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# Rotate each ring of matrix anticlockwise by K elements

• Difficulty Level : Hard
• Last Updated : 06 Jul, 2022

Given a matrix of order M*N and a value K, the task is to rotate each ring of the matrix anticlockwise by K elements. If in any ring elements are less than and equal K then don’t rotate it.

Examples:

```Input : k = 3
mat = {{1, 2, 3, 4},
{5, 6, 7, 8},
{9, 10, 11, 12},
{13, 14, 15, 16}}
Output: 4 8  12 16
3 10  6 15
2 11  7 14
1  5  9 13

Input : k = 2
mat = {{1, 2, 3, 4},
{10, 11, 12, 5},
{9, 8, 7, 6}}
Output: 3 4  5  6
2 11 12 7
1 10  9 8```

The idea is to traverse matrix in spiral form. Here is the algorithm to solve this problem :

• Make an auxiliary array temp[] of size M*N.
• Start traversing matrix in spiral form and store elements of current ring in temp[] array. While storing the elements in temp, keep track of starting and ending positions of current ring.
• For every ring that is being stored in temp[], rotate that subarray temp[]
• Repeat this process for each ring of matrix.
• In last traverse matrix again spirally and copy elements of temp[] array to matrix.

Below is C++ implementation of above steps.

## CPP

 `// C++ program to rotate individual rings by k in` `// spiral order traversal.` `#include` `#define MAX 100` `using` `namespace` `std;`   `// Fills temp array into mat[][] using spiral order` `// traversal.` `void` `fillSpiral(``int` `mat[][MAX], ``int` `m, ``int` `n, ``int` `temp[])` `{` `    ``int` `i, k = 0, l = 0;`   `    ``/*  k - starting row index` `        ``m - ending row index` `        ``l - starting column index` `        ``n - ending column index  */` `    ``int` `tIdx  = 0;  ``// Index in temp array` `    ``while` `(k < m && l < n)` `    ``{` `        ``/* first row from the remaining rows */` `        ``for` `(``int` `i = l; i < n; ++i)` `            ``mat[k][i] = temp[tIdx++];` `        ``k++;`   `        ``/* last column from the remaining columns */` `        ``for` `(``int` `i = k; i < m; ++i)` `            ``mat[i][n-1] = temp[tIdx++];` `        ``n--;`   `        ``/* last row from the remaining rows */` `        ``if` `(k < m)` `        ``{` `            ``for` `(``int` `i = n-1; i >= l; --i)` `                ``mat[m-1][i] = temp[tIdx++];` `            ``m--;` `        ``}`   `        ``/* first column from the remaining columns */` `        ``if` `(l < n)` `        ``{` `            ``for` `(``int` `i = m-1; i >= k; --i)` `                ``mat[i][l] = temp[tIdx++];` `            ``l++;` `        ``}` `    ``}` `}`   `// Function to spirally traverse matrix and` `// rotate each ring of matrix by K elements` `// mat[][] --> matrix of elements` `// M     --> number of rows` `// N    --> number of columns` `void` `spiralRotate(``int` `mat[][MAX], ``int` `M, ``int` `N, ``int` `k)` `{` `    ``// Create a temporary array to store the result` `    ``int` `temp[M*N];`   `    ``/*      s - starting row index` `            ``m - ending row index` `            ``l - starting column index` `            ``n - ending column index;  */` `    ``int` `m = M, n = N, s = 0, l = 0;`   `    ``int` `*start = temp;  ``// Start position of current ring` `    ``int` `tIdx = 0;  ``// Index in temp` `    ``while` `(s < m && l < n)` `    ``{` `        ``// Initialize end position of current ring` `        ``int` `*end = start;`   `        ``// copy the first row from the remaining rows` `        ``for` `(``int` `i = l; i < n; ++i)` `        ``{` `            ``temp[tIdx++] = mat[s][i];` `            ``end++;` `        ``}` `        ``s++;`   `        ``// copy the last column from the remaining columns` `        ``for` `(``int` `i = s; i < m; ++i)` `        ``{` `            ``temp[tIdx++] = mat[i][n-1];` `            ``end++;` `        ``}` `        ``n--;`   `        ``// copy the last row from the remaining rows` `        ``if` `(s < m)` `        ``{` `            ``for` `(``int` `i = n-1; i >= l; --i)` `            ``{` `                ``temp[tIdx++] = mat[m-1][i];` `                ``end++;` `            ``}` `            ``m--;` `        ``}`   `        ``/* copy the first column from the remaining columns */` `        ``if` `(l < n)` `        ``{` `            ``for` `(``int` `i = m-1; i >= s; --i)` `            ``{` `                ``temp[tIdx++] = mat[i][l];` `                ``end++;` `            ``}` `            ``l++;` `        ``}`   `        ``// if elements in current ring greater than` `        ``// k then rotate elements of current ring` `        ``if` `(end-start > k)` `        ``{` `            ``// Rotate current ring using reversal` `            ``// algorithm for rotation` `            ``reverse(start, start+k);` `            ``reverse(start+k, end);` `            ``reverse(start, end);`   `            ``// Reset start for next ring` `            ``start = end;` `        ``}` `    ``}`   `    ``// Fill temp array in original matrix.` `    ``fillSpiral(mat, M, N, temp);` `}`   `// Driver program to run the case` `int` `main()` `{` `    ``// Your C++ Code` `    ``int` `M = 4, N = 4, k = 3;` `    ``int` `mat[][MAX]= {{1, 2, 3, 4},` `                     ``{5, 6, 7, 8},` `                     ``{9, 10, 11, 12},` `                     ``{13, 14, 15, 16} };`   `    ``spiralRotate(mat, M, N, k);`   `    ``// print modified matrix` `    ``for` `(``int` `i=0; i

Output

```4 8 12 16
3 10 6 15
2 11 7 14
1 5 9 13
```

Time Complexity : O(M*N)  as we are using nested loops to traverse the matrix.
Auxiliary space : O(M*N)  as we are using extra space for matrix.

This article is contributed by Shashank Mishra ( Gullu ). If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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