Skip to content
Related Articles

Related Articles

Check if matrix can be converted to another matrix by transposing square sub-matrices
  • Difficulty Level : Hard
  • Last Updated : 18 Apr, 2019

Given two N X M matrices of integers. In an operation, we can transpose any square sub-matrix in matrix1. The task is to check if matrix1 can be converted to matrix2 with the given operation.

Examples:

Input: matrix1[][] = {
{1, 2, 3},
{4, 5, 6},
{7, 8, 9}},
matrix2[][] = {
{1, 4, 7},
{2, 5, 6},
{3, 8, 9}}
Output: Yes
Transpose the whole matrix and then we transpose the
sub-matrix with corners in cells (2, 2) and (3, 3).

Input: matrix1[][] = {
{1, 2},
{3, 4}},
matrix2[][] = {
{1, 4},
{3, 8}}
Output: No

Approach: Sort the diagonals of both the matrices and compare them. If both matrix are equal after sorting diagonals, then matrix1 can be converted to matrix2. The following method is correct because of the fact that while transposing the numbers can only be transposed to any of its diagonal.



Below is the implementation of the above approach.

C++




// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
#define n 3
#define m 3
  
// Function that returns true if matrix1
// can be converted to matrix2
// with the given operation
bool check(int a[n][m], int b[n][m])
{
  
    // Traverse all the diagonals
    // starting at first column
    for (int i = 0; i < n; i++) {
        vector<int> v1, v2;
        int r = i;
        int col = 0;
  
        // Traverse in diagonal
        while (r >= 0 && col < m) {
  
            // Store the diagonal elements
            v1.push_back(a[r][col]);
            v2.push_back(b[r][col]);
  
            // Move up
            r--;
            col++;
        }
  
        // Sort the elements
        sort(v1.begin(), v1.end());
        sort(v2.begin(), v2.end());
  
        // Check if they are same
        for (int i = 0; i < v1.size(); i++) {
            if (v1[i] != v2[i])
                return false;
        }
    }
  
    // Traverse all the diagonals
    // starting at last row
    for (int j = 1; j < m; j++) {
        vector<int> v1, v2;
        int r = n - 1;
        int col = j;
  
        // Traverse in the diagonal
        while (r >= 0 && col < m) {
  
            // Store diagonal elements
            v1.push_back(a[r][col]);
            v2.push_back(b[r][col]);
            r--;
            col++;
        }
  
        // Sort all elements
        sort(v1.begin(), v1.end());
        sort(v2.begin(), v2.end());
  
        // Check for same
        for (int i = 0; i < v1.size(); i++) {
            if (v1[i] != v2[i])
                return false;
        }
    }
  
    // If every element matches
    return true;
}
  
// Driver code
int main()
{
    int a[n][m] = { { 1, 2, 3 }, { 4, 5, 6 }, { 7, 8, 9 } };
    int b[n][m] = { { 1, 4, 7 }, { 2, 5, 6 }, { 3, 8, 9 } };
  
    if (check(a, b))
        cout << "Yes";
    else
        cout << "No";
  
    return 0;
}

Java




// Java implementation of the approach
import java.util.*;
  
class GFG
{
  
    static final int n = 3;
    static final int m = 3;
  
    // Function that returns true if matrix1
    // can be converted to matrix2
    // with the given operation
    static boolean check(int a[][], int b[][])
    {
  
        // Traverse all the diagonals
        // starting at first column
        for (int i = 0; i < n; i++) 
        {
            Vector<Integer> v1 = new Vector<Integer>();
            Vector<Integer> v2 = new Vector<Integer>();
            int r = i;
            int col = 0;
  
            // Traverse in diagonal
            while (r >= 0 && col < m)
            {
  
                // Store the diagonal elements
                v1.add(a[r][col]);
                v2.add(b[r][col]);
  
                // Move up
                r--;
                col++;
            }
  
            // Sort the elements
            Collections.sort(v1);
            Collections.sort(v2);
  
