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
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")) |
gee geek geeks geeksf geeksfo geeksfor geeksforge geeksforgee geeksforgeek geeksforgeeks 10
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