Count all the occurrences of the prefix string

Question

Find the alphabet present at that location and determine the no of occurrences of the same alphabet preceding that location n .

Input:

length = 9

string = "abababbsa"

n = 9

Output:

3

Explanation:

Find the alphabet at nth position 9 ie a . Count all the occurrences before that index which is 3 .

Code:

length = int(input())
string = list(input())
n = int(input())
str1 = string[n-1]
print(string[0:n-1].count(str1))

The above code gives TLE. How can I optimize this?

Answer 1

Well It becomes easy if you use re ( regex )

import re
length = int(input())
my_string = input('Enter your string')
n = int(input())
print(len(re.findall(my_string[n-1],my_string[:n-1])))

So what is happening is we are finding all occurences of the character before n by slicing the string at n-1 and just getting the length of the list ( all occurences ).

Answer 2

One way to optimize is to use string as usual in spite of converting it into a list:

string = input()

The complexity would now be O(n) where n is the input position, whereas, in your case, because the string was explicitly converted to list, the complexity was O(length of the string).