Find the Surface area of a 3D figure

Given a N*M matrix A[][] representing a 3D figure. The height of the building at is . 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

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 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; }

/div>

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

 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 ?>

Output :

60