Minimum number of rotations required to convert given Matrix to another

Last Updated : 02 Jun, 2022

Given two 2D arrays arr[][] and brr[][] of equal size. The task is to find the minimum number of rotations required either clockwise or anticlockwise to convert arr[][] to brr[][], in the following manner:

+x means x rotation in clockwise direction
-x means x rotation in anti-clockwise direction
0 means no rotation needed, and
NA means rotation is not possible

Examples

Input: arr[][] = {{2, 3}, {4, 5}}, brr[][] = {{4, 2}, {5, 3}}
Output: +1
Explanation: arr[][] can be converted to brr[][] by one 90 degree rotation clockwise.

Input: arr[][] = {{1, 2}, {3, 4}}, brr[][] = {{1, 3}, {2, 4}}
Output: NA

Approach: The given problem is implementation-based and similar to Rotate matrix with 90 degrees. Follow the steps below to solve the given problem.

Create a function say rotate(), that rotates a matrix by 90 degrees in the clockwise direction.
Create a function say isEqual(), to check if two matrices are equal or not.
At first, check if both the arrays are already equal or not. if equal print 0 and return, as 0 operations are needed in that case.
Rotate arr[][] by 90 degrees 4 times and each time check whether it became equal to brr or not.
If at any point arr[][] becomes equal to brr[][] then check if taking clockwise or anti-clockwise is giving the minimum number or rotation to reach brr[][].
Print the result found above.
If after 4 clockwise rotations arr[][] never became equal to brr[][] then print NA at the end.

Below is the implementation of the approach:

C++

// C++ program for above approach
#include <bits/stdc++.h>
#include <vector>
using namespace std;

// Function to check if two matrices
// are equal or not
bool isEqual(vector<vector<int> >& arr,
             vector<vector<int> >& brr)
{
    int n = arr.size();
    int m = arr[0].size();

    for (int i = 0; i < n; i++) {
        for (int j = 0; j < m; j++) {
            if (arr[i][j] != brr[i][j])
                return false;
        }
    }

    return true;
}

// Function to rotate matrix by 90 degrees clockwise
void rotate(vector<vector<int> >& m)
{
    int n = m.size();

    for (int i = 0; i < n; i++)
        for (int j = 0; j < i; j++)
            swap(m[i][j], m[j][i]);

    for (int i = 0; i < n; i++)
        reverse(m[i].begin(), m[i].end());
}

// Find Minimum rotation to reach the desired matrix
void findRotation(vector<vector<int> > arr,
                  vector<vector<int> > brr)
{

    if (isEqual(arr, brr)) {
        cout << 0;
        return;
    }

    for (int i = 1; i < 4; i++) {

        // Rotate by 90 degrees clockwise
        rotate(arr);

        // Checking if both arr[][] and brr[]
        // are now equal or not
        if (isEqual(arr, brr)) {
            if (i < 4 - i) {
                cout << "+" << i;
            }
            else
                cout << "-" << 4 - i;
            return;
        }
    }

    // If converting brr[][] is not possible
    cout << "NA";
}

// Driver Code
int main()
{

    vector<vector<int> > arr, brr;

    arr = { { 2, 3 }, { 4, 5 } };
    brr = { { 4, 2 }, { 5, 3 } };

    // Function Call
    findRotation(arr, brr);

    return 0;
}

Java

// Java program for the above approach
import java.io.*;
import java.lang.*;
import java.util.*;

class GFG {

    // Function to check if two matrices
    // are equal or not
   static Boolean isEqual(int[ ][ ] arr, int[ ][ ] brr)
    {
        int n = arr.length;
        int m = arr[0].length;

        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                if (arr[i][j] != brr[i][j])
                    return false;
            }
        }

        return true;
    }

    // Function to rotate matrix by 90 degrees clockwise
    static void rotate(int [ ][ ] m)
    {
        int n = m.length;

        for (int i = 0; i < n; i++)
            for (int j = 0; j < i; j++){
                      //swap
                      int temp = m[i][j];
                      m[i][j] = m[j][i];
                      m[j][i] = temp;
            }

        for (int i = 0; i < n; i++)
           Collections.reverse(Arrays.asList(m[i]));
    }

    // Find Minimum rotation to reach the desired matrix
    static void findRotation(int[ ][ ] arr,  int[ ][ ] brr)
    {

        if (isEqual(arr, brr) == true) {
            System.out.print("0");
              return;
        }

        for (int i = 1; i < 4; i++) {

            // Rotate by 90 degrees clockwise
            rotate(arr);

            // Checking if both arr[][] and brr[]
            // are now equal or not
            if (!isEqual(arr, brr)) {
                if (i < 4 - i) {
                  System.out.print("+" + i );
                }
                else
                      System.out.print("-" + (4 - i));
                return;
            }
        }

        // If converting brr[][] is not possible
        System.out.print("NA");
    }

    // Driver Code
    public static void main (String[] args) {

        int [ ][ ] arr = { { 2, 3 }, { 4, 5 } };
        int [ ][ ] brr = { { 4, 2 }, { 5, 3 } };
      
