Today we are going to solve HackerRank Day 11 : 2D arrays 30 days of code solution in C, C++ , Java , Python & Javascript.
Objective
Today, we are building on our knowledge of arrays by adding another dimension.
Context
Given a 6 x 6 2D Array, A:
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
We define an hourglass in A to be a subset of values with indices falling in this pattern in A‘s graphical representation:
a b c
d
e f g
There are 16 hourglasses in A, and an hourglass sum is the sum of an hourglass’ values.
Task
Calculate the hourglass sum for every hourglass in A, then print the maximum hourglass sum.
Example
In the array shown above, the maximum hourglass sum is 7 for the hourglass in the top left corner.
Input Format
There are 6 lines of input, where each line contains 6 space-separated integers that describe the 2D Array A.
Constraints
- -9 <= A[i][j] <= 9
- 0 <= i, j <= 5
Output Format
Print the maximum hourglass sum in A.
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
A contains the following hourglasses:
1 1 1 1 1 0 1 0 0 0 0 0
1 0 0 0
1 1 1 1 1 0 1 0 0 0 0 0
0 1 0 1 0 0 0 0 0 0 0 0
1 1 0 0
0 0 2 0 2 4 2 4 4 4 4 0
1 1 1 1 1 0 1 0 0 0 0 0
0 2 4 4
0 0 0 0 0 2 0 2 0 2 0 0
0 0 2 0 2 4 2 4 4 4 4 0
0 0 2 0
0 0 1 0 1 2 1 2 4 2 4 0
The hourglass with the maximum sum (19) is:
2 4 4
2
1 2 4
HackerRank Day 11 : 2D arrays 30 days of code solution
2D arrays HackerRank Solution in C
#include <math.h> #include <stdio.h> #include <string.h> #include <stdlib.h> #include <assert.h> #include <limits.h> #include <stdbool.h> int soln(int arr[6][6]) { int i, j, sum,temp_sum; temp_sum = 0; sum = -9999999; for(i=0; i<6-2;i++) { for(j=0; j<6-2;j++){ temp_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]; //printf("%d %d %d\n", arr[i][j], arr[i][j+1], arr[i][j+2]); //printf(" %d \n", arr[i+1][j+1]); //printf("%d %d %d\n", arr[i+2][j] ,arr[i+2][j+1] ,arr[i+2][j+2]); //printf("temp sum: %d \n\n", temp_sum); if(temp_sum > sum) { sum = temp_sum; } } } return sum; } int main(){ int arr[6][6]; for(int arr_i = 0; arr_i < 6; arr_i++){ for(int arr_j = 0; arr_j < 6; arr_j++){ scanf("%d",&arr[arr_i][arr_j]); } } printf("%d\n", soln(arr)); return 0; }
2D arrays HackerRank Solution in C ++
#include <map> #include <set> #include <list> #include <cmath> #include <ctime> #include <deque> #include <queue> #include <stack> #include <string> #include <bitset> #include <cstdio> #include <limits> #include <vector> #include <climits> #include <cstring> #include <cstdlib> #include <fstream> #include <numeric> #include <sstream> #include <iostream> #include <algorithm> using namespace std; int add(int arr[6][6], int arr_i, int arr_j){ int sum = arr[arr_i][arr_j]; sum += arr[arr_i][arr_j +1]; sum += arr[arr_i][arr_j +2]; sum += arr[arr_i +1][arr_j +1]; sum += arr[arr_i +2][arr_j]; sum += arr[arr_i +2][arr_j +1]; sum += arr[arr_i +2][arr_j +2]; return sum; } int main(){ int sum = -9 * 7; int arr[6][6]; for(int arr_i = 0;arr_i < 6;arr_i++){ for(int arr_j = 0;arr_j < 6;arr_j++){ cin >> arr[arr_i][arr_j]; } } for(int arr_i = 0;arr_i < 4;arr_i++){ for(int arr_j = 0;arr_j < 4;arr_j++){ int t = add(arr, arr_i, arr_j); if(t > sum) sum = t; } } cout << sum << endl; return 0; }
2D arrays HackerRank Solution in Java
import java.io.*; import java.util.*; import java.text.*; import java.math.*; import java.util.regex.*; public class Solution { public static void main(String[] args) { Scanner in = new Scanner(System.in); int arr[][] = new int[6][6]; for(int i=0; i < 6; i++){ for(int j=0; j < 6; j++){ arr[i][j] = in.nextInt(); } } int maxval = -9*6; for(int i=0; i<4; i++) { for(int j=0; j<4; j++) { int sum = arr[i][j] + arr[i][j+1] + arr[i][j+2]; sum += arr[i+1][j+1]; sum += arr[i+2][j] + arr[i+2][j+1] + arr[i+2][j+2]; if(sum>maxval) { maxval = sum; } } } System.out.println(maxval); } }
2D arrays HackerRank Solution in Python 3
#!/bin/python3 import math import os import random import re import sys if __name__ == '__main__': arr = [] for _ in range(6): arr.append(list(map(int, input().rstrip().split()))) sum = 0 tarr = [] for l in range(0,4): for k in range(0,4): for i in range(l,l+3): for j in range(k,k+3): if i == l+1 and ( j == k or j == k+2): continue else: sum += arr[i][j] tarr.append(sum) sum = 0 print(max(tarr))
2D arrays HackerRank Solution in JavaScript
process.stdin.resume(); process.stdin.setEncoding('ascii'); var input_stdin = ""; var input_stdin_array = ""; var input_currentline = 0; process.stdin.on('data', function (data) { input_stdin += data; }); process.stdin.on('end', function () { input_stdin_array = input_stdin.split("\n"); main(); }); function readLine() { return input_stdin_array[input_currentline++]; } /////////////// ignore above this line //////////////////// function main() { var arr = []; for(arr_i = 0; arr_i < 6; arr_i++){ arr[arr_i] = readLine().split(' '); arr[arr_i] = arr[arr_i].map(Number); } /* row i, column j * 1 1 1 * 1 1 1 * 1 1 1 */ var arrs = [] for (var i = 1; i < arr.length - 1;i++){ for (var j = 1; j < arr[i].length - 1; j++){ var sum = 0; // top sum = parseInt(arr[i-1][j-1]) + parseInt(arr[i-1][j]) + parseInt(arr[i-1][j+1]); // middle sum = sum + parseInt(arr[i][j]); // bottom sum = sum + parseInt(arr[i+1][j-1]) + parseInt(arr[i+1][j]) + parseInt(arr[i+1][j+1]); arrs.push(sum); } } //console.log(arrs); console.log(Math.max.apply(null, arrs)); }
NEXT : HackerRank Day 12 : Inheritance 30 days of code solution
30 Days of Code HackerRank Solutions List – Day 0 to Day 29
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