            // Check if they are same
            for (int j = 0; j < v1.size(); j++)
            {
                if (v1.get(j) != v2.get(j))
                {
                    return false;
                }
            }
        }
  
        // Traverse all the diagonals
        // starting at last row
        for (int j = 1; j < m; j++) 
        {
            Vector<Integer> v1 = new Vector<Integer>();
            Vector<Integer> v2 = new Vector<Integer>();
            int r = n - 1;
            int col = j;
  
            // Traverse in the diagonal
            while (r >= 0 && col < m) 
            {
  
                // Store diagonal elements
                v1.add(a[r][col]);
                v2.add(b[r][col]);
                r--;
                col++;
            }
  
            // Sort all elements
            Collections.sort(v1);
            Collections.sort(v2);
  
            // Check for same
            for (int i = 0; i < v1.size(); i++) 
            {
                if (v1.get(i) != v2.get(i)) 
                {
                    return false;
                }
            }
        }
  
        // If every element matches
        return true;
    }
  
    // Driver code
    public static void main(String[] args) 
    {
        int a[][] = {{1, 2, 3}, {4, 5, 6}, {7, 8, 9}};
        int b[][] = {{1, 4, 7}, {2, 5, 6}, {3, 8, 9}};
  
        if (check(a, b)) 
        {
            System.out.println("Yes");
        
        else 
        {
            System.out.println("No");
        }
    }
}
  
// This code contributed by Rajput-Ji

Python3




# Python 3 implementation of the approach
n = 3
m = 3
  
# Function that returns true if matrix1
# can be converted to matrix2
# with the given operation
def check(a, b):
      
    # Traverse all the diagonals
    # starting at first column
    for i in range(n):
        v1 = []
        v2 = []
        r = i
        col = 0
  
        # Traverse in diagonal
        while (r >= 0 and col < m):
              
            # Store the diagonal elements
            v1.append(a[r][col])
            v2.append(b[r][col])
  
            # Move up
            r -= 1
            col += 1
  
        # Sort the elements
        v1.sort(reverse = False)
        v2.sort(reverse = False)
  
        # Check if they are same
        for i in range(len(v1)):
            if (v1[i] != v2[i]):
                return False
  
    # Traverse all the diagonals
    # starting at last row
    for j in range(1, m):
        v1 = []
        v2 = []
        r = n - 1
        col = j
  
        # Traverse in the diagonal
        while (r >= 0 and col < m):
              
            # Store diagonal elements
            v1.append(a[r][col])
            v2.append(b[r][col])
            r -= 1
            col += 1
  
        # Sort all elements
        v1.sort(reverse = False)
        v2.sort(reverse = False)
  
        # Check for same
        for i in range(len(v1)):
            if (v1[i] != v2[i]):
                return False
                  
    # If every element matches
    return True
  
# Driver code
if __name__ == '__main__':
    a = [[1, 2, 3], [4, 5, 6], [7, 8, 9]]
    b = [[1, 4, 7], [2, 5, 6], [3, 8, 9]]
  
    if (check(a, b)):
        print("Yes")
    else:
        print("No")
  
# This code is contributed by
# Surendra_Gangwar

C#




// C# implementation of the approach
using System;
using System.Collections.Generic; 
  
class GFG
{
  
    static readonly int n = 3;
    static readonly int m = 3;
  
    // Function that returns true if matrix1
    // can be converted to matrix2
    // with the given operation
    static bool check(int [,]a, int [,]b)
    {
  
        // Traverse all the diagonals
        // starting at first column
        for (int i = 0; i < n; i++) 
        {
            List<int> v1 = new List<int>();
            List<int> v2 = new List<int>();
            int r = i;
            int col = 0;
  
            // Traverse in diagonal
            while (r >= 0 && col < m)
            {
  
                // Store the diagonal elements
                v1.Add(a[r, col]);
                v2.Add(b[r, col]);
  
