Given a 2D matrix, the task is to find the maximum sum of an hourglass.
An hour glass is made of 7 cells
in following form.
A B C
D
E F G
Examples:
Input : 1 1 1 0 0
0 1 0 0 0
1 1 1 0 0
0 0 0 0 0
0 0 0 0 0
Output : 7
Below is the hour glass with
maximum sum:
1 1 1
1
1 1 1
Input : 0 3 0 0 0
0 1 0 0 0
1 1 1 0 0
0 0 2 4 4
0 0 0 2 4
Output : 11
Below is the hour glass with
maximum sum
1 0 0
4
0 2 4
Approach:
It is evident from the definition of the hourglass that the number of rows and number of columns must be equal to 3. If we count the total number of hourglasses in a matrix, we can say that the count is equal to the count of possible top left cells in an hourglass. The number of top-left cells in an hourglass is equal to (R-2)*(C-2). Therefore, in a matrix total number of an hourglass is (R-2)*(C-2).
mat[][] = 2 3 0 0 0
0 1 0 0 0
1 1 1 0 0
0 0 2 4 4
0 0 0 2 0
Possible hour glass are :
2 3 0 3 0 0 0 0 0
1 0 0
1 1 1 1 1 0 1 0 0
0 1 0 1 0 0 0 0 0
1 1 0
0 0 2 0 2 4 2 4 4
1 1 1 1 1 0 1 0 0
0 2 4
0 0 0 0 0 2 0 2 0
Consider all top left cells of hourglasses one by one. For every cell, we compute the sum of the hourglass formed by it. Finally, return the maximum sum.
Below is the implementation of the above idea:
C++
// C++ program to find maximum sum of hour// glass in matrix#include<bits/stdc++.h>using namespace std;const int R = 5;const int C = 5;// Returns maximum sum of hour glass in ar[][]int findMaxSum(int mat[R][C]){ if (R<3 || C<3){ cout << "Not possible" << endl; exit(0); } // Here loop runs (R-2)*(C-2) times considering // different top left cells of hour glasses. int max_sum = INT_MIN; for (int i=0; i<R-2; i++) { for (int j=0; j<C-2; j++) { // Considering mat[i][j] as top left cell of // hour glass. int sum = (mat[i][j]+mat[i][j+1]+mat[i][j+2])+ (mat[i+1][j+1])+ (mat[i+2][j]+mat[i+2][j+1]+mat[i+2][j+2]); // If previous sum is less than current sum then // update new sum in max_sum max_sum = max(max_sum, sum); } } return max_sum;}// Driver codeint main(){ int mat[][C] = {{1, 2, 3, 0, 0}, {0, 0, 0, 0, 0}, {2, 1, 4, 0, 0}, {0, 0, 0, 0, 0}, {1, 1, 0, 1, 0}}; int res = findMaxSum(mat); cout << "Maximum sum of hour glass = "<< res << endl; return 0;}//Code is modified by Susobhan Akhuli |
C
/* C program to find the maximum sum of hour glass in a Matrix */#include <stdio.h>#include <stdlib.h>// Fixing the size of the matrix // ( Here it is of the order 6// x 6 )#define R 5#define C 5// Function to find the maximum// sum of the hour glassint MaxSum(int arr[R][C]){ int i, j, sum; if (R<3 || C<3){ printf("Not Possible"); exit(0); } int max_sum = -500000; /* Considering the matrix also contains negative values , so initialized with -50000. It can be any value but very smaller.*/ // int max_sum=0 -> Initialize with 0 only if your // matrix elements are positive // Here loop runs (R-2)*(C-2) times considering // different top left cells of hour glasses. for (i = 0; i < R - 2; i++) { for (j = 0; j < C - 2; j++) { // Considering arr[i][j] as top left cell of // hour glass. sum = (arr[i][j] + arr[i][j + 1] + arr[i][j + 2]) + (arr[i + 1][j + 1]) + (arr[i + 2][j] + arr[i + 2][j + 1] + arr[i + 2][j + 2]); // If previous sum is less than current sum then // update new sum in max_sum if (sum > max_sum) max_sum = sum; else continue; } } return max_sum;}// Driver Codeint main(){ int arr[][C] = { { 1, 2, 3, 0, 0 }, { 0, 0, 0, 0, 0 }, { 2, 1, 4, 0, 0 }, { 0, 0, 0, 0, 0 }, { 1, 1, 0, 1, 0 } }; int res = MaxSum(arr); printf("Maximum sum of hour glass = %d", res); return 0;}// This code is written by Akshay Prakash// Code is modified by Susobhan Akhuli |
