Python – Substring Suffix Frequency

Photo of Namachila Namachila Send an email 11 hours ago
0 0 4 minutes read

Given a String and substring, count all the substitutes from string that can be used to complete the substring.

Input : test_str = “Gfg is good . Gfg is good . Gfg is better . Gfg is good .”, substr = “Gfg is” 

Output : {‘good’: 3, ‘better’: 1} 

Explanation : good occurs 3 times as suffix after substring in string hence 3. and so on. 

Input : test_str = “Gfg is good . Gfg is good . Gfg is good . Gfg is good .”, substr = “Gfg is” 

Output : {‘good’: 4} 

Explanation : good occurs 4 times as suffix after substring in string hence 4. and so on.

Method #1 : Using regex() + defaultdict() + loop

This is one of the ways in which this task can be performed. In this, we construct regex for getting all the matching elements for substring. Then check all possible occurrences in String, frequency count using defaultdict().

Python3




# Python3 code to demonstrate working of
# Substring substitutes frequency
# Using regex() + defaultdict() + loop
from collections import defaultdict
import re
 
# initializing string
test_str = "Gfg is good . Gfg is best . Gfg is better . Gfg is good ."
 
# printing original string
print("The original string is : " + str(test_str))
 
# initializing substring
substr = "Gfg is"
 
# initializing regex
temp = re.findall(substr + " (\w+)", test_str, flags = re.IGNORECASE)
 
# adding values to form frequencies
res = defaultdict(int)
for idx in temp:
   res[idx] += 1
 
# printing result
print("Frequency of replacements : " + str(dict(res)))
 



















Output


The original string is : Gfg is good . Gfg is best . Gfg is better . Gfg is good .
Frequency of replacements : {'good': 2, 'best': 1, 'better': 1}




Method #2 : Using Counter() + regex()


This is yet another way in which this task can be performed. In this, we compute elements frequency using Counter().




Python3


















 






 




 



























# Python3 code to demonstrate working of


# Substring substitutes frequency


# Using Counter() + regex()


import re


from collections import Counter


 


# initializing string


test_str = "Gfg is good . Gfg is best . Gfg is better . Gfg is good ."


 


# printing original string


print("The original string is : " + str(test_str))


 


# initializing substring


substr = "Gfg is"


 


# initializing regex


temp = re.findall(substr + " (\w+)", test_str, flags = re.IGNORECASE)


 


# adding values to form frequencies


res = dict(Counter(temp))


 


# printing result


print("Frequency of replacements : " + str(res))










































Output




The original string is : Gfg is good . Gfg is best . Gfg is better . Gfg is good .
Frequency of replacements : {'good': 2, 'best': 1, 'better': 1}




Method #3 : Using split(),find(),count(),strip() methods




Python3


















 






 




 



























# Python3 code to demonstrate working of


# Substring substitutes frequency


 


# initializing string


test_str = "Gfg is good . Gfg is good . Gfg is better . Gfg is good ."


 


# printing original string


print("The original string is : " + str(test_str))


 


# initializing substring


substr = "Gfg is"


 


x=test_str.split(".")


y=[]


for i in x:


    if(i.find(substr)!=-1):


        i=i.strip().split(" ")


        y.append(i[-1])


y1=list(set(y))


d=dict()


for i in y1:


    d[i]=y.count(i)


 


# printing result


print("Frequency of replacements : " + str(d))










































Output


The original string is : Gfg is good . Gfg is good . Gfg is better . Gfg is good .
Frequency of replacements : {'good': 3, 'better': 1}




The Time and Space Complexity for all the methods are the same:


Time Complexity: O(n)


Space Complexity: O(n)


Method #4 : Using operator.countOf() methods




Python3


















 






 




 



























# Python3 code to demonstrate working of


# Substring substitutes frequency


import operator as op


# initializing string


test_str = "Gfg is good . Gfg is good . Gfg is better . Gfg is good ."


 


# printing original string


print("The original string is : " + str(test_str))


 


# initializing substring


substr = "Gfg is"


 


x=test_str.split(".")


y=[]


for i in x:


    if(i.find(substr)!=-1):


        i=i.strip().split(" ")


        y.append(i[-1])


y1=list(set(y))


d=dict()


for i in y1:


    d[i]=op.countOf(y,i)


 


# printing result


print("Frequency of replacements : " + str(d))










































Output


The original string is : Gfg is good . Gfg is good . Gfg is better . Gfg is good .
Frequency of replacements : {'better': 1, 'good': 3}




Time Complexity: O(n)


Auxiliary Space: O(n)


Method #5: Using list comprehension and dictionary comprehension


Step by step approach:


    

  1. Split the original string into a list of sentences using the split() method and the period character as the separator.
    2. 

  2. Filter out the non-matching sentences using a list comprehension that checks if the substring is present in each sentence using the in keyword.
    3. 

  3. Strip the whitespace from each sentence using the strip() method.
    4. 

  4. Extract the last word of each matching sentence using a list comprehension that splits each sentence into words using the split() method and gets the last element using the -1 index.
    5. 

  5. Create a set of the unique last words using the set() function.
    6. 

  6. Count the frequency of each last word using a dictionary comprehension that uses the count() method of the list of last words to count the occurrences of each unique word.
    7. 

  7. Print the result.
    8. 





Python3


















 






 




 



























# initializing string


test_str = "Gfg is good . Gfg is good . Gfg is better . Gfg is good ."


 


# initializing substring


substr = "Gfg is"


 


# split the string into sentences and filter out non-matching sentences


sentences = [s.strip() for s in test_str.split('.') if substr in s]


 


# extract the last word of each matching sentence


last_words = [s.split()[-1] for s in sentences]


 


# count the frequency of each last word


freq = {w: last_words.count(w) for w in set(last_words)}


 


# print the result


print("Frequency of replacements : " + str(freq))










































Output


Frequency of replacements : {'good': 3, 'better': 1}




The time complexity of this approach is O(n), where n is the number of characters in the input string.


The space complexity of this approach is O(m), where m is the number of unique last words in the input string. 








			
		


				

			

				

						
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