# Rotate each ring of matrix anticlockwise by K elements

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[4][4] = {{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[3][4] = {{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
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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.

 `// 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 ` `// traveral. ` `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 contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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