Java
// Java program to find maximum // sum of hour glass in matriximport java.io.*;class GFG { static int R = 5;static int C = 5;// Returns maximum sum of // hour glass in ar[][]static int findMaxSum(int [][]mat){ if (R < 3 || C < 3){ System.out.println("Not possible"); System.exit(0); } // Here loop runs (R-2)*(C-2) // times considering different // top left cells of hour glasses. int max_sum = Integer.MIN_VALUE; for (int i = 0; i < R - 2; i++) { for (int j = 0; j < C - 2; j++) { // Considering mat[i][j] as top // left cell of hour glass. int sum = (mat[i][j] + mat[i][j + 1] + mat[i][j + 2]) + (mat[i + 1][j + 1]) + (mat[i + 2][j] + mat[i + 2][j + 1] + mat[i + 2][j + 2]); // If previous sum is less than // current sum then update // new sum in max_sum max_sum = Math.max(max_sum, sum); } } return max_sum;} // Driver code static public void main (String[] args) { int [][]mat = {{1, 2, 3, 0, 0}, {0, 0, 0, 0, 0}, {2, 1, 4, 0, 0}, {0, 0, 0, 0, 0}, {1, 1, 0, 1, 0}}; int res = findMaxSum(mat); System.out.println("Maximum sum of hour glass = "+ res); } }// This code is contributed by vt_m .// Code is modified by Susobhan Akhuli |
Python3
# Python 3 program to find the maximum# sum of hour glass in a Matrix# Fixing the size of the Matrix. # Here it is of order 6 x 6R = 5C = 5# Function to find the maximum sum of hour glassdef MaxSum(arr): # Considering the matrix also contains max_sum = -50000 # Negative values , so initialized with # -50000. It can be any value but very # smaller. # max_sum=0 -> Initialize with 0 only if your # matrix elements are positive if(R < 3 or C < 3): print("Not possible") exit() # Here loop runs (R-2)*(C-2) times considering # different top left cells of hour glasses. for i in range(0, R-2): for j in range(0, C-2): # Considering arr[i][j] as top # left cell of hour glass. SUM = (arr[i][j] + arr[i][j + 1] + arr[i][j + 2]) + (arr[i + 1][j + 1]) + (arr[i + 2][j] + arr[i + 2][j + 1] + arr[i + 2][j + 2]) # If previous sum is less # then current sum then # update new sum in max_sum if(SUM > max_sum): max_sum = SUM else: continue return max_sum# Driver Codearr = [[1, 2, 3, 0, 0], [0, 0, 0, 0, 0], [2, 1, 4, 0, 0], [0, 0, 0, 0, 0], [1, 1, 0, 1, 0]]res = MaxSum(arr)print(f"Maximum sum of hour glass = {res}")# This code is written by Akshay Prakash# Code is modified by Susobhan Akhuli |
C#
