Given a string of words, extract the count of words with a specific letters.
Input : test_str = ‘neveropen is best for Lazyroar’, letter = “g”
Output : 2
Explanation : “g” occurs in 2 words.Input : test_str = ‘neveropen is best for Lazyroar’, letter = “s”
Output : 4
Explanation : “s” occurs in 4 words.
Method #1 : Using list comprehension + len() + split()
This is one of the ways in which this task can be performed. In this, we perform task of extracting words from string using split() and loop is used to iterate words to check for letter existence of letter and len() is used to fetch the number of words with letter.
Python3
# Python3 code to demonstrate working of# Count of Words with specific letter# Using list comprehension + len() + split()# initializing stringtest_str = 'neveropen is best for Lazyroar'# printing original stringprint("The original string is : " + str(test_str))# initializing letterletter = "e"# extracting desired count using len()# list comprehension is used as shorthandres = len([ele for ele in test_str.split() if letter in ele])# printing resultprint("Words count : " + str(res)) |
The original string is : neveropen is best for Lazyroar Words count : 3
Time Complexity: O(n)
Auxiliary Space: O(1)
Method #2 : Using filter() + lambda + len() + split()
This is yet another way in which this task can be performed. In this, we perform the task of filtering using filter() + lambda.
Python3
# Python3 code to demonstrate working of# Count of Words with specific letter# Using filter() + lambda + len() + split()# initializing stringtest_str = 'neveropen is best for Lazyroar'# printing original stringprint("The original string is : " + str(test_str))# initializing letterletter = "e"# extracting desired count using len()# filter() used to check for letter existenceres = len(list(filter(lambda ele : letter in ele, test_str.split())))# printing resultprint("Words count : " + str(res)) |
The original string is : neveropen is best for Lazyroar Words count : 3
Time Complexity: O(n)
Auxiliary Space: O(1)
Method #3 : Using len() + for loop
We use len() and for loop to iterate over the string to solve the problem.
Python3
# Python3 code to get count of Words with specific letter# initializing stringtest_str = 'neveropen is best for Lazyroar'# printing original stringprint("The original string is : " + str(test_str))# initializing letterletter = "e"""" As we have to only count one letter per word we have to check if we have checked that word or not and to do so we use variable word_count that become odd if we founded the letter in the word and even when not"""word_count = 0# count of words with specific letterword = 0# For loop to iterate over stringfor i in range(len(test_str)): if word_count % 2 == 0 and test_str[i] == letter: word += 1 word_count = 1 # set word_count an odd number so we know to not check the current word. elif test_str[i] == " ": """ New word starts with " " """ word_count = 0 # set word_count an even number so we know to check the current word.print("Words count :", word)# This code is contributed by Shivesh Kumar Dwivedi |
The original string is : neveropen is best for Lazyroar Words count : 3
Time Complexity: O(n)
Auxiliary Space: O(1)
Method #4 : Using split() and find()
Python3
# Python3 code to demonstrate working of# Count of Words with specific letter# Using split() and find()# initializing stringtest_str = 'neveropen is best for Lazyroar'# printing original stringprint("The original string is : " + str(test_str))# initializing letterletter = "e"# extracting words with specific letterres=0x=test_str.split()for i in x: if(i.find(letter)!=-1): res+=1# printing resultprint("Words count : " + str(res)) |
The original string is : neveropen is best for Lazyroar Words count : 3
Time Complexity: O(n)
Auxiliary Space: O(n)
Method #5 : Using split(),replace() and len() methods
Python3
# Python3 code to demonstrate working of# Count of Words with specific letter# initializing stringtest_str = 'neveropen is best for Lazyroar'# printing original stringprint("The original string is : " + str(test_str))# initializing letterletter = "e"# extracting words with specific letterres = 0x = test_str.split()for i in x: a = i.replace(letter, "") if(len(a) < len(i)): res += 1# printing resultprint("Words count : " + str(res)) |
The original string is : neveropen is best for Lazyroar Words count : 3
The Time and Space Complexity for all the methods are the same
Time Complexity: O(n)
Auxiliary Space: O(n)
Method #6: Using Counter() function
Python3
