Given two n * m matrices A[][] and B[][], the task is to make both the matrices strictly increasing (both rows and columns) only by swapping two elements in different matrices if they are located in the corresponding position i.e. A[i][j] can only be swapped with B[i][j]. If possible then print Yes otherwise, No.
Examples:
Input:
A[][] = {{2, 10}, B[][] = {{9, 4},
{11, 5}} {3, 12}}
Output: Yes
Swap 2 with 9 and 5 with 12 then the resulting
matrices will be strictly increasing.
Input:
A[][] = {{1, 3}, B[][] = {{3, 1},
{2, 4}, {3, 6},
{5, 10}} {4, 8}}
Output: No
Approach: We can solve this problem using greedy technique. Swap A[i][j] with B[i][j] if A[i][j] > B[i][j]. At the end for every i and j we have A[i][j] ? B[i][j].
If the resultant matrices are strictly increasing then print Yes otherwise, print No.
Below is the implementation of the above approach:
C++
#include<bits/stdc++.h>using namespace std;// Function to check whether the matrices // can be made strictly increasing // with the given operationstring Check(int a[][2], int b[][2], int n, int m){ // Swap only when a[i][j] > b[i][j] for (int i = 0; i < n; i++) for (int j = 0; j < m; j++) if (a[i][j] > b[i][j]) swap(a[i][j], b[i][j]); // Check if rows are strictly increasing for (int i = 0; i < n; i++) for (int j = 0; j < m - 1; j++) if(a[i][j] >= a[i][j + 1] || b[i][j] >= b[i][j + 1]) return "No"; // Check if columns are strictly increasing for (int i = 0; i < n - 1; i++) for (int j = 0; j < m ;j++) if (a[i][j] >= a[i + 1][j] || b[i][j] >= b[i + 1][j]) return "No"; return "Yes";}// Driver codeint main(){ int n = 2, m = 2; int a[][2] = {{2, 10}, {11, 5}}; int b[][2] = {{9, 4}, {3, 12}}; cout << (Check(a, b,n,m));}// This code is contributed by chitranayal |
Java
class GFG{ // Function to check whether the matrices // can be made strictly increasing // with the given operationpublic static String Check(int a[][], int b[][], int n, int m){ // Swap only when a[i][j] > b[i][j] for(int i = 0; i < n; i++) for(int j = 0; j < m; j++) if (a[i][j] > b[i][j]) { int temp = a[i][j]; a[i][j] = b[i][j]; b[i][j] = temp; } // Check if rows are strictly increasing for(int i = 0; i < n; i++) for(int j = 0; j < m - 1; j++) if (a[i][j] >= a[i][j + 1] || b[i][j] >= b[i][j + 1]) return "No"; // Check if columns are strictly increasing for(int i = 0; i < n - 1; i++) for(int j = 0; j < m ;j++) if (a[i][j] >= a[i + 1][j] || b[i][j] >= b[i + 1][j]) return "No"; return "Yes";} // Driver codepublic static void main(String[] args){ int n = 2, m = 2; int a[][] = { { 2, 10 }, { 11, 5 } }; int b[][] = { { 9, 4 }, { 3, 12 } }; System.out.print(Check(a, b, n, m));}}// This code is contributed by divyeshrabadiya07 |
Python3
# Function to check whether the matrices # can be made strictly increasing # with the given operationdef Check(a, b): # Swap only when a[i][j] > b[i][j] for i in range(n): for j in range(m): if a[i][j]>b[i][j]: a[i][j], b[i][j]= b[i][j], a[i][j] # Check if rows are strictly increasing for i in range(n): for j in range(m-1): if(a[i][j]>= a[i][j + 1] or b[i][j]>= b[i][j + 1]): return "No" # Check if columns are strictly increasing for i in range(n-1): for j in range(m): if (a[i][j]>= a[i + 1][j] or b[i][j]>= b[i + 1][j]): return "No" return "Yes"# Driver codeif __name__=="__main__": n, m = 2, 2 a =[[2, 10], [11, 5]] b =[[9, 4], [3, 12]] print(Check(a, b)) |
C#
