Given an N x N matrix and two integers S and K, the task is to find whether there exists a K x K sub-matrix with sum equal to S.
Examples:
Input: K = 2, S = 14, mat[][] = {
{ 1, 2, 3, 4 },
{ 5, 6, 7, 8 },
{ 9, 10, 11, 12 },
{ 13, 14, 15, 16 }}
Output: Yes
1 2
5 6
is the required 2 x 2 sub-matrix with element sum = 14Input: K = 1, S = 5, mat[][] = {{1, 2}, {7, 8}}
Output: No
Approach:
Dynamic programming can be used to solve this problem,
- Create an array dp[N + 1][N + 1] where dp[i][j] stores the sum of all the elements with row between 1 to i and column between 1 to j.
- Once the 2-D matrix is generated, now suppose we wish to find sum of square starting with (i, j) to (i + x, j + x). The required sum will be dp[i + x][j + x] – dp[i][j + x] – dp[i + x][j] + dp[i][j] where,
- First term denotes the sum of all the elements present in rows between 1 to i + x and columns between 1 to j + x. This area has our required square.
- Second two terms is to remove the area which is outside our required region but inside the region calculated in the first step.
- Sum of elements of rows between 1 to i and columns between 1 to j is subtracted twice in the second step, so it is added once.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach #include <bits/stdc++.h> using namespace std; #define ll long long int #define N 4 // Function to return the sum of the sub-matrix int getSum( int r1, int r2, int c1, int c2, int dp[N + 1][N + 1]) { return dp[r2][c2] - dp[r2][c1] - dp[r1][c2] + dp[r1][c1]; } // Function that returns true if it is possible // to find the sub-matrix with required sum bool sumFound( int K, int S, int grid[N][N]) { // 2-D array to store the sum of // all the sub-matrices int dp[N + 1][N + 1]; // Filling of dp[][] array for ( int i = 0; i < N; i++) for ( int j = 0; j < N; j++) dp[i + 1][j + 1] = dp[i + 1][j] + dp[i][j + 1] - dp[i][j] + grid[i][j]; // Checking for each possible sub-matrix of size k X k for ( int i = 0; i < N; i++) for ( int j = 0; j < N; j++) { int sum = getSum(i, i + K, j, j + K, dp); if (sum == S) return true ; } // Sub-matrix with the given sum not found return false ; } // Driver code int main() { int grid[N][N] = { { 1, 2, 3, 4 }, { 5, 6, 7, 8 }, { 9, 10, 11, 12 }, { 13, 14, 15, 16 } }; int K = 2; int S = 14; // Function call if (sumFound(K, S, grid)) cout << "Yes" << endl; else cout << "No" << endl; } // Modified by Kartik Verma |
Java
// Java implementation of the approach class GfG { static int N = 4 ; // Function to return the sum of the sub-matrix static int getSum( int r1, int r2, int c1, int c2, int dp[][]) { return dp[r2][c2] - dp[r2][c1] - dp[r1][c2] + dp[r1][c1]; } // Function that returns true if it is possible // to find the sub-matrix with required sum static boolean sumFound( int K, int S, int grid[][]) { // 2-D array to store the sum of // all the sub-matrices int dp[][] = new int [N + 1 ][N + 1 ]; // Filling of dp[][] array for ( int i = 0 ; i < N; i++) { for ( int j = 0 ; j < N; j++) { dp[i + 1 ][j + 1 ] = dp[i + 1 ][j] + dp[i][j + 1 ] - dp[i][j] + grid[i][j]; } } // Checking for each possible sub-matrix of size k X // k for ( int i = 0 ; i < N; i++) { for ( int j = 0 ; j < N; j++) { int sum = getSum(i, i + K, j, j + K, dp); if (sum == S) { return true ; } } } // Sub-matrix with the given sum not found return false ; } // Driver code public static void main(String[] args) { int grid[][] = { { 1 , 2 , 3 , 4 }, { 5 , 6 , 7 , 8 }, { 9 , 10 , 11 , 12 }, { 13 , 14 , 15 , 16 } }; int K = 2 ; int S = 14 ; // Function call if (sumFound(K, S, grid)) { System.out.println( "Yes" ); } else { System.out.println( "No" ); } } } // This code contributed by Rajput-Ji // Modified by Kartik Verma |
Python3
# Python implementation of the approach N = 4 # Function to return the sum of the sub-matrix def getSum(r1, r2, c1, c2, dp): return dp[r2][c2] - dp[r2][c1] - dp[r1][c2] + dp[r1][c1] # Function that returns true if it is possible # to find the sub-matrix with required sum def sumFound(K, S, grid): # 2-D array to store the sum of # all the sub-matrices dp = [[ 0 for i in range (N + 1 )] for j in range (N + 1 )] # Filling of dp[][] array for i in range (N): for j in range (N): dp[i + 1 ][j + 1 ] = dp[i + 1 ][j] + \ dp[i][j + 1 ] - dp[i][j] + grid[i][j] # Checking for each possible sub-matrix of size k X k for i in range ( 0 , N): for j in range ( 0 , N): sum = getSum(i, i + K, j, j + K, dp) if ( sum = = S): return True # Sub-matrix with the given sum not found return False # Driver code grid = [[ 1 , 2 , 3 , 4 ], [ 5 , 6 , 7 , 8 ], [ 9 , 10 , 11 , 12 ], [ 13 , 14 , 15 , 16 ]] K = 2 S = 14 # Function call if (sumFound(K, S, grid)): print ( "Yes" ) else : print ( "No" ) # This code is contributed by ankush_953 # Modified by Kartik Verma |
C#
