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Find the Surface area of a 3D figure
  • Difficulty Level : Easy
  • Last Updated : 12 Apr, 2021

Given a N*M matrix A[][] representing a 3D figure. The height of the building at (i, j)  is A[i][j]  . Find the surface area of the figure.
Examples : 
 

Input : N = 1, M = 1   A[][] = { {1} }
Output : 6

Explanation : 
The total surface area is 6 i.e 6 side of 
the figure and each are of height 1.

Input : N = 3, M = 3   A[][] = { {1, 3, 4},
                                 {2, 2, 3},
                                 {1, 2, 4} }
Output : 60

 

Approach : To find the surface area we need to consider the contribution of all the six sides of the given 3D figure. We will solve the questions in part to make it easy. The base of the Figure will always contribute N*M to the total surface area of the figure, and same N*M area will be contributed by the top of the figure. Now, to calculate the area contributed by the walls, we will take out the absolute difference between the height of two adjacent wall. The difference will be the contribution in the total surface area.
Below is the implementation of the above idea : 
 

C++




// CPP program to find the Surface area of a 3D figure
#include <bits/stdc++.h>
using namespace std;
 
// Declaring the size of the matrix
const int M = 3;
const int N = 3;
 
// Absolute Difference between the height of
// two consecutive blocks
int contribution_height(int current, int previous)
{
    return abs(current - previous);
}
 
// Function To calculate the Total surfaceArea.
int surfaceArea(int A[N][M])
{
    int ans = 0;
 
    // Traversing the matrix.
    for (int i = 0; i < N; i++) {
        for (int j = 0; j < M; j++) {
 
            /* If we are traveling the topmost row in the
            matrix, we declare the wall above it as 0
            as there is no wall above it. */
            int up = 0;
 
            /* If we are traveling the leftmost column in the
            matrix, we declare the wall left to it as 0
            as there is no wall left it. */
            int left = 0;
 
            // If its not the topmost row
            if (i > 0)
                up = A[i - 1][j];
 
            // If its not the leftmost column
            if (j > 0)
                left = A[i][j - 1];
 
            // Summing up the contribution of by
            // the current block
            ans += contribution_height(A[i][j], up)
                    + contribution_height(A[i][j], left);
 
            /* If its the rightmost block of the matrix
               it will contribute area equal to its height
               as a wall on the right of the figure */
            if (i == N - 1)
                ans += A[i][j];
 
            /* If its the lowest block of the matrix it will
               contribute area equal to its height as a wall
               on the bottom of the figure */
            if (j == M - 1)
                ans += A[i][j];
        }
    }
 
    // Adding the contribution by the base and top of the figure
    ans += N * M * 2;
    return ans;
}
 
// Driver program
int main()
{
    int A[N][M] = { { 1, 3, 4 },
                    { 2, 2, 3 },
                    { 1, 2, 4 } };
    cout << surfaceArea(A) << endl;
    return 0;
}

Java




// Java program to find the Surface
// area of a 3D figure
 
class GFG
{
    // Declaring the size of the matrix
    static final int M=3;
    static final int N=3;
     
    // Absolute Difference between the height of
    // two consecutive blocks
    static int contribution_height(int current, int previous)
    {
        return Math.abs(current - previous);
    }
     
    // Function To calculate the Total surfaceArea.
    static int surfaceArea(int A[][])
    {
        int ans = 0;
     
        // Traversing the matrix.
        for (int i = 0; i < N; i++)
        {
            for (int j = 0; j < M; j++) {
     
                /* If we are traveling the topmost
                row in the matrix, we declare the
                wall above it as 0 as there is no
                wall above it. */
                int up = 0;
     
                /* If we are traveling the leftmost
                column in the matrix, we declare the
                wall left to it as 0as there is no
                wall left it. */
                int left = 0;
     
                // If its not the topmost row
                if (i > 0)
                    up = A[i - 1][j];
     
                // If its not the leftmost column
                if (j > 0)
                    left = A[i][j - 1];
     
                // Summing up the contribution of by
                // the current block
                ans += contribution_height(A[i][j], up)
                       + contribution_height(A[i][j], left);
     
