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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
Updated on: 18-Jan-2021
123 Views
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