// C# program to find maximum // sum of hour glass in matrixusing System;class GFG { static int R = 5;static int C = 5;// Returns maximum sum of // hour glass in ar[][]static int findMaxSum(int [,]mat){ if (R < 3 || C < 3){ Console.WriteLine("Not possible"); Environment.Exit(0); } // Here loop runs (R-2)*(C-2) // times considering different // top left cells of hour glasses. int max_sum = int.MinValue; for (int i = 0; i < R - 2; i++) { for (int j = 0; j < C - 2; j++) { // Considering mat[i][j] as top // left cell of hour glass. int sum = (mat[i, j] + mat[i, j + 1] + mat[i, j + 2]) + (mat[i + 1, j + 1]) + (mat[i + 2, j] + mat[i + 2, j + 1] + mat[i + 2, j + 2]); // If previous sum is less than // current sum then update // new sum in max_sum max_sum = Math.Max(max_sum, sum); } } return max_sum;} // Driver code static public void Main(String[] args) { int [,]mat = {{1, 2, 3, 0, 0}, {0, 0, 0, 0, 0}, {2, 1, 4, 0, 0}, {0, 0, 0, 0, 0}, {1, 1, 0, 1, 0}}; int res = findMaxSum(mat); Console.WriteLine("Maximum sum of hour glass = "+ res); } }// This code is contributed by vt_m .// Code is modified by Susobhan Akhuli |
PHP
<?php// PHP program to find maximum sum // of hour glass in matrix$R = 5;$C = 5;// Returns maximum sum // of hour glass in ar[][]function findMaxSum($mat){ global $R; global $C; if ($R < 3 || $C < 3){ exit("Not possible"); } // Here loop runs (R-2)*(C-2) times considering // different top left cells of hour glasses. $max_sum = PHP_INT_MIN; for ($i = 0; $i < ($R - 2); $i++) { for ($j = 0; $j < ($C - 2); $j++) { // Considering mat[i][j] as // top left cell of hour glass. $sum = ($mat[$i][$j] + $mat[$i][$j + 1] + $mat[$i][$j + 2]) + ($mat[$i + 1][$j + 1]) + ($mat[$i + 2][$j] + $mat[$i + 2][$j + 1] + $mat[$i + 2][$j + 2]); // If previous sum is less than current sum // then update new sum in max_sum $max_sum = max($max_sum, $sum); } } return $max_sum;}// Driver code$mat = array(array(1, 2, 3, 0, 0), array(0, 0, 0, 0, 0), array(2, 1, 4, 0, 0), array(0, 0, 0, 0, 0), array(1, 1, 0, 1, 0));$res = findMaxSum($mat);echo "Maximum sum of hour glass = ",$res, "\n";// This code is contributed by ajit.// Code is modified by Susobhan Akhuli?> |
Javascript
<script> // Javascript program to find maximum // sum of hour glass in matrix let R = 5; let C = 5; // Returns maximum sum of // hour glass in ar[][] function findMaxSum(mat) { if (R < 3 || C < 3){ document.write("Not possible"); process.exit(0); } // Here loop runs (R-2)*(C-2) // times considering different // top left cells of hour glasses. let max_sum = Number.MIN_VALUE; for (let i = 0; i < R - 2; i++) { for (let j = 0; j < C - 2; j++) { // Considering mat[i][j] as top // left cell of hour glass. let sum = (mat[i][j] + mat[i][j + 1] + mat[i][j + 2]) + (mat[i + 1][j + 1]) + (mat[i + 2][j] + mat[i + 2][j + 1] + mat[i + 2][j + 2]); // If previous sum is less than // current sum then update // new sum in max_sum max_sum = Math.max(max_sum, sum); } } return max_sum; } let mat = [[1, 2, 3, 0, 0], [0, 0, 0, 0, 0], [2, 1, 4, 0, 0], [0, 0, 0, 0, 0], [1, 1, 0, 1, 0]]; let res = findMaxSum(mat); document.write("Maximum sum of hour glass = " + res); </script> |
Output
Maximum sum of hour glass = 13
Time complexity: O(R x C).
Auxiliary Space: O(1)
Reference :
http://stackoverflow.com/questions/38019861/hourglass-sum-in-2d-array
This article is contributed by AKSHAY PRAKASH. If you like neveropen and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the neveropen main page and help other Geeks.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
