Given two strings S1 and S2(all characters are in lower-case). The task is to check if S2 can be formed from S1 using given constraints:
1. Characters of S2 is there in S1 if there are two ‘a’ in S2, then S1 should have two ‘a’ also.
2. If any character of S2 is not present in S1, check if the previous two ASCII characters are there in S1. e.g., if ‘e’ is there in S2 and not in S1, then ‘c’ and ‘d’ can be used from S1 to make ‘e’.
Note: All characters from S1 can be used only once.
Input: S= abbat, W= cat
‘c’ is formed from ‘a’ and ‘b’, ‘a’ and ‘t’ is present in S1.
Input: S= abbt, W= cat
‘c’ is formed from ‘a’ and ‘b’, but to form the next character
‘a’ in S2, there is no more unused ‘a’ left in S1.
Approach: The above problem can be solved using hashing. The count of all the characters in S1 is stored in a hash-table. Traverse in the string, and check if the character in S2 is there in the hash-table, reduce the count of that particular character in the hash-table. If the character is not there in the hash-table, check if the previous two ASCII characters are there in the hash-table, then reduce the count of the previous two ASCII characters in the hash-table. If all the characters can be formed from S1 using the given constraints, the string S2 can be formed from S1, else it cannot be formed.
Below is the implementation of the above approach:
# Python3 program to Check if a given string
# can be formed from another string using
# given constraints
from collections import defaultdict
# Function to check if S2 can
# be formed of S1
def check(S1, S2):
# length of strings
n1 = len(S1)
n2 = len(S2)
# hash-table to store count
mp = defaultdict(lambda:0)
# store count of each character
for i in range(0, n1):
mp[S1[i]] += 1
# traverse and check for every character
for i in range(0, n2):
# if the character of s2 is
# present in s1
mp[S2[i]] -= 1
# if the character of s2 is not present
# in S1, then check if previous two ASCII
# characters are present in S1
elif (mp[chr(ord(S2[i]) – 1)] and
mp[chr(ord(S2[i]) – 2)]):
mp[chr(ord(S2[i]) – 1)] -= 1
mp[chr(ord(S2[i]) – 2)] -= 1
# Driver Code
if __name__ == “__main__”:
S1 = “abbat”
S2 = “cat”
# Calling function to check
if check(S1, S2):
# This code is contributed by Rituraj Jain
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Improved By : rituraj_jain