Given a N*M matrix A[][] representing a 3D figure. The height of the building at 
is ![A[i][j]](https://geeksforgeeks.org/wp-content/uploads/2023/10/rubhabha_quicklatex.com-85663c31349dfd7dfc0beb1e3219e0fa_l3.png "Rendered by QuickLaTeX.com")
. 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 matrixconst int M = 3;const int N = 3;// Absolute Difference between the height of// two consecutive blocksint 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 programint 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 figureclass 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 matrixM = 3;N = 3;# Absolute Difference # between the height of# two consecutive blocksdef 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 CodeA = [[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 figureusing 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 blocksfunction 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
Time Complexity: O(N*M), as the above code is been iterating over through two loops.
Auxiliary Space: O(1)
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
