The Rainy Season

hacker_earth_python.py

'''
The solution for Hackerearth problem: The Rainy Season.
https://www.hackerearth.com/problem/algorithm/bfs-waali-7409c2ca-c1be890b/description/

The problem is not hard, but due to the limitation of recursion limit
in Python, we sometimes need to make different approaches.

Here, I've tried an iterative approach.
'''
# Write your code here
from collections import deque

m, n = map(int, input().split())
data = []
for _ in range(m):
    data.append(list(input()))

def find_volume(i, j):
    dq = deque()
    dq.append((i, j))

    res = 0

    while(dq):
        x, y = dq.pop()
        if x < 0 or y < 0 or x >= m or y >= n:
            # Water will flow-out
            res += float('inf')

        elif data[x][y] != '*':
            # The cell is not a tower
            data[x][y] = '*' # make it visited

            dq.append((x,y+1))
            dq.append((x,y-1))
            dq.append((x+1,y))
            dq.append((x-1,y))
            
            res += 1
    
    return res if res != float('inf') else 0

total_volume = 0

for i in range(m):
    for j in range(n):
        if(data[i][j] == '*'): continue
        total_volume += find_volume(i, j)

print(total_volume)

original_recursive.py

# Write your code here

m, n = map(int, input().split())
data = []
for _ in range(m):
  data.append(list(input()))

def find_volume(i, j):
  if i < 0 or j < 0 or i >= m or y >= n:
    return float('inf')
  res = 0
  if data[i][j] == '*':
    data[i][j] = '*'
    
    res += find_volume((i, j+1))
    res += find_volume((i, j-1))
    res += find_volume((i+1, j))
    res += find_volume((i-1, j))
  return res

total_volume = 0

for i in range(m):
  for j in range(n):
    if(data[i][j] == '*'): continue
    curr_volume = find_volume(i, j)
    total_volume += (curr_volume if curr_volume != float('inf') else 0)
    
print(total_volume)