Check if matrix can be converted to another matrix by transposing square sub-matrices
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 matrices 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 operationbool 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 codeint 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 approachimport 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 approachn = 3m = 3# Function that returns true if matrix1# can be converted to matrix2# with the given operationdef 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 codeif __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 approachusing 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 operationfunction 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?> |
Javascript
<script>// JavaScript implementation of the approachvar n = 3var m = 3// Function that returns true if matrix1// can be converted to matrix2// with the given operationfunction check(a, b){ // Traverse all the diagonals // starting at first column for (var i = 0; i < n; i++) { var v1 = [], v2 = []; var r = i; var col = 0; // Traverse in diagonal while (r >= 0 && col < m) { // Store the diagonal elements v1.push(a[r][col]); v2.push(b[r][col]); // Move up r--; col++; } // Sort the elements v1.sort(); v2.sort(); // Check if they are same for (var i = 0; i < v1.length; i++) { if (v1[i] != v2[i]) return false; } } // Traverse all the diagonals // starting at last row for (var j = 1; j < m; j++) { var v1 = [], v2 = []; var r = n - 1; var col = j; // Traverse in the diagonal while (r >= 0 && col < m) { // Store diagonal elements v1.push(a[r][col]); v2.push(b[r][col]); r--; col++; } // Sort all elements v1.sort(); v2.sort(); // Check for same for (var i = 0; i < v1.length; i++) { if (v1[i] != v2[i]) return false; } } // If every element matches return true;}// Driver codevar a = [ [ 1, 2, 3 ], [ 4, 5, 6 ], [ 7, 8, 9 ] ];var b = [ [ 1, 4, 7 ], [ 2, 5, 6 ], [ 3, 8, 9 ] ];if (check(a, b)) document.write( "Yes");else document.write( "No");</script> |
Output:
Yes
Time Complexity: O(max(N *M*logM, N*M * logN)) which can be written as O(N*M*max(logN,logM)), as we are using a loop to traverse N times and we are using sort function on an array of size M which will cost O(M*logM) and we are also traversing M times and sorting an array of size N which will cost O(N*logN).
Auxiliary Space: O(max(N,M)), as we are using extra space for the array v1 and v2.
Please Login to comment...