# Python3 code to demonstrate working of# Count of Words with specific letterfrom collections import Counter# initializing stringtest_str = 'neveropen is best for Lazyroar'# printing original stringprint("The original string is : " + str(test_str))# initializing letterletter = "e"# extracting words with specific letterres = 0x = test_str.split()for i in x: freq = Counter(i) if letter in freq.keys(): res += 1# printing resultprint("Words count : " + str(res)) |
The original string is : neveropen is best for Lazyroar Words count : 3
Time Complexity: O(n)
Auxiliary Space: O(n)
Method #7: Using operator.countOf() method
Python3
# Python3 code to demonstrate working of# Count of Words with specific letterimport operator as op# initializing stringtest_str = 'neveropen is best for Lazyroar'# printing original stringprint("The original string is : " + str(test_str))# initializing letterletter = "e"# extracting words with specific letterres = 0x = test_str.split()for i in x: if op.countOf(i, letter) > 0: res += 1# printing resultprint("Words count : " + str(res)) |
The original string is : neveropen is best for Lazyroar Words count : 3
Time Complexity: O(N)
Auxiliary Space : O(N)
Method #8 : Using split() and set():
Python3
# Python3 code to demonstrate working of# Count of Words with specific letter# Using set + split()# initializing stringtest_str = 'neveropen is best for Lazyroar'# printing original stringprint("The original string is : " + str(test_str))# initializing letterletter = "e"# Convert the input string to a set of unique wordswords = set(test_str.split())# Initialize count to keep track of the number of words that contain the specific lettercount = 0# Iterate over the unique words in the setfor word in words: # Check if the specific letter is in the current word if letter in word: # If the letter is found, increment the count count +=1# printing resultprint("Words count : " + str(count)) |
The original string is : neveropen is best for Lazyroar Words count : 3
Time Complexity: O(N)
Space Complexity : O(1)
Method#9: Using Recursion method.
Python3
# Python3 code to demonstrate working of# Count of Words with specific letter# Using recursive methoddef count_words_with_letter(string, letter, count=0): if not string: return count if letter in string.split()[0]: count += 1 return count_words_with_letter(string[len(string.split()[0]) + 1:], letter, count)# initializing stringtest_str = 'neveropen is best for Lazyroar'# printing original stringprint("The original string is : " + str(test_str))# initializing letterletter = "e"result = count_words_with_letter(test_str, letter)# printing resultprint("Words count:", result)#this code contributed by tvsk |
The original string is : neveropen is best for Lazyroar Words count: 3
Time Complexity: O(n)
Auxiliary Space: O(1)
Method #10 : Using re module
This approach uses the re (regular expression) module in python to find all occurrences of the given letter in the string and counting the number of matches.
Python3
import redef count_words_with_letter(test_str, letter): # Split the string into words words = test_str.split() # Use re.match to check if the specified letter is present in each word count = sum([1 for word in words if re.match(f'.*{letter}.*', word)]) return count# Test the function with the example inputstest_str = 'neveropen is best for Lazyroar'letter = 'e'print(count_words_with_letter(test_str, letter)) |
3
Time Complexity: O(n * m), where n is the number of words in the input string and m is the length of each word on average. This is because the re.match function performs a search operation on each word in the input string, and the length of each search operation is proportional to the length of the word.
Auxiliary Space: O(1)
Method #11 : Using operator.contains() method
Approach
- Split the string and it turns to list
- Initiated a for loop to traverse the list
- Checked whether the letter is present in each element of list using operator.contains()
- If yes increment the result variable
- Display the result variable
Python3
# Python3 code to demonstrate working of# Count of Words with specific letter# Using split() and find()# initializing stringtest_str = 'neveropen is best for Lazyroar'# printing original stringprint("The original string is : " + str(test_str))# initializing letterletter = "e"# extracting words with specific letterres=0x=test_str.split()import operatorfor i in x: if(operator.contains(i,letter)): res+=1# printing resultprint("Words count : " + str(res)) |
The original string is : neveropen is best for Lazyroar Words count : 3
Time Complexity : O(N)
Auxiliary Space : O(1)
