Given a string s (containing lowercase letters only), we have to find the minimum cost to construct the given string. The cost can determined using the following operations:
1. Appending a single character cost 1 unit
2. A sub-string of new string(intermediate string) can be appended without any cost
Note* Intermediate string is the string formed so far.
Input : "geks" Output : cost: 4 Explanation: appending 'g' cost 1, string "g" appending 'e' cost 1, string "ge" appending 'k' cost 1, string "gek" appending 's' cost 1, string "geks" Hence, Total cost to construct "geks" is 4 Input : "abab" Output : cost: 2 Explanation: Appending 'a' cost 1, string "a" Appending 'b' cost 1, string "ab" Appending "ab" cost nothing as it is substring of intermediate. Hence, Total cost to construct "abab" is 2
Naive Approach: Check if there is sub-string in the remaining string to be constructed which is also a sub-string in the intermediate string, if there is then append it at no cost and if not then append it at the cost of 1 unit per character.
In the above example when intermediate string was “ab” and we need to construct “abab” then remaining string was “ab”. Hence there is a sub-string in remaining string which is also a sub-string of intermediate string (i.e. “ab”) and therefore cost us nothing.
Better Approach: We will use hashing technique, to maintain that whether we have seen a character or not. If we have seen the character, then there is no cost to append the character and if not, then it cost us 1 unit.
Now in this approach we take one character at a time and not a string. This is because if “ab” is substring of “abab”, so is ‘a’ and ‘b’ alone and hence make no difference.
This also leads us to the conclusion that the cost to construct a string is never more than 26 in case the string contains all the alphabets (a-z).
# Python 3 Program to find minimum cost to
# construct a string
# Initially all characters are un-seen
alphabets = [False for i in range(26)]
# Marking seen characters
for i in range(len(s)):
alphabets[ord(s[i]) – 97] = True
# Count total seen character, and that
# is the cost
count = 0
for i in range(26):
count += 1
# Driver Code
if __name__ == ‘__main__’:
# s is the string that needs to
# be constructed
s = “geeksforgeeks”
print(“Total cost to construct”, s,
# This code is contributed by
Total cost to construct geeksforgeeks is 7
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