Given a N*M matrix A[][] representing a 3D figure. The height of the building at
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 :
// 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 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 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# 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 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 ?> |
<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)