Problem 11: Count Character Frequency

Hey everyone! 👋

Today, we're tackling a string manipulation problem: Counting Character Frquency.

The Problem

The goal is to write a function that returns a dictionary with the frequency of each character in a string. The function should be case-insensitive and ignore spaces.

Example:

• char_frequency("hello world") should return {'h': 1, 'e': 1, 'l': 3, 'o': 2, 'w': 1, 'r': 1, 'd': 1}

The Solution

Here is the Python implementation:

def char_frequency(s):
    """
    Returns a dictionary with the frequency of each character in a string.
    Ignores spaces and converts to lowercase.
    """
    freq_dict = {}
    s = s.lower().replace(" ", "")

    for char in s:
        if char in freq_dict:
            freq_dict[char] += 1
        else:
            freq_dict[char] = 1
    return freq_dict

# Test case
print(char_frequency("hello world"))
# Output: {'h': 1, 'e': 1, 'l': 3, 'o': 2, 'w': 1, 'r': 1, 'd': 1}

Code Breakdown

Let's walk through the code line by line:

1. def char_frequency(s):

   • Defines a function named char_frequency that takes one parameter s (a string).

2. freq_dict = {}

   • Creates an empty dictionary called freq_dict to store the characters and their counts.

3. s = s.lower().replace(" ", "")

   • Converts the input string s to lowercase using .lower() to ensure case-insensitivity.
   • Removes all spaces from the string using .replace(" ", "") so they aren't counted.

4. for char in s:

   • Starts a loop that iterates through each character in the processed string s.

5. if char in freq_dict:

   • Checks if the current character char is already a key in the freq_dict.

6. freq_dict[char] += 1

   • If the character is already in the dictionary, increments its count by 1.

7. else: freq_dict[char] = 1

   • If the character is not in the dictionary (it's the first time we've seen it), adds it to the dictionary with a count of 1.

8. return freq_dict

   • Returns the final dictionary containing the character frequencies.

Example Walkthrough with "hello world"

1. Preprocessing:

   • Input: "hello world"
   • Lowercase & Remove Spaces: "helloworld"

2. Iteration:

   • char = 'h': Not in dict → freq_dict = {'h': 1}
   • char = 'e': Not in dict → freq_dict = {'h': 1, 'e': 1}
   • char = 'l': Not in dict → freq_dict = {'h': 1, 'e': 1, 'l': 1}
   • char = 'l': In dict → freq_dict = {'h': 1, 'e': 1, 'l': 2}
   • char = 'o': Not in dict → freq_dict = {'h': 1, 'e': 1, 'l': 2, 'o': 1}
   • char = 'w': Not in dict → freq_dict = {'h': 1, 'e': 1, 'l': 2, 'o': 1, 'w': 1}
   • char = 'o': In dict → freq_dict = {'h': 1, 'e': 1, 'l': 2, 'o': 2, 'w': 1}
   • char = 'r': Not in dict → freq_dict = {'h': 1, 'e': 1, 'l': 2, 'o': 2, 'w': 1, 'r': 1}
   • char = 'l': In dict → freq_dict = {'h': 1, 'e': 1, 'l': 3, 'o': 2, 'w': 1, 'r': 1}
   • char = 'd': Not in dict → freq_dict = {'h': 1, 'e': 1, 'l': 3, 'o': 2, 'w': 1, 'r': 1, 'd': 1}

Final result: {'h': 1, 'e': 1, 'l': 3, 'o': 2, 'w': 1, 'r': 1, 'd': 1}

---

Happy coding! 💻