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# Python | Count all prefixes in given string with greatest frequency

• Difficulty Level : Medium
• Last Updated : 06 Jun, 2018

Given a string, print and count all prefixes in which first alphabet has greater frequency than second alphabet.

Take two alphabets from the user and compare them. The prefixes in which the alphabet given first has greater frequency than the second alphabet, such prefixes are printed, else the result will be 0.

Examples :

```Input : string1 = "geek",
alphabet1 = "e", alphabet2 = "k"
Output :
ge
gee
geek
3

Input : string1 = "geek",
alphabet1 = "k", alphabet2 = "e"
Output :
0
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach : Take an empty string to store the string values of all the prefixes formed. Then check for the alphabet with greater frequency than the second alphabet. If no such case is found then the result will be 0 prefixes.

Below is the implementation :

 `# Python program to Count all``# prefixes in given string with``# greatest frequency`` ` `# Function to print the prefixes``def` `prefix(string1, alphabet1, alphabet2):``    ``count ``=` `0``    ``non_empty_string ``=` `""``     ` `    ``string2 ``=` `list``(string1)``     ` `    ``# Loop for iterating the length of``    ``# the string and print the prefixes ``    ``# and the count of query prefixes.``    ``for` `i ``in` `range``(``0``, ``len``(string2)):``        ``non_empty_string ``=` `non_empty_string ``+` `(string2[i])``         ` `        ``if` `(non_empty_string.count(alphabet1) >``            ``non_empty_string.count(alphabet2)):``                 ` `            ``# prints all required prefixes``            ``print``(non_empty_string)``             ` `            ``# increment count``            ``count ``+``=` `1``             ` `    ``# returns count of the``    ``# required prefixes``    ``return``(count)``     ` `# Driver Code``print``(prefix(``"geeksforgeeks"``, ``"e"``, ``"g"``))`
Output :
```gee
geek
geeks
geeksf
geeksfo
geeksfor
geeksforge
geeksforgee
geeksforgeek
geeksforgeeks
10
```

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