Syntax error Check if it is possible to convert one string into another with given constraints in Python

Check if it is possible to convert one string into another with given constraints in Python

Suppose we have two strings s and t with only three characters 'A', 'B' and '#'. We have to check whether it is possible to convert s into t by performing these operations on s.

  • 'A' can move only to the left hand side
  • 'B' can move only to the right hand side
  • Neither 'A' nor 'B' can cross each other

So, if the input is like s = "##AB##B" t = "A###B#B", then the output will be True as in s A can move easily to the left most position, and middle B can move one step to the right

To solve this, we will follow these steps −

  • s := a list by taking characters from s
  • t := a list by taking characters from t
  • if size of s is not same as size of t, then
    • return False
  • if count of 'A' in s and t are different or count of 'B' in s and t are different or, then
    • return False
  • for i in range 0 to size of s - 1, do
    • if s[i] is not same as '#', then
      • for j in range 0 to size of t - 1, do
        • if (t[j] is not same as s[i]) and t[j] is not same as '#', then
          • return False
        • if t[j] is same as s[i], then
          • t[j] := '#'
          • if s[i] is same as 'A' and i < j, then
            • return False
          • if s[i] is same as 'B' and i > j, then
            • return False
          • come out from loop
  • return True

Example

Let us see the following implementation to get better understanding −

 Live Demo

def solve(s, t):
   s = list(s)
   t = list(t)
   if len(s) != len(t):
      return False
   if s.count('A') != t.count('A') or s.count('B') != t.count('B'):
      return False
   for i in range(len(s)):
      if s[i] != '#':
         for j in range(len(t)):
            if (t[j] != s[i]) and t[j] != '#':
               return False
            if t[j] == s[i]:
               t[j] = '#'
               if s[i] == 'A' and i < j:
                  return False
               if s[i] == 'B' and i > j:
                  return False
               break
   return True
s = "##AB##B"
t = "A###B#B"
print (solve(s, t))

Input

"##AB##B", "A###B#B"

Output

True
Updated on: 2021-01-18T13:00:36+05:30

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