                /* If its the rightmost block of the matrix
                it will contribute area equal to its height
                as a wall on the right of the figure */
                if (i == N - 1)
                    ans += A[i][j];
     
                /* If its the lowest block of the
                matrix it will contribute area equal
                to its height as a wall on
                 the bottom of the figure */
                if (j == M - 1)
                    ans += A[i][j];
            }
        }
     
        // Adding the contribution by
        // the base and top of the figure
        ans += N * M * 2;
        return ans;
    }
     
    // Driver code
    public static void main (String[] args)
    {
        int A[][] = {{ 1, 3, 4 },
                     { 2, 2, 3 },
                     { 1, 2, 4 } };
        System.out.println(surfaceArea(A));
    }
}
 
// This code is contributed By Anant Agarwal.

Python3




# Python3 program to find the
# Surface area of a 3D figure
 
 
# Declaring the size
# of the matrix
M = 3;
N = 3;
 
# Absolute Difference
# between the height of
# two consecutive blocks
def contribution_height(current, previous):
    return abs(current - previous);
 
# Function To calculate
# the Total surfaceArea.
def surfaceArea(A):
    ans = 0;
 
    # Traversing the matrix.
    for i in range(N):
        for j in range(M):
 
            # If we are traveling the
            # topmost row in the matrix,
            # we declare the wall above it
            # as 0 as there is no wall
            # above it.
            up = 0;
 
            # If we are traveling the
            # leftmost column in the
            # matrix, we declare the wall
            # left to it as 0 as there is
            # no wall left it.
            left = 0;
 
            # If its not the topmost row
            if (i > 0):
                up = A[i - 1][j];
 
            # If its not the
            # leftmost column
            if (j > 0):
                left = A[i][j - 1];
 
            # Summing up the
            # contribution of by
            # the current block
            ans += contribution_height(A[i][j], up)+contribution_height(A[i][j], left);
             
            # If its the rightmost block
            # of the matrix it will contribute
            # area equal to its height as a
            # wall on the right of the figure */
            if (i == N - 1):
                ans += A[i][j];
 
            # If its the lowest block
            # of the matrix it will
            # contribute area equal to
            # its height as a wall on
            # the bottom of the figure
            if (j == M - 1):
                ans += A[i][j];
 
    # Adding the contribution by
    # the base and top of the figure
    ans += N * M * 2;
    return ans;
 
# Driver Code
A = [[1, 3, 4],[2, 2, 3],[1, 2, 4]];
print(surfaceArea(A));
 
# This code is contributed By mits

C#




// C# program to find the
// Surface area of a 3D figure
using System;
 
class GFG
{
    // Declaring the size of the matrix
    static int M=3;
    static int N=3;
     
    // Absolute Difference between the
    // height of two consecutive blocks
    static int contribution_height(int current, int previous)
    {
        return Math.Abs(current - previous);
    }
     
    // Function To calculate the
    // Total surfaceArea.
    static int surfaceArea(int [,]A)
    {
        int ans = 0;
     
    // Traversing the matrix.
        for (int i = 0; i < N; i++)
        {
            for (int j = 0; j < M; j++) {
     
    // If we are traveling the topmost
    // row in the matrix, we declare the
    // wall above it as 0 as there is no
    // wall above it.
                int up = 0;
     
    // If we are traveling the leftmost
    // column in the matrix, we declare
    // the wall left to it as 0as there
    // is no wall left it.
                int left = 0;
     
    // If its not the topmost row
                if (i > 0)
                    up = A[i - 1,j];
     
    // If its not the leftmost column
                if (j > 0)
                    left = A[i,j - 1];
     
    // Summing up the contribution 
    // of by the current block
            ans += contribution_height(A[i,j], up)
                + contribution_height(A[i,j], left);
     
    // If its the rightmost block of the
    // matrix it will contribute area equal
    // to its height as a wall on the right
    // of the figure
                if (i == N - 1)
                    ans += A[i,j];
     
    // If its the lowest block of the
    // matrix it will contribute area 
    // equal to its height as a wall
    // on the bottom of the figure
                if (j == M - 1)
                    ans += A[i,j];
            }
        }
     
    // Adding the contribution by the
    // base and top of the figure
        ans += N * M * 2;
        return ans;
    }
     
    // Driver code
    public static void Main ()
    {
        int [,]A = {{ 1, 3, 4 },
                    { 2, 2, 3 },
                    { 1, 2, 4 } };
        Console.WriteLine(surfaceArea(A));
    }
}
 
// This code is contributed By vt_m.

