3D Surface Area – HackerRank Solution

In this post, we will solve 3D Surface Area HackerRank Solution. This problem (3D Surface Area) is a part of HackerRank Problem Solving series.

Task

Madison is a little girl who is fond of toys. Her friend Mason works in a toy manufacturing factory . Mason has a 2D board A of size H x W  with H rows and W columns. The board is divided into cells of size 1 x 1 with each cell indicated by its coordinate (i, j). The cell (i, j) has an integer A_ij written on it. To create the toy Mason stacks A_ij number of cubes of size 1 x 1 x 1 on the cell (i, j).

Given the description of the board showing the values of A_ij and that the price of the toy is equal to the 3d surface area find the price of the toy.

Input Format

The first line contains two space-separated integers H and W the height and the width of the board respectively.

The next H lines contains W space separated integers. The j^th integer in i^th line denotes A_ij.

Constraints

• 1 <= H, W <= 100
• 1 <= A_ij <= 100

Output Format

Print the required answer, i.e the price of the toy, in one line.

Sample Input 0

1 1
1

Sample Output 0

6

Explanation 0

The surface area of 1 x 1 x 1 cube is 6.

Sample Input 1

3 3
1 3 4
2 2 3
1 2 4

Sample Output 1

60

Explanation 1

The object is rotated so the front row matches column 1 of the input, heights 1, 2, and 1.

• The front face is 1 + 2 + 1 = 4 units in area.
• The top is 3 units.
• The sides are 4 units.
• None of the rear faces are exposed.
• The underside is 3 units.

The front row contributes 4 + 3 + 4 + 3 = 14 units to the surface area.

Solution – 3D Surface Area – HackerRank Solution

C++

#include <iostream>
#include <string>
#include <cmath>
using namespace std;

int T[200][200];

int main() {
    int H, W;
    while(cin >> H >> W) {
        for(int i=0; i<H; i++) 
            for(int j=0; j<W; j++) 
                cin >> T[i][j];
                
        int total = 2*H*W;
        for(int i=0; i<H; i++) {
            total += T[i][0];
            for(int j=1; j<W; j++)
                total += abs(T[i][j] - T[i][j-1]);
            total += T[i][W-1];
        }
        for(int i=0; i<W; i++) {
            total += T[0][i];
            for(int j=1; j<H; j++)
                total += abs(T[j][i] - T[j-1][i]);
            total += T[H-1][i];
        }
        
        cout << total << endl;
    }
}

Python

import sys

def surfaceArea(A):
    H, W = len(A), len(A[0])
    area = 2*H*W
    #print("H = {} W = {} area = {}".format(len(A), len(A[0]), area))
    
    for ind in range(H):
        for jnd in range(W):
            #print("i = {} j = {} A = {}".format(ind, jnd, A[ind][jnd]))
            # ind-1, jnd
            if ind-1 >= 0:
                #print("adding part {}".format(max(0, A[ind][jnd] - A[ind-1][jnd])))
                area += max(0, A[ind][jnd] - A[ind-1][jnd])
            else:
                #print("adding full {}".format(A[ind][jnd]))
                area += A[ind][jnd]
            # ind, jnd-1
            if jnd-1 >= 0:
                #print("adding part {}".format(max(0, A[ind][jnd] - A[ind][jnd-1])))
                area += max(0, A[ind][jnd] - A[ind][jnd-1])
            else:
                #print("adding full {}".format(A[ind][jnd]))
                area += A[ind][jnd]
            # ind+1, jnd
            if ind+1 < H:
                #print("adding part {}".format(max(0, A[ind][jnd] - A[ind+1][jnd])))
                area += max(0, A[ind][jnd] - A[ind+1][jnd])
            else:
                #print("adding full {}".format(A[ind][jnd]))
                area += A[ind][jnd]
            # ind, jnd+1
            if jnd+1 < W:
                #print("adding part {}".format(max(0, A[ind][jnd] - A[ind][jnd+1])))
                area += max(0, A[ind][jnd] - A[ind][jnd+1])
            else:
                #print("adding full {}".format(A[ind][jnd]))
                area += A[ind][jnd]
    return area

if __name__ == "__main__":
    H, W = input().strip().split(' ')
    H, W = [int(H), int(W)]
    A = []
    for A_i in range(H):
        A_t = [int(A_temp) for A_temp in input().strip().split(' ')]
        A.append(A_t)
    result = surfaceArea(A)
    print(result)

Java

import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;

public class Solution {

    static int surfaceArea(int[][] A) {
        int cost = 0 ;
        int m = A.length-2;
        int k = A[0].length-2;
        for(int i=1;i<=m;i++){
            for(int j=1;j<=k;j++){
                cost+=2;
                cost+=(A[i-1][j]<A[i][j]?(A[i][j] - A[i-1][j]):0);
                cost+=(A[i+1][j]<A[i][j]?(A[i][j] - A[i+1][j]):0);
                cost+=(A[i][j-1]<A[i][j]?(A[i][j] - A[i][j-1]):0);
                cost+=(A[i][j+1]<A[i][j]?(A[i][j] - A[i][j+1]):0);
                
            }
            
        }
        return cost;
    }

    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int H = in.nextInt();
        int W = in.nextInt();
        int[][] A = new int[H+2][W+2];
        for(int A_i = 1; A_i <= H; A_i++){
            for(int A_j = 1; A_j <= W; A_j++){
                A[A_i][A_j] = in.nextInt();
            }
        }
        int result = surfaceArea(A);
        System.out.println(result);
        in.close();
    }
}

Note: This problem (3D Surface Area) is generated by HackerRank but the solution is provided by CodingBroz. This tutorial is only for Educational and Learning purpose.