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Shuffle the given Matrix K times by reversing row and columns alternatively in sequence

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  • Last Updated : 03 Feb, 2022

Given a matrix arr[][] of size M * N, where M is the number of rows and N is the number of columns, and an integer K. The task is to shuffle this matrix in K steps such that in each step, the rows and columns of the matrix are reversed alternatively i.e start with reversing the 1st row first step, then the 1st column in the second step, then the 2nd row in the third step and so on. 

Note: If K exceeds the M or N then reversal starts from the first row or column again.

Examples:

Input: arr[][] = {{3,   4,   1,   8},
                        {11, 23, 43, 21},
                        {12, 17, 65, 91},
                        {71, 56, 34, 24}}
K = 4
Output: {{71, 56, 4, 3},
              {21, 17, 23, 12}, 
              {11, 43, 65, 91}, 
              {8, 1, 34, 24}}
Explanation: Steps to be followed:
Reverse the first row
Reverse the first column
Reverse the second row
Reverse the second column

Input: arr[][] = {{11, 23, 43, 21}, 
                         {12, 17, 65, 91}, 
                         {71, 56, 34, 24}}
K = 8
Output: {{11, 43, 56, 21}, 
             {91, 65, 17, 12}, 
             {24, 34, 23, 71}} 

 

Approach: The problem can be solved by simply running two loops for traversing rows and columns alternatively, till K becomes 0. In the end, print the resultant matrix. Follow the steps mentioned below:

  • Run a loop for traversing the matrix.
  • In each iteration reverse the proper row and column.
  • Do this operation K times.
  • Print the final loop.

Below is the implementation of the above approach

C++




// C++ code to implement the above approach
#include <bits/stdc++.h>
using namespace std;
const int M = 3;
const int N = 4;
 
// Print matrix elements
void showArray(int arr[][N])
{
    for (int i = 0; i < M; i++) {
        for (int j = 0; j < N; j++)
            cout << arr[i][j] << " ";
        cout << endl;
    }
}
 
// Function to shuffle matrix
void reverseAlternate(int arr[][N], int K)
{
    int turn = 0;
    int row = 0, col = 0;
    while (turn < K) {
         
        // Reverse the row
        if (turn % 2 == 0) {
            int start = 0, end = N - 1, temp;
            while (start < end) {
                temp = arr[row % M][start];
                arr[row % M][start] =
                    arr[row % M][end];
                arr[row % M][end] = temp;
                start += 1;
                end -= 1;
            }
            row++;
            turn++;
        }
         
        // Reverse the column
        else {
            int start = 0, end = M - 1, temp;
            while (start < end) {
                temp = arr[start][col % N];
                arr[start][col % N] =
                    arr[end][col % N];
                arr[end][col % N] = temp;
                start += 1;
                end -= 1;
            }
            col++;
            turn++;
        }
    }
}
 
// Driver code
int main()
{
 
    int matrix[M][N] = { { 11, 23, 43, 21 },
                         { 12, 17, 65, 91 },
                         { 71, 56, 34, 24 } };
    int K = 8;
    reverseAlternate(matrix, K);
    showArray(matrix);
    return 0;
}

Java




// Java code to implement the above approach
import java.io.*;
 
class GFG {
 
  public static int M = 3;
  public static int N = 4;
 
  // Print matrix elements
  public static void showArray(int arr[][])
  {
    for (int i = 0; i < M; i++) {
      for (int j = 0; j < N; j++)
        System.out.print(arr[i][j] + " ");
      System.out.println();
    }
  }
 
  // Function to shuffle matrix
  public static void reverseAlternate(int arr[][], int K)
  {
    int turn = 0;
    int row = 0, col = 0;
    while (turn < K) {
 
      // Reverse the row
      if (turn % 2 == 0) {
        int start = 0, end = N - 1, temp;
        while (start < end) {
          temp = arr[row % M][start];
          arr[row % M][start] =
            arr[row % M][end];
          arr[row % M][end] = temp;
          start += 1;
          end -= 1;
        }
        row++;
        turn++;
      }
 
      // Reverse the column
      else {
        int start = 0, end = M - 1, temp;
        while (start < end) {
          temp = arr[start][col % N];
          arr[start][col % N] =
            arr[end][col % N];
          arr[end][col % N] = temp;
          start += 1;
          end -= 1;
        }
        col++;
        turn++;
      }
    }
  }
 
  // Driver code
  public static void main (String[] args)
  {
 
    int matrix[][] = { { 11, 23, 43, 21 },
                      { 12, 17, 65, 91 },
                      { 71, 56, 34, 24 } };
    int K = 8;
    reverseAlternate(matrix, K);
    showArray(matrix);
  }
}
 