// C# implementation of the // above approach using System;using System.Collections;class GfG{ // Function to check // whether the matrices // can be made strictly // increasing with the // given operationstatic string Check(int [,]a, int [,]b, int n, int m){ // Swap only when // a[i][j] > b[i][j] for (int i = 0; i < n; i++) for (int j = 0; j < m; j++) if (a[i, j] > b[i, j]) { int tmp = a[i, j]; a[i, j] = b[i, j]; b[i, j] = tmp; } // Check if rows are // strictly increasing for (int i = 0; i < n; i++) for (int j = 0; j < m - 1; j++) if(a[i, j] >= a[i, j + 1] || b[i, j] >= b[i, j + 1]) return "No"; // Check if columns are // strictly increasing for (int i = 0; i < n - 1; i++) for (int j = 0; j < m ; j++) if (a[i, j] >= a[i + 1, j] || b[i, j] >= b[i + 1, j]) return "No"; return "Yes";} // Driver Codepublic static void Main(string []arg){ int n = 2, m = 2; int [,]a = {{2, 10}, {11, 5}}; int [,]b = {{9, 4}, {3, 12}}; Console.Write(Check(a, b, n, m));}}// This code is contributed by Rutvik_56 |
PHP
<?php// PHP implementation of the approach // Function to check whether the matrices // can be made strictly increasing // with the given operation function Check($a, $b, $n, $m){ // Swap only when a[i][j] > b[i][j] for ($i = 0; $i < $n; $i++) { for ($j= 0; $j < $n; $j++) { if ($a[$i][$j] > $b[$i][$j]) { $temp = $a[$i][$j]; $a[$i][$j] = $b[$i][$j]; $b[$i][$j] = $temp; } } } // Check if rows are strictly increasing for ($i = 0; $i < $n; $i++) { for ($j= 0;$j < $m-1 ; $j++) { if($a[$i][$j] >= $a[$i][$j + 1] or $b[$i][$j] >= $b[$i][$j + 1]) return "No"; } } // Check if columns are strictly increasing for ($i = 0; $i < $n - 1; $i++) { for ($j = 0; $j < $m; $j++) { if ($a[$i][$j] >= $a[$i + 1][$j] or $b[$i][$j] >= $b[$i + 1][$j]) return "No"; } } return "Yes";}// Driver code $n = 2; $m = 2;$a = array(array(2, 10), array(11, 5)); $b = array(array(9, 4), array(3, 12)); print(Check($a, $b, $n, $m));// This code is contributed by AnkitRai01?> |
Javascript
<script> // Function to check whether the matrices// can be made strictly increasing// with the given operationfunction Check(a, b, n, m){ // Swap only when a[i][j] > b[i][j] for(let i = 0; i < n; i++) for(let j = 0; j < m; j++) if (a[i][j] > b[i][j]) { let temp = a[i][j]; a[i][j] = b[i][j]; b[i][j] = temp; } // Check if rows are strictly increasing for(let i = 0; i < n; i++) for(let j = 0; j < m - 1; j++) if (a[i][j] >= a[i][j + 1] || b[i][j] >= b[i][j + 1]) return "No"; // Check if columns are strictly increasing for(let i = 0; i < n - 1; i++) for(let j = 0; j < m ;j++) if (a[i][j] >= a[i + 1][j] || b[i][j] >= b[i + 1][j]) return "No"; return "Yes";} // Driver Code let n = 2, m = 2; let a = [[ 2, 10 ], [ 11, 5 ]]; let b = [[ 9, 4 ], [3, 12 ]]; document.write(Check(a, b, n, m)); </script> |
Output:
Yes
Time Complexity: O(N*M), as we are traversing over the matrix.
Auxiliary Space: O(1), as we are not using any extra space.
