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 :



title-img


                            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:






def convert(x):
    res1=[]
    for i in range(len(x)):
        res1.append(int(x[i]))
        
    return res1

def cal_values(matrix):
    a0 = matrix[0]
    a1 = matrix[1]
    a2 = matrix[2]
    res3=[]
    for kk in range(4):
        temp1 = 0
        temp1 = sum(a0[kk:kk+3])
        temp1 += a1[kk + 1]
        temp1 += sum(a2[kk:kk+3])
        res3.append(temp1)

    return max(res3)


ss=[]
result=[]
for i in range(6):
    temp=input().split()
    res2=convert(temp)    
    ss.append(res2)

for j in range(4):
     ee=cal_values(ss[j:j+3])
     result.append(ee)
        
print(max(result))
                        








View More Similar Problems

Tree Coordinates

We consider metric space to be a pair, , where is a set and such that the following conditions hold: where is the distance between points and . Let's define the product of two metric spaces, , to be such that: , where , . So, it follows logically that is also a metric space. We then define squared metric space, , to be the product of a metric space multiplied with itself: . For

View Solution →

Array Pairs

Consider an array of n integers, A = [ a1, a2, . . . . an] . Find and print the total number of (i , j) pairs such that ai * aj <= max(ai, ai+1, . . . aj) where i < j. Input Format The first line contains an integer, n , denoting the number of elements in the array. The second line consists of n space-separated integers describing the respective values of a1, a2 , . . . an .

View Solution →

Self Balancing Tree

An AVL tree (Georgy Adelson-Velsky and Landis' tree, named after the inventors) is a self-balancing binary search tree. In an AVL tree, the heights of the two child subtrees of any node differ by at most one; if at any time they differ by more than one, rebalancing is done to restore this property. We define balance factor for each node as : balanceFactor = height(left subtree) - height(righ

View Solution →

Array and simple queries

Given two numbers N and M. N indicates the number of elements in the array A[](1-indexed) and M indicates number of queries. You need to perform two types of queries on the array A[] . You are given queries. Queries can be of two types, type 1 and type 2. Type 1 queries are represented as 1 i j : Modify the given array by removing elements from i to j and adding them to the front. Ty

View Solution →

Median Updates

The median M of numbers is defined as the middle number after sorting them in order if M is odd. Or it is the average of the middle two numbers if M is even. You start with an empty number list. Then, you can add numbers to the list, or remove existing numbers from it. After each add or remove operation, output the median. Input: The first line is an integer, N , that indicates the number o

View Solution →

Maximum Element

You have an empty sequence, and you will be given N queries. Each query is one of these three types: 1 x -Push the element x into the stack. 2 -Delete the element present at the top of the stack. 3 -Print the maximum element in the stack. Input Format The first line of input contains an integer, N . The next N lines each contain an above mentioned query. (It is guaranteed that each

View Solution →