Python – Frequency of K in sliced String

Given a String, find the frequency of certain characters in the index range.

Input : test_str = ‘geeksforgeeks is best for geeks’, i = 3, j = 9, K = ‘e’ 
Output : 0 
Explanation : No occurrence of ‘e’ between 4th [s] and 9th element

Input : test_str = ‘geeksforgeeks is best for geeks’, i = 0, j = 9, K = ‘e’ 
Output : 2 
Explanation : e present as 2nd and 3rd element. 

Method #1: Using slicing and count()

In this, we perform slicing of required string using slice operation, then count() is used to get count of K in that sliced String.

Python3

# Python3 code to demonstrate working of
# Frequency of K in sliced String
# Using slicing + count()

# initializing strings
test_str = 'geeksforgeeks is best for geeks'

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

# initializing i, j
i, j = 3, 20

# initializing K
K = 'e'

# slicing String
slc = test_str[i : j]

# using count() to get count of K
res = slc.count(K)

# printing result
print("The required Frequency : " + str(res))

Output

The original string is : geeksforgeeks is best for geeks
The required Frequency : 3

Time complexity: O(n), where n is the length of the sliced string, as the code needs to traverse the entire sliced string to count the frequency of K. 
Auxiliary space: O(1), as the code, only uses a constant amount of memory regardless of the length of the input string.

Method #2 : Using Counter() + slicing 

In this, we perform the task of getting count using Counter(), and slicing is used to perform slice of ranges.

Python3

# Python3 code to demonstrate working of
# Frequency of K in sliced String
# Using Counter() + slicing
from collections import Counter

# initializing strings
test_str = 'geeksforgeeks is best for geeks'

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

# initializing i, j
i, j = 3, 20

# initializing K
K = 'e'

# slicing String
slc = test_str[i : j]

# Counter() is used to get count
res = Counter(slc)[K]

# printing result
print("The required Frequency : " + str(res))

Output

The original string is : geeksforgeeks is best for geeks
The required Frequency : 3

Time Complexity: O(n) -> (slicing)
Auxiliary Space: O(n)

Method #3: Using operator.countOf()

Python3

# Python3 code to demonstrate working of
# Frequency of K in sliced String
# Using slicing + operator.countOf()
import operator as op
# initializing strings
test_str = 'geeksforgeeks is best for geeks'

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

# initializing i, j
i, j = 3, 20

# initializing K
K = 'e'

# slicing String
slc = test_str[i : j]

# using operator.countOf() to get count of K
res = op.countOf(slc,K)

# printing result
print("The required Frequency : " + str(res))

Output

The original string is : geeksforgeeks is best for geeks
The required Frequency : 3

Time Complexity: O(N), where n is the length of the given string
Auxiliary Space: O(N)

Method #4 : Using for loop

Python3

# Python3 code to demonstrate working of
# Frequency of K in sliced String

# initializing strings
test_str = 'geeksforgeeks is best for geeks'

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

# initializing i, j
i, j = 3, 20

# initializing K
K = 'e'
res=0
for p in range(i,j):
    if(test_str[p]==K):
        res+=1
# printing result
print("The required Frequency : " + str(res))

Output

The original string is : geeksforgeeks is best for geeks
The required Frequency : 3

Time Complexity: O(N), where n is the length of the given string
Auxiliary Space: O(N)

Method #5: Using regex findall()

This program aims to count the frequency of a given character K in a sliced string slc of the input string test_str. It does so by using the re.findall() method from the re module to find all occurrences of the character K in the sliced string slc, and then assigning the length of the resulting list to the res variable. Finally, the program prints the frequency of character K in the sliced string slc.

Step-by-step approach:

1. Import the re module to work with regular expressions.
2. Initialize the test_str variable with the input string.
3. Print the original string using the print() function.
4. Initialize variables i and j to slice the string. The slice will be from index i up to but not including index j.
5. Initialize the K variable with the character whose frequency we want to count.
6. Slice the input string using string slicing notation and assign it to the slc variable.
7. Use the findall() method from the re module to find all occurrences of the character K in the sliced string slc.
8. Assign the length of the list returned by findall() to the res variable.
9. Print the frequency of character K in the sliced string slc.

Below is the implementation of the above approach:

Python3

import re

# initializing strings
test_str = 'geeksforgeeks is best for geeks'

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

# initializing i, j
i, j = 3, 20

# initializing K
K = 'e'

# slicing String
slc = test_str[i:j]

# using regex findall() to get count of K
res = len(re.findall(K, slc))

# printing result
print("The required Frequency : " + str(res))

Output

The original string is : geeksforgeeks is best for geeks
The required Frequency : 3

Time Complexity: O(n), where n is the length of the sliced string.
Auxiliary Space: O(m), where m is the number of matches found by findall().

Method #6: Using the built-in function filter() with lambda function

Step-by-step approach:

• Define a lambda function that takes a character as input and returns True if it’s equal to K, False otherwise.
• Use the filter() function to filter out the characters in the sliced string that satisfy the condition in the lambda function.
• Find the length of the resulting filtered object, which gives the frequency of K in the sliced string.

Below is the implementation of the above approach:

Python3

# Python3 code to demonstrate working of
# Frequency of K in sliced String

# initializing strings
test_str = 'geeksforgeeks is best for geeks'

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

# initializing i, j
i, j = 3, 20

# initializing K
K = 'e'

# using filter() with lambda function to count frequency
filtered_str = filter(lambda x: x == K, test_str[i:j])
res = len(list(filtered_str))

# printing result
print("The required Frequency : " + str(res))

Output

The original string is : geeksforgeeks is best for geeks
The required Frequency : 3

Time complexity: O(n), where n is the length of the sliced string.
Auxiliary space: O(k), where k is the number of occurrences of K in the sliced string.