Given a string str consisting of lowercase English alphabets, the task is to find the total number of palindromic sub-strings present in the sorted form of str.
Input: str = “acbbd”
All palindromic sub-string in it’s sorted form (“abbcd”) are “a”, “b”, “b”, “bb”, “c” and “d”.
Input: str = “abbabdbd”
Naive approach: One way is to sort the given string and then count the total number of sub-strings present which are palindromes. For finding number of palindromic sub-strings this approach can be used which has time complexity of O(n^2).
Optimized approach: An efficient way is to count the frequency of each character and then for each frequency total number of palindromes will (n*(n+1))/2 as all the palindromic sub-strings of a sorted string will consist of the same character.
For example, palindromic sub-string for the string “aabbbcd” will be “a”, “aa”, …, “bbb”, “c”, … etc. Time complexity for this approach will be O(n).
- Create a hash table for storing the frequencies of each character of the string str.
- Traverse the hash table and for each non-zero frequency add (hash[i] * (hash[i]+1)) / 2 to the sum.
- Print the sum in the end.
Below is the implementation of the above approach:
# Python3 program to find the count of
# palindromic sub-string of a string
# in it’s ascending form
MAX_CHAR = 26
# function to return count of
# palindromic sub-string
n = len (str)
sum = 0
# calculate frequency
hashTable =  * MAX_CHAR
for i in range(n):
ord(‘a’)] += 1
# calculate count of palindromic
for i in range(26) :
sum += (hashTable[i] *
(hashTable[i] + 1) // 2)
# return result
# Driver Code
if __name__ == “__main__”:
str = “ananananddd”
# This code is contributed by ita_c
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