *
The idea is simple. Transform each row of source matrix into required column of final image. We will use an auxiliary buffer to transform the image. From the above picture, we can observe that
first row of source ------> last column of destination second row of source ------> last but-one column of destination so ... on last row of source ------> first column of destination
In pictorial form, we can represent the above transformations of an (m x n) matrix into (n x m) matrix,
If you have not attempted, atleast try your pseudo code now. It will be easy to write our pseudo code. In C/C++ we will usually traverse matrix on row major order. Each row is transformed into different column of final image. We need to construct columns of final image. See the following algorithm (transformation)
for (r = 0; r < m; r++)
{
for (c = 0; c < n; c++)
{
// Hint: Map each source element indices into
// indices of destination matrix element.
dest_buffer [ c ] [ m - r - 1 ] = source_buffer [ r ] [ c ];
}
}
Note that there are various ways to implement the algorithm based on traversal of matrix, row major or column major order. We have two matrices and two ways (row and column major) to traverse each matrix. Hence, there can atleast be 4 different ways of transformation of source matrix into final matrix.
C++
// C++ program to turn an // image by 90 Degree #include <bits/stdc++.h>using namespace std;void displayMatrix(unsigned int const *p, unsigned int row, unsigned int col); void rotate(unsigned int *pS, unsigned int *pD, unsigned int row, unsigned int col); void displayMatrix(unsigned int const *p, unsigned int r, unsigned int c) { unsigned int row, col; cout << ""; for (row = 0; row < r; row++) { for (col = 0; col < c; col++) cout << * (p + row * c + col) << " "; cout << ""; } cout << ""; } void rotate(unsigned int *pS, unsigned int *pD, unsigned int row, unsigned int col) { unsigned int r, c; for (r = 0; r < row; r++) { for (c = 0; c < col; c++) { *(pD + c * row + (row - r - 1)) = *(pS + r * col + c); } } } // Driver Code int main() { // declarations unsigned int image[][4] = {{1, 2, 3, 4}, {5, 6, 7, 8}, {9, 10, 11, 12}}; unsigned int *pSource; unsigned int *pDestination; unsigned int m, n; // setting initial values // and memory allocation m = 3, n = 4, pSource = (unsigned int *)image; pDestination = (unsigned int *)malloc (sizeof(int) * m * n); // process each buffer displayMatrix(pSource, m, n); rotate(pSource, pDestination, m, n); displayMatrix(pDestination, n, m); free(pDestination); return 0; } // This code is contributed by rathbhupendra |
Output :
1 2 3 4 5 6 7 8 9 10 11 12 9 5 1 10 6 2 11 7 3 12 8 4
Time Complexity: O(n*m) where n and m are matrix dimensions
Auxiliary Space: O(1)
Please refer complete article on Turn an image by 90 degree for more details!
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