Problem Description

You are given a string s consisting of lowercase English letters. Your task is to find the maximum difference between the frequency of two characters in the string such that:

• One of the characters has an even frequency in the string.
• The other character has an odd frequency in the string.

Return the maximum difference, calculated as the frequency of the character with an odd frequency minus the frequency of the character with an even frequency.

Intuition

The goal is to identify two characters within the string such that the difference between their frequencies maximizes the condition where one has an odd frequency and the other has an even frequency.

To solve this, we can use a frequency counter to tally the occurrences of each character in s. The strategy involves two main steps:

1. Tracking Frequencies: First, collect the frequencies of each character using a hash table or an array. This helps identify how many times each character appears in the string.

2. Determine Maximum and Minimum Frequencies: Next, iterate through the frequency values to find:

   • The maximum frequency among characters that appear an odd number of times (a).
   • The minimum frequency among characters that appear an even number of times (b).

Finally, return the result of the subtraction a - b, representing the maximum frequency difference as per the defined criteria.

Solution Approach

To find the maximum difference between the frequencies of two characters in s such that one has an odd frequency and the other has an even frequency, the following approach can be used:

1. Use a Frequency Counter: Utilize a hash table (specifically, Python's Counter from the collections module) to count the occurrences of each character in the string s. This helps track how many times each character appears.

2. Initialize Tracking Variables:

   • Let a represent the maximum frequency of characters with odd counts. Start with 0 as the initial value since it needs to be maximized.
   • Let b represent the minimum frequency of characters with even counts. Begin with positive infinity (inf) as the initial value because it needs to be minimized.

3. Iterate Through Frequencies:

   • For each frequency value v from the frequency counter:
     • If v is odd, update a to be the maximum of its current value and v.
     • If v is even, update b to be the minimum of its current value and v.

4. Compute the Result: Calculate the difference a - b and return it. This difference represents the maximum difference between character frequencies with the desired properties.

The code implements this approach effectively:

class Solution:
    def maxDifference(self, s: str) -> int:
        cnt = Counter(s)  # Count frequencies of each character
        a, b = 0, inf  # Initialize max odd frequency and min even frequency
        for v in cnt.values():
            if v % 2:
                a = max(a, v)  # Update max odd frequency
            else:
                b = min(b, v)  # Update min even frequency
        return a - b  # Return the calculated difference

This method efficiently uses counting and comparison operations to achieve the desired outcome.

Example Walkthrough

Let's take a small example string s = "abacacbba". Our goal is to find the maximum difference between the frequency of two characters such that one has an even frequency and the other has an odd frequency.

1. Use a Frequency Counter: Count the occurrences of each character.

   • 'a': 4 times
   • 'b': 3 times
   • 'c': 2 times

2. Initialize Tracking Variables:

   • a = 0: This will track the maximum frequency among characters with odd frequencies.
   • b = inf: This will track the minimum frequency among characters with even frequencies.

3. Iterate Through Frequencies:

   • For 'a' with a frequency of 4 (even), set b = min(inf, 4), so b = 4.
   • For 'b' with a frequency of 3 (odd), set a = max(0, 3), so a = 3.
   • For 'c' with a frequency of 2 (even), set b = min(4, 2), so b = 2.

4. Compute the Result: The maximum difference is a - b = 3 - 2 = 1.

Thus, for the string s = "abacacbba", the maximum difference between the frequencies of a character with an odd frequency and a character with an even frequency is 1.

Solution Implementation

Time and Space Complexity

The time complexity of the code is O(n), where n is the length of the string s. This is because the code involves iterating over the string once to count the occurrences of each character, and then iterating over the counted values.

The space complexity is O(|Σ|