                // Move up
                r--;
                col++;
            }
  
            // Sort the elements
            v1.Sort();
            v2.Sort();
  
            // Check if they are same
            for (int j = 0; j < v1.Count; j++)
            {
                if (v1[j] != v2[j])
                {
                    return false;
                }
            }
        }
  
        // Traverse all the diagonals
        // starting at last row
        for (int j = 1; j < m; j++) 
        {
            List<int> v1 = new List<int>();
            List<int> v2 = new List<int>();
            int r = n - 1;
            int col = j;
  
            // Traverse in the diagonal
            while (r >= 0 && col < m) 
            {
  
                // Store diagonal elements
                v1.Add(a[r, col]);
                v2.Add(b[r, col]);
                r--;
                col++;
            }
  
            // Sort all elements
            v1.Sort();
            v2.Sort();
  
            // Check for same
            for (int i = 0; i < v1.Count; i++) 
            {
                if (v1[i] != v2[i]) 
                {
                    return false;
                }
            }
        }
  
        // If every element matches
        return true;
    }
  
    // Driver code
    public static void Main() 
    {
        int [,]a = {{1, 2, 3}, {4, 5, 6}, {7, 8, 9}};
        int [,]b = {{1, 4, 7}, {2, 5, 6}, {3, 8, 9}};
  
        if (check(a, b)) 
        {
            Console.WriteLine("Yes");
        
        else
        {
            Console.WriteLine("No");
        }
    }
}
  
/* This code contributed by PrinciRaj1992 */

PHP




<?php
// PHP implementation of the approach
$n = 3;
$m = 3;
  
// Function that returns true if matrix1
// can be converted to matrix2
// with the given operation
function check($a, $b)
{
    global $n, $m;
  
    // Traverse all the diagonals
    // starting at first column
    for ($i = 0; $i < $n; $i++) 
    {
        $v1 = array();
        $v2 = array();
        $r = $i;
        $col = 0;
  
        // Traverse in diagonal
        while ($r >= 0 && $col < $m)
        {
  
            // Store the diagonal elements
            array_push($v1, $a[$r][$col]);
            array_push($v2, $b[$r][$col]);
  
            // Move up
            $r--;
            $col++;
        }
  
        // Sort the elements
        sort($v1);
        sort($v2);
  
        // Check if they are same
        for ($i = 0; $i < count($v1); $i++) 
        {
            if ($v1[$i] != $v2[$i])
                return false;
        }
    }
  
    // Traverse all the diagonals
    // starting at last row
    for ($j = 1; $j < $m; $j++)
    {
        $v1 = array();
        $v2 = array();
        $r = $n - 1;
        $col = $j;
  
        // Traverse in the diagonal
        while ($r >= 0 && $col < $m)
        {
  
            // Store diagonal elements
            array_push($v1, $a[$r][$col]);
            array_push($v2, $b[$r][$col]);
            $r--;
            $col++;
        }
  
        // Sort all elements
        sort($v1);
        sort($v2);
  
        // Check for same
        for ($i = 0; $i < count($v1); $i++) 
        {
            if ($v1[$i] != $v2[$i])
                return false;
        }
    }
  
    // If every element matches
    return true;
}
  
// Driver code
$a = array(array( 1, 2, 3 ), 
           array( 4, 5, 6 ), 
           array( 7, 8, 9 ));
$b = array(array( 1, 4, 7 ), 
           array( 2, 5, 6 ), 
           array( 3, 8, 9 ));
  
if (check($a, $b))
    echo "Yes";
else
    echo "No";
  
// This code is contributed by mits
?>
Output:
Yes

Time Complexity: O(N * M * log (max(N, M)))

Attention reader! Don’t stop learning now. Get hold of all the important mathematical concepts for competitive programming with the Essential Maths for CP Course at a student-friendly price. To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

My Personal Notes arrow_drop_up
Recommended Articles
Page :