// This code is contributed by Shubham Singh

Python3




# Python code to implement the above approach
M = 3
N = 4
 
# Print matrix elements
def showArray(arr):
     
    for i in range(M):
        for j in range(N):
            print(arr[i][j], end= " ")
        print()
 
# Function to shuffle matrix
def reverseAlternate(arr, K):
     
    turn = 0
    row = 0
    col = 0
    while (turn < K):
         
        # Reverse the row
        if (turn % 2 == 0):
            start = 0
            end = N - 1
            temp = 0
            while (start < end):
                temp = arr[row % M][start]
                arr[row % M][start] = arr[row % M][end]
                arr[row % M][end] = temp
                start += 1
                end -= 1
             
            row += 1
            turn += 1
         
        #Reverse the column
        else:
            start = 0
            end = M - 1
            temp = 0
            while (start < end):
                temp = arr[start][col % N]
                arr[start][col % N] = arr[end][col % N]
                arr[end][col % N] = temp
                start += 1
                end -= 1
                 
            col += 1
            turn += 1
             
# Driver code
matrix = [[11, 23, 43, 21] ,
            [12, 17, 65, 91] ,
            [71, 56, 34, 24]]
K = 8
reverseAlternate(matrix, K)
showArray(matrix)
         
# This code is contributed by Shubham Singh

C#




// C# code to implement the above approach
using System;
 
class GFG {
 
  public static int M = 3;
  public static int N = 4;
 
  // Print matrix elements
  public static void showArray(int[, ] arr)
  {
    for (int i = 0; i < M; i++) {
      for (int j = 0; j < N; j++)
        Console.Write(arr[i, j] + " ");
      Console.WriteLine();
    }
  }
 
  // Function to shuffle matrix
  public static void reverseAlternate(int[, ] arr, int K)
  {
    int turn = 0;
    int row = 0, col = 0;
    while (turn < K) {
 
      // Reverse the row
      if (turn % 2 == 0) {
        int start = 0, end = N - 1, temp;
        while (start < end) {
          temp = arr[row % M, start];
          arr[row % M, start] = arr[row % M, end];
          arr[row % M, end] = temp;
          start += 1;
          end -= 1;
        }
        row++;
        turn++;
      }
 
      // Reverse the column
      else {
        int start = 0, end = M - 1, temp;
        while (start < end) {
          temp = arr[start, col % N];
          arr[start, col % N] = arr[end, col % N];
          arr[end, col % N] = temp;
          start += 1;
          end -= 1;
        }
        col++;
        turn++;
      }
    }
  }
 
  // Driver code
  public static void Main(string[] args)
  {
 
    int[, ] matrix = { { 11, 23, 43, 21 },
                      { 12, 17, 65, 91 },
                      { 71, 56, 34, 24 } };
    int K = 8;
    reverseAlternate(matrix, K);
    showArray(matrix);
  }
}
 
// This code is contributed by ukasp.

Javascript




<script>
       // JavaScript code for the above approach
       let M = 3;
       let N = 4;
 
       // Print matrix elements
       function showArray(arr) {
           for (let i = 0; i < M; i++) {
               for (let j = 0; j < N; j++)
                   document.write(arr[i][j] + " ");
               document.write('<br>');
           }
       }
 
       // Function to shuffle matrix
       function reverseAlternate(arr, K) {
           let turn = 0;
           let row = 0, col = 0;
           while (turn < K) {
 
               // Reverse the row
               if (turn % 2 == 0) {
                   let start = 0, end = N - 1, temp;
                   while (start < end) {
                       temp = arr[row % M][start];
                       arr[row % M][start] =
                           arr[row % M][end];
                       arr[row % M][end] = temp;
                       start += 1;
                       end -= 1;
                   }
                   row++;
                   turn++;
               }
 
               // Reverse the column
               else {
                   let start = 0, end = M - 1, temp;
                   while (start < end) {
                       temp = arr[start][col % N];
                       arr[start][col % N] =
                           arr[end][col % N];
                       arr[end][col % N] = temp;
                       start += 1;
                       end -= 1;
                   }
                   col++;
                   turn++;
               }
           }
       }
 
       // Driver code
       let matrix = [[11, 23, 43, 21],
       [12, 17, 65, 91],
       [71, 56, 34, 24]];
       let K = 8;
       reverseAlternate(matrix, K);
       showArray(matrix);
 
 // This code is contributed by Potta Lokesh
   </script>

 
 

Output
11 43 56 21 
91 65 17 12 
24 34 23 71 

Time Complexity: O(K * max(M, N))
Auxiliary Space: O(1) 


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