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-matrixint 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 sumbool 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 codeint 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 approachclass 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 approachN = 4# Function to return the sum of the sub-matrixdef 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 sumdef 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 codegrid = [[1, 2, 3, 4], [5, 6, 7, 8], [9, 10, 11, 12], [13, 14, 15, 16]]K = 2S = 14# Function callif (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 approachusing 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; $S = 14; // Function callif (sumFound($K, $S, $grid)) echo "Yes"; else echo "No";// This code is contributed by Ryuga//Modified by Kartik Verma?> |
Javascript
<script>// Javascript implementation of the approachvar N = 4// Function to return the sum of the sub-matrixfunction 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 sumfunction 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 codevar grid = [ [ 1, 2, 3, 4 ], [ 5, 6, 7, 8 ], [ 9, 10, 11, 12 ], [ 13, 14, 15, 16 ] ];var K = 2;var S = 14;// Function callif (sumFound(K, S, grid)) document.write( "Yes");else document.write( "No" );</script> |
Output
Yes
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