Program to get final position of moving animals when they stops in Python

Python Server Side Programming Programming

Suppose we have a string s that is representing the initial conditions of some animals. Each animal can take one of three values: L, indicates the animal moved to left. R, indicates the animal moved to right. @, indicates the animal is standing still. Animals moving on a direction will pick up other animals unless the animal receives a force from the opposite direction. Then, it will stand still. We have to find the orientation of each animal when the animal stop moving.

So, if the input is like s = "@@L@R@@@@L", then the output will be "LLL@RRRLLL"

To solve this, we will follow these steps −

levels := a list of size same as s and fill with -1
q := a double ended queue
for idx is in range 0 to size of s, do
- if s[idx] is same as "R" or s[idx] is same as "L", then
  - insert (idx, 0, s[idx]) at the end of q
l := a new list of characters of s
while q is not empty, do
- (idx, new_level, dir) := left element of q, and delete it from q
- if levels[idx] is same as -1, then
  - levels[idx] := new_level
  - l[idx] := dir
  - if dir is same as "R" and idx + 1 < size of l , then
    - insert (idx + 1, new_level + 1, dir) at the end of q
  - otherwise when dir is same as "L" and idx - 1 >= 0, then
    - insert (idx - 1, new_level + 1, dir) at the end of q
- otherwise when levels[idx] is same as new_level, then
  - if l[idx] is not same as dir, then
    - l[idx] := "@"
return a string by joining elements of l

Example

Let us see the following implementation to get a better understanding −

Live Demo

from collections import deque
class Solution:
   def solve(self, s):
      levels = [-1 for i in s]
      q = deque()
      for idx in range(len(s)):
         if s[idx] == "R" or s[idx] == "L":
            q.append((idx, 0, s[idx]))
      l = list(s)
      while q:
         idx, new_level, dir = q.popleft()
         if levels[idx] == -1:
            levels[idx] = new_level
            l[idx] = dir
            if dir == "R" and idx + 1 < len(l):
               q.append((idx + 1, new_level + 1, dir))
            elif dir == "L" and idx - 1 >= 0:
               q.append((idx - 1, new_level + 1, dir))
         elif levels[idx] == new_level:
            if l[idx] != dir:
               l[idx] = "@"
      return "".join(l)
ob = Solution()
s = "@@L@R@@@@L"
print(ob.solve(s))

Input

"@@L@R@@@@L"

Output

LLL@RRRLLL

Arnab Chakraborty

Updated on: 22-Dec-2020

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