## Algorithm

Problem Name: Data Structures - 2D Array - DS

In this HackerRank in Data Structures - 2D Array - DS

Given a 6 * 6 2D Arrayarr:

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 arrs 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 , 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):

• int arr[6][6]: an array of integers

Returns

• 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.

Sample Input

1 1 1 0 0 0
0 1 0 0 0 0
1 1 1 0 0 0
0 0 2 4 4 0
0 0 0 2 0 0
0 0 1 2 4 0


Sample Output

19


Explanation

arr ontains the following hourglasses:

The hourglass with the maximum sum (19) is:
2 4 4
2
1 2 4

## Code Examples

### #1 Code Example with C Programming

Code - C Programming


#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>

int main() {

/* Enter your code here. Read input from STDIN. Print output to STDOUT */
int a[6][6];

int i, j;
for (i=0; i < 6; i++){
for (j=0; j < 6; j++) {
scanf("%d", &a[i][j]);
}
}

int out=-100, sum=0;

for (i=0; i < 4; i++){
for (j=0; j < 4; j++) {
sum = 0;
sum += a[i][j];
sum += a[i][j+1];
sum += a[i][j+2];
sum += a[i+1][j+1];
sum += a[i+2][j];
sum += a[i+2][j+1];
sum += a[i+2][j+2];
if (sum > out){
out = sum;
}
}
}

printf("%d", out);

return 0;
}

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### #2 Code Example with C++ Programming

Code - C++ Programming


#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;

int arr[7][7];
int sum(int stx , int sty){
return arr[stx][sty] + arr[stx][sty+1] + arr[stx][sty+2] + arr[stx+1][sty+1] + arr[stx+2][sty] + arr[stx+2][sty+1] + arr[stx+2][sty+2];

}
int main() {
int ans = -100;
for(int i = 0 ; i  <  6 ; i++){
for(int j = 0 ; j  <  6 ; j++){
cin >> arr[i][j];
}
}

for(int i = 0 ; i  <  4 ; i++){
for(int j = 0 ; j  <  4 ; j++){
ans = max(ans , sum(i,j));
}
}

cout << ans << endl;
return 0;
}

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### #4 Code Example with Java Programming

Code - Java Programming


import java.util.Scanner;

public class Intro2dArray {

public static void main(String[] args) {

Scanner sc = new Scanner(System.in);
int multiDimArr[][] = new int[6][6];
for(int row = 0; row  <  6; row++){
for(int col = 0; col  <  6; col++){
multiDimArr[row][col] = sc.nextInt();
}
}
System.out.println(Solve(multiDimArr));
}
static int Solve(int arr[][]){
int max = Integer.MIN_VALUE;
int total = 0;
for(int row = 0; row  <  4; row++){

for(int col = 0; col  <  4; col++ ){

total = arr[row][col] + arr[row][col + 1] + arr[row][col + 2];
total += arr[row+1][col+1];
total += arr[row + 2][col] + arr[row+2][col + 1] + arr[row+2][col + 2];
max = total>max?total:max;
}
}
return max;

}

}

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### #5 Code Example with Javascript Programming

Code - Javascript Programming


'use strict';

const fs = require('fs');

process.stdin.resume();
process.stdin.setEncoding('utf-8');

let inputString = '';
let currentLine = 0;

process.stdin.on('data', function(inputStdin) {
inputString += inputStdin;
});

process.stdin.on('end', function() {
inputString = inputString.split('\n');

main();
});

return inputString[currentLine++];
}

/*
* Complete the 'hourglassSum' function below.
*
* The function is expected to return an INTEGER.
* The function accepts 2D_INTEGER_ARRAY arr as parameter.
*/

function hourglassSum(arr) {
let maxSum = ''
for (let step1 = 0; step1  <  arr.length-2; step1++){
for (let step2 = 0; step2  <  arr.length-2; step2++){
const currentSum = arr[step1][step2] + arr[step1][step2+1]
+ arr[step1][step2+2] + arr[step1+1][step2+1]
+ arr[step1+2][step2] + arr[step1+2][step2+1]
+ arr[step1+2][step2+2]

if(currentSum <= 0){
const temp = currentSum
if (typeof maxSum == 'string' ){
maxSum = temp
} else if(temp == 0 && maxSum  < = 0>{
maxSum = temp
}
else if(temp > maxSum){
maxSum = temp
}
} else if(currentSum > maxSum){
maxSum = currentSum
}
}
}
return maxSum
}

function main() {
const ws = fs.createWriteStream(process.env.OUTPUT_PATH);

let arr = Array(6);

for (let i = 0; i < 6; i++) {
arr[i] = readLine().replace(/\s+\$/g, '').split(' '>.map(arrTemp => parseInt(arrTemp, 10));
}

const result = hourglassSum(arr);

ws.write(result + '\n');

ws.end();
}

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### #6 Code Example with Python Programming

Code - Python Programming


n = 6
m = []
for i in range(n):
m.append(list(map(int, input().split()[:n])))

def sum_glass(m, i, j):
"""Assumes hour-glass is in bounds of m!"""
return sum(m[i][j:j + 3]) + m[i + 1][j + 1] + sum(m[i + 2][j:j + 3])

best = float("-inf")
for i in range(4):
for j in range(4):
s = sum_glass(m, i, j)
if s > best:
best = s

print (best)

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