Program to generate all possible strings by taking the choices in python

Suppose we have a string s of lowercase alphabet characters, other characters like "[", "|", and "]". Here "[a|b|c]" indicates either "a", "b", or "c" can be chosen as a possibility. We have to find a list of strings containing all possible values that s can represent. Here "[]" cannot be nested and may have any number of choices.

So, if the input is like s = "[d|t|l]im[e|s]", then the output will be ['dime', 'dims', 'lime', 'lims', 'time', 'tims']

To solve this, we will follow these steps:

• if s is empty, then
  • return a list with blank string
• n := size of s
• seq := a new list, res := a new list
• Define a function helper() . This will take pos
  • if pos is same as n, then
    • join each element present in seq and insert into res
  • otherwise,
    • if "[" in substring of s[from index pos to end], then
      • start := pos + index of "[" in substring of s[from index pos to end]
      • end := pos + index of "]" in substring of s[from index pos to end]
      • for each option in the substring of s from start to end splitted by "|", do
        • insert s[from index pos to start - 1] at the end of seq
        • insert option at the end of seq
        • helper(end + 1)
        • delete last two elements from seq
  • otherwise,
    • insert s[from index pos to end] at the end of seq
    • helper(n)
    • delete last element from seq
• From the main method do the following:
• helper(0)
• return res in sorted order

Let us see the following implementation to get better understanding:

Example

Live Demo

class Solution:
   def solve(self, s):
      if not s:
         return [""]
      n = len(s)

      def helper(pos):
         if pos == n:
            res.append("".join(seq))
         else:
            if "[" in s[pos:]:
               start = pos + s[pos:].index("[")
               end = pos + s[pos:].index("]")
               for option in s[start + 1 : end].split("|"):
                  seq.append(s[pos:start])
                  seq.append(option)
                  helper(end + 1)
                  seq.pop()
                  seq.pop()
            else:
               seq.append(s[pos:])
               helper(n)
               seq.pop()

         seq = []
         res = []
         helper(0)
         return sorted(res)

ob = Solution()
s = "[d|t|l]im[e|s]"
print(ob.solve(s))

Input

"[d|t|l]im[e|s]"

Output

['dime', 'dims', 'lime', 'lims', 'time', 'tims']

Arnab Chakraborty

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Updated on: 26-Nov-2020

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