        // Function Call
        findRotation(arr, brr);
    }
}

// This code is contributed by hrithikgarg03188

Python3

# python3 program for above approach

# Function to check if two matrices
# are equal or not
def isEqual(arr, brr):
    n = len(arr)
    m = len(arr[0])

    for i in range(0, n):
        for j in range(0, m):
            if (arr[i][j] != brr[i][j]):
                return False

    return True

# Function to rotate matrix by 90 degrees clockwise
def rotate(m):
    n = len(m)

    for i in range(0, n):
        for j in range(0, i):
            temp = m[i][j]
            m[i][j] = m[j][i]
            m[j][i] = temp

    for i in range(0, n):
        m[i].reverse()

# Find Minimum rotation to reach the desired matrix
def findRotation(arr, brr):
    if (isEqual(arr, brr)):
        print(0)
        return

    for i in range(1, 4):

        # Rotate by 90 degrees clockwise
        rotate(arr)

        # Checking if both arr[][] and brr[]
        # are now equal or not
        if (isEqual(arr, brr)):
            if (i < 4 - i):
                print(f"+{i}")

            else:
                print(f"-{4-i}")
            return

    # If converting brr[][] is not possible
    print("NA")

# Driver Code
if __name__ == "__main__":

    arr = [[2, 3], [4, 5]]
    brr = [[4, 2], [5, 3]]

    # Function Call
    findRotation(arr, brr)

    # This code is contributed by rakeshsahni

using System;
using System.Linq;

class GFG {

  static void reverse(int [,]arr, int N, int M)
  {

    // Traverse each row of [,]arr
    for (int i = 0; i < M; i++) {

      // Initialise start and end index
      int start = 0;
      int end = N - 1;

      // Till start < end, swap the element
      // at start and end index
      while (start < end) {

        // Swap the element
        int temp = arr[i,start];
        arr[i, start] = arr[i, end];
        arr[i, end] = temp;

        // Increment start and decrement
        // end for next pair of swapping
        start++;
        end--;
      }
    }
  }

  // Function to check if two matrices
  // are equal or not
  static bool isEqual(int [,] arr, int [,] brr, int n, int m)
  {
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < m; j++) {
        if (arr[i,j] != brr[i,j])
          return false;
      }
    }

    return true;
  }

  // Function to rotate matrix by 90 degrees clockwise
  static void rotate(int [,] arr, int n, int m)
  {
    for (int i = 0; i < n; i++)
      for (int j = 0; j < i; j++)
      {
        int curr = arr[i,j];
        arr[i,j] = arr[j,i];
        arr[j,i] = curr;
      }

    reverse(arr,n,m);
  }

  // Find Minimum rotation to reach the desired matrix
  static void findRotation(int [,] arr, int [,] brr, int n, int m)
  {

    if (isEqual(arr, brr, n, m)) {
      Console.Write(0);
      return;
    }

    for (int i = 1; i < 4; i++) {

      // Rotate by 90 degrees clockwise
      rotate(arr,n,m);

      // Checking if both arr[][] and brr[]
      // are now equal or not
      if (isEqual(arr, brr,n,m)) {
        if (i < 4 - i) {
          Console.Write( "+" + i);
        }
        else
          Console.Write("-" + (4 - i));
        return;
      }
    }

    // If converting brr[][] is not possible
    Console.Write("NA");
  }


  /* Driver program to test above
    functions */
  public static void Main()
  {
    int [,] arr = new int[,] { { 2, 3 }, { 4, 5 } };
    int [,] brr = new int[,] { { 4, 2 }, { 5, 3 } };
    int N = 2;
    int M = 2;
    // Function Call
    findRotation(arr, brr, N, M);
  }
}

// This code is contributed by Aarti_Rathi

JavaScript

    <script>
        // JavaScript code for the above approach 

        // Function to check if two matrices
        // are equal or not
        function isEqual(arr, brr)
        {
            let n = arr.length;
            let m = arr[0].length;

            for (let i = 0; i < n; i++) {
                for (let j = 0; j < m; j++) {
                    if (arr[i][j] != brr[i][j])
                        return false;
                }
            }

            return true;
        }

        // Function to rotate matrix by 90 degrees clockwise
        function rotate(m) {
            let n = m.length;

            for (let i = 0; i < n; i++) {
                for (let j = 0; j < i; j++) {
                    let temp = m[i][j]
                    m[i][j] = m[j][i]
                    m[j][i] = temp
                }
            }
            for (let i = 0; i < n; i++)
                m[i].reverse()
        }

        // Find Minimum rotation to reach the desired matrix
        function findRotation(arr,
            brr) {

            if (isEqual(arr, brr)) {
                document.write(0)
                return;
            }

            for (let i = 1; i < 4; i++) {

                // Rotate by 90 degrees clockwise
                rotate(arr);

                // Checking if both arr[][] and brr[]
                // are now equal or not
                if (isEqual(arr, brr)) {
                    if (i < 4 - i) {
                        document.write("+" + i);
                    }
                    else
                        document.write("-" + (4 - i));
                    return;
                }
            }

            // If converting brr[][] is not possible
            document.write("NA");
        }

        // Driver Code
        let arr, brr;

        arr = [[2, 3], [4, 5]];
        brr = [[4, 2], [5, 3]];

        // Function Call
        findRotation(arr, brr);

  // This code is contributed by Potta Lokesh
    </script>

Output

+1

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

Minimum number of rotations required to convert given Matrix to another

manasvviaggarwal

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Minimum number of rotations required to convert given Matrix to another

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