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# Rotate each ring of matrix anticlockwise by K elements

• Difficulty Level : Hard
• Last Updated : 20 Oct, 2021

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 revarsal``            ``// algorithm for rotation``            ``reverse(start, start+k);``            ``reverse(start+k, end);``            ``reverse(start, end);` `            ``// Reset start for next ring``            ``start = end;``        ``}``        ``else` `// There are less than k elements in ring``            ``break``;``    ``}` `    ``// Fill tenp 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)
Auxiliary space : O(M*N)
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.