**2D Array-DS**

### Problem Statement :

Given a 6*6 2D Array, arr: 1 1 1 0 0 0 0 1 0 0 0 0 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 An hourglass in A is a subset of values with indices falling in this pattern in arr's graphical representation: a b c d e f g There are 16 hourglasses in arr. An hourglass sum is the sum of an hourglass' values. Calculate the hourglass sum for every hourglass in arr, then print the maximum hourglass sum. The array will always be 6*6. Example: arr= -9 -9 -9 1 1 1 0 -9 0 4 3 2 -9 -9 -9 1 2 3 0 0 8 6 6 0 0 0 0 -2 0 0 0 0 1 2 4 0 The 16 hourglass sums are: -63, -34, -9, 12, -10, 0, 28, 23, -27, -11, -2, 10, 9, 17, 25, 18 The highest hourglass sum is 28 from the hourglass beginning at row 1, column 2: 0 4 3 1 8 6 6 Note: If you have already solved the Java domain's Java 2D Array challenge, you may wish to skip this challenge. Function Description Complete the function hourglassSum in the editor below. hourglassSum has the following parameter(s): 1. int arr[6][6]: an array of integers Returns 2 .int: the maximum hourglass sum Input Format: Each of the 6 lines of inputs arr[i] contains 6 space-separated integers arr[i][j]. Constraints: -9<=arr[i][j]<=9 0<=i,j<=5 Output Format: Print the largest (maximum) hourglass sum found in arr.

### Solution :

` ````
Solution in C :
In C:
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
int main() {
int i,j,k;
int arr[6][6],temp=-9999,a,b;
for(i=0;i<6;i++)
for(j=0;j<6;j++)
scanf("%d",&arr[i][j]);
for(i=0;i<=3;i++)
for(j=0;j<=3;j++)
{
a = 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(temp < a)
temp = a ;
}
printf("%d",temp);
return 0;
}
In C++:
#include <cmath>
#include <cstdio>
#include <vector>
#include <climits>
#include <iostream>
#include <algorithm>
using namespace std;
int main() {
int a[6][6],s;
int m=INT_MIN;
for(int i=0;i<6;i++)
{
for(int j=0;j<6;j++)
{
cin>>a[i][j];
}
}
for(int i=0;i<4;i++)
{
for(int j=0;j<4;j++)
{
s=(a[i][j]+a[i][j+1]+a[i][j+2]+a[i+1][j+1]+a[i+2][j]+a[i+2][j+1]+a[i+2][j+2]);
if(s>m)
m=s;
}
}
cout<<m;
return 0;
}
In Java:
import java.io.*;
import java.util.*;
public class Solution {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int[][] array = new int[6][6];
for (int y = 0; y < 6; y++){
for (int x =0; x<6; x++){
array[x][y] = sc.nextInt();
}
}
int maxHourglass = getHourglass(array, 1,1);
for (int y=1; y<5; y++){
for (int x=1; x<5; x++){
int hourres = getHourglass(array, x, y);
if (hourres > maxHourglass){
maxHourglass = hourres;
}
}
}
System.out.println(maxHourglass);
}
public static int getHourglass(int[][] array, int x, int y) {
return array[x][y] + array[x-1][y-1] + array[x][y-1] + array[x+1][y-1] + array[x-1][y+1]
+ array[x][y+1] + array[x+1][y+1];
}
}
In Python 3:
```

## View More Similar Problems

## Costly Intervals

Given an array, your goal is to find, for each element, the largest subarray containing it whose cost is at least k. Specifically, let A = [A1, A2, . . . , An ] be an array of length n, and let be the subarray from index l to index r. Also, Let MAX( l, r ) be the largest number in Al. . . r. Let MIN( l, r ) be the smallest number in Al . . .r . Let OR( l , r ) be the bitwise OR of the

View Solution →## The Strange Function

One of the most important skills a programmer needs to learn early on is the ability to pose a problem in an abstract way. This skill is important not just for researchers but also in applied fields like software engineering and web development. You are able to solve most of a problem, except for one last subproblem, which you have posed in an abstract way as follows: Given an array consisting

View Solution →## Self-Driving Bus

Treeland is a country with n cities and n - 1 roads. There is exactly one path between any two cities. The ruler of Treeland wants to implement a self-driving bus system and asks tree-loving Alex to plan the bus routes. Alex decides that each route must contain a subset of connected cities; a subset of cities is connected if the following two conditions are true: There is a path between ever

View Solution →## Unique Colors

You are given an unrooted tree of n nodes numbered from 1 to n . Each node i has a color, ci. Let d( i , j ) be the number of different colors in the path between node i and node j. For each node i, calculate the value of sum, defined as follows: Your task is to print the value of sumi for each node 1 <= i <= n. Input Format The first line contains a single integer, n, denoti

View Solution →## Fibonacci Numbers Tree

Shashank loves trees and math. He has a rooted tree, T , consisting of N nodes uniquely labeled with integers in the inclusive range [1 , N ]. The node labeled as 1 is the root node of tree , and each node in is associated with some positive integer value (all values are initially ). Let's define Fk as the Kth Fibonacci number. Shashank wants to perform 22 types of operations over his tree, T

View Solution →## Pair Sums

Given an array, we define its value to be the value obtained by following these instructions: Write down all pairs of numbers from this array. Compute the product of each pair. Find the sum of all the products. For example, for a given array, for a given array [7,2 ,-1 ,2 ] Note that ( 7 , 2 ) is listed twice, one for each occurrence of 2. Given an array of integers, find the largest v

View Solution →