Rotate each ring of matrix anticlockwise by K elements

4.2

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

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<bits/stdc++.h>
#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<M; i++)
    {
        for (int j=0; j<N; j++)
            cout << mat[i][j] << " ";
        cout << endl;
    }
    return 0;
}

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.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

GATE CS Corner    Company Wise Coding Practice

Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.

Recommended Posts:



4.2 Average Difficulty : 4.2/5.0
Based on 23 vote(s)










Writing code in comment? Please use ide.geeksforgeeks.org, generate link and share the link here.