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# 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

- if (t[j] is not same as s[i]) and t[j] is not same as '#', then

- for j in range 0 to size of t - 1, do

- if s[i] is not same as '#', then
- return True

## Example

Let us see the following implementation to get better understanding −

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

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