# Check if a string can be formed from another string using given constraints in Python

Suppose we have two strings lowercase strings s and t. We have to check whether t can be generated from s using following constraints or not −

• Characters of t is there in s for example if there are two 'a' in t, then s should also have two 'a's.

• When any character in t is not in s, check whether the previous two characters (previous two ASCII values) are there in s or not. For example, if 'f' is in t but not in s, then 'd' and 'e' can be used from s to make 'f'.

So, if the input is like s = "pghn" t = "pin", then the output will be True as we can make 'i' from 'g' and 'h' to make "pin".

To solve this, we will follow these steps −

• freq := frequency of each character in s
• for i in range 0 to size of t, do
• if freq[t[i]] is non-zero, then
• freq[t[i]] := freq[t[i]] - 1
• otherwise when freq[first previous character of t[i]] and freq[second previous character of t[i]] is non-zero, then
• decrease freq[first previous character of t[i]] by 1
• decrease freq[second previous character of t[i]] by 1
• otherwise,
• return False
• return True

Let us see the following implementation to get better understanding −

## Example

Live Demo

from collections import defaultdict
def solve(s, t):
freq = defaultdict(lambda:0)
for i in range(0, len(s)):
freq[s[i]] += 1
for i in range(0, len(t)):
if freq[t[i]]:
freq[t[i]] -= 1
elif (freq[chr(ord(t[i]) - 1)] and freq[chr(ord(t[i]) - 2)]):
freq[chr(ord(t[i]) - 1)] -= 1
freq[chr(ord(t[i]) - 2)] -= 1
else:
return False
return True
s = "pghn"
t = "pin"
print(solve(s, t))

## Input

"pghn", "pin"

## Output

True