PHP




<?php
// PHP program to find the
// Surface area of a 3D figure
 
 
// Declaring the size
// of the matrix
$M = 3;
$N = 3;
 
// Absolute Difference
// between the height of
// two consecutive blocks
function contribution_height($current,
                             $previous)
{
    return abs($current - $previous);
}
 
// Function To calculate
// the Total surfaceArea.
function surfaceArea($A)
{
    global $M;
    global $N;
    $ans = 0;
 
    // Traversing the matrix.
    for ($i = 0; $i < $N; $i++)
    {
        for ($j = 0; $j < $M; $j++)
        {
 
            /* If we are traveling the
            topmost row in the matrix,
            we declare the wall above it
            as 0 as there is no wall
            above it. */
            $up = 0;
 
            /* If we are traveling the
            leftmost column in the
            matrix, we declare the wall
            left to it as 0 as there is
            no wall left it. */
            $left = 0;
 
            // If its not the topmost row
            if ($i > 0)
                $up = $A[$i - 1][$j];
 
            // If its not the
            // leftmost column
            if ($j > 0)
                $left = $A[$i][$j - 1];
 
            // Summing up the
            // contribution of by
            // the current block
            $ans += contribution_height($A[$i][$j], $up) +
                    contribution_height($A[$i][$j], $left);
             
            /* If its the rightmost block
            of the matrix it will contribute
            area equal to its height as a
            wall on the right of the figure */
            if ($i == $N - 1)
                $ans += $A[$i][$j];
 
            /* If its the lowest block
               of the matrix it will
               contribute area equal to
               its height as a wall on
               the bottom of the figure */
            if ($j == $M - 1)
                $ans += $A[$i][$j];
        }
    }
 
    // Adding the contribution by
    // the base and top of the figure
    $ans += $N * $M * 2;
    return $ans;
}
 
// Driver Code
$A = array(array(1, 3, 4),
           array(2, 2, 3),
           array(1, 2, 4));
echo surfaceArea($A);
 
// This code is contributed By mits
?>

Javascript




<script>
 
// JavaScript program to find the Surface
// area of a 3D figure
 
// Declaring the size of the matrix
    let M=3;
    let N=3;
       
    // Absolute Difference between the height of
    // two consecutive blocks
    function contribution_height(current, previous)
    {
        return Math.abs(current - previous);
    }
       
    // Function To calculate the Total surfaceArea.
    function surfaceArea( A)
    {
        let ans = 0;
       
        // Traversing the matrix.
        for (let i = 0; i < N; i++)
        {
            for (let j = 0; j < M; j++) {
       
                /* If we are traveling the topmost
                row in the matrix, we declare the
                wall above it as 0 as there is no
                wall above it. */
                let up = 0;
       
                /* If we are traveling the leftmost
                column in the matrix, we declare the
                wall left to it as 0as there is no
                wall left it. */
                let left = 0;
       
                // If its not the topmost row
                if (i > 0)
                    up = A[i - 1][j];
       
                // If its not the leftmost column
                if (j > 0)
                    left = A[i][j - 1];
       
                // Summing up the contribution of by
                // the current block
                ans += contribution_height(A[i][j], up)
                       + contribution_height(A[i][j], left);
       
                /* If its the rightmost block of the matrix
                it will contribute area equal to its height
                as a wall on the right of the figure */
                if (i == N - 1)
                    ans += A[i][j];
       
                /* If its the lowest block of the
                matrix it will contribute area equal
                to its height as a wall on
                 the bottom of the figure */
                if (j == M - 1)
                    ans += A[i][j];
            }
        }
       
        // Adding the contribution by
        // the base and top of the figure
        ans += N * M * 2;
        return ans;
    }
 
// Driver code
         
        let A = [[ 1, 3, 4 ],
                     [ 2, 2, 3 ],
                     [ 1, 2, 4 ]];
        document.write(surfaceArea(A));
                   
</script>

Output : 

60

 

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