// C# implementation of the approach using System; class GfG { static int N = 4; // Function to return the sum of the sub-matrix static int getSum( int r1, int r2, int c1, int c2, int [, ] dp) { return dp[r2, c2] - dp[r2, c1] - dp[r1, c2] + dp[r1, c1]; } // Function that returns true if it is possible // to find the sub-matrix with required sum static bool sumFound( int K, int S, int [, ] grid) { // 2-D array to store the sum of // all the sub-matrices int [, ] dp = new int [N + 1, N + 1]; // Filling of dp[,] array for ( int i = 0; i < N; i++) { for ( int j = 0; j < N; j++) { dp[i + 1, j + 1] = dp[i + 1, j] + dp[i, j + 1] - dp[i, j] + grid[i, j]; } } // Checking for each possible sub-matrix of size k X // k for ( int i = 0; i < N; i++) { for ( int j = 0; j < N; j++) { int sum = getSum(i, i + K, j, j + K, dp); if (sum == S) { return true ; } } } // Sub-matrix with the given sum not found return false ; } // Driver code public static void Main(String[] args) { int [, ] grid = { { 1, 2, 3, 4 }, { 5, 6, 7, 8 }, { 9, 10, 11, 12 }, { 13, 14, 15, 16 } }; int K = 2; int S = 14; // Function call if (sumFound(K, S, grid)) { Console.WriteLine( "Yes" ); } else { Console.WriteLine( "No" ); } } } // This code has been contributed by 29AjayKumar // Modified by Kartik Verma |
PHP
<?php // PHP implementation of the approach $GLOBALS [ 'N' ] = 4; // Function to return the sum of // the sub-matrix function getSum( $r1 , $r2 , $c1 , $c2 , $dp ) { return $dp [ $r2 ][ $c2 ] - $dp [ $r2 ][ $c1 ] - $dp [ $r1 ][ $c2 ] + $dp [ $r1 ][ $c1 ]; } // Function that returns true if it is // possible to find the sub-matrix with // required sum function sumFound( $K , $S , $grid ) { // 2-D array to store the sum of // all the sub-matrices $dp = array ( array ()); for ( $i = 0; $i < $GLOBALS [ 'N' ]; $i ++) for ( $j = 0; $j < $GLOBALS [ 'N' ]; $j ++) $dp [ $i ][ $j ] = 0 ; // Filling of dp[][] array for ( $i = 0; $i < $GLOBALS [ 'N' ]; $i ++) for ( $j = 0; $j < $GLOBALS [ 'N' ]; $j ++) $dp [ $i + 1][ $j + 1] = $dp [ $i + 1][ $j ] + $dp [ $i ][ $j + 1] - $dp [ $i ][ $j ] + $grid [ $i ][ $j ]; // Checking for each possible sub-matrix // of size k X k for ( $i = 0; $i < $GLOBALS [ 'N' ]; $i ++) for ( $j = 0; $j < $GLOBALS [ 'N' ]; $j ++) { $sum = getSum( $i , $i + $K , $j , $j + $K , $dp ); if ( $sum == $S ) return true; } // Sub-matrix with the given // sum not found return false; } // Driver code $grid = array ( array (1, 2, 3, 4), array (5, 6, 7, 8), array (9, 10, 11, 12), array (13, 14, 15, 16)); $K = 2; $S = 14; // Function call if (sumFound( $K , $S , $grid )) echo "Yes" ; else echo "No" ; // This code is contributed by Ryuga //Modified by Kartik Verma ?> |
Javascript
<script> // Javascript implementation of the approach var N = 4 // Function to return the sum of the sub-matrix function getSum(r1, r2, c1, c2, dp) { return dp[r2][c2] - dp[r2][c1] - dp[r1][c2] + dp[r1][c1]; } // Function that returns true if it is possible // to find the sub-matrix with required sum function sumFound(K, S, grid) { // 2-D array to store the sum of // all the sub-matrices var dp = Array.from(Array(N+1), ()=> Array(N+1).fill(0)); // Filling of dp[][] array for ( var i = 0; i < N; i++) for ( var j = 0; j < N; j++) dp[i + 1][j + 1] = dp[i + 1][j] + dp[i][j + 1] - dp[i][j] + grid[i][j]; // Checking for each possible sub-matrix of size k X k for ( var i = 0; i < N; i++) for ( var j = 0; j < N; j++) { var sum = getSum(i, i + K, j, j + K, dp); if (sum == S) return true ; } // Sub-matrix with the given sum not found return false ; } // Driver code var grid = [ [ 1, 2, 3, 4 ], [ 5, 6, 7, 8 ], [ 9, 10, 11, 12 ], [ 13, 14, 15, 16 ] ]; var K = 2; var S = 14; // Function call if (sumFound(K, S, grid)) document.write( "Yes" ); else document.write( "No" ); </script> |
Output
Yes
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