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Python | Count all prefixes in a given string with the highest frequency

| |

Take two alphabets from the user and compare them. Prefixes in which the alphabet specified first has a higher frequency than the second alphabet, such prefixes are printed, otherwise 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 generated prefixes ... Then check the alphabet with a higher frequency than the second alphabet. If no such case is found, the result is 0 prefixes.

The implementation is as follows:

# Python program for counting all
# prefixes in this line with
# highest frequency

 
# Function for printing prefixes

def prefix (string1, alphabet1, alphabet2):

  count = 0

non_empty_string = ""

 

  string2 = list (string1)

 

  # Loop to iterate length

# line and print prefixes

# and number of request 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 any required prefixes

print (non_empty_string)

 

# number of increments

count + = 1

 

# returns counter

  # required prefixes

  return (count)

 
Driver code

print (prefix ( "pythonengineering" , "e" , "g" ))

Output:

 gee geek geeks geeksf geeksfo geeksfor geeksforge geeksforgee geeksforgeek pythonengineering 10 

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