Suppose we have a N*M matrix A, this is the representation of 3D figure. The height of the building at point (i, j) is A[i][j]. We have to find the surface area of the figure.
So, if the input is like N = 3, M = 3, A = [[1, 4, 5],[3, 3, 4],[1, 3, 5]], then the output will be 72.
To solve this, we will follow these steps −
res := 0
for i in range 0 to N, do
for j in range 0 to M, do
up_side := 0
left_side := 0
if i > 0, then
up_side := array[i - 1, j]
if j > 0, then
left_side := array[i, j - 1]
res := res + |array[i][j] - up_side| + |array[i][j] - left_side|
if i is same as N - 1, then
res := res + array[i, j]
if j is same as M - 1, then
res := res + array[i, j]
res := res + N * M * 2
return res
Example
Let us see the following implementation to get better understanding −
M = 3 N = 3 def get_surface_area(array): res = 0 for i in range(N): for j in range(M): up_side = 0 left_side = 0 if (i > 0): up_side = array[i - 1][j] if (j > 0): left_side = array[i][j - 1] res += abs(array[i][j] - up_side)+abs(array[i][j] - left_side) if (i == N - 1): res += array[i][j] if (j == M - 1): res += array[i][j] res += N * M * 2 return res array = [[1, 4, 5],[3, 3, 4],[1, 3, 5]] print(get_surface_area(array))
Input
[[1, 4, 5],[3, 3, 4],[1, 3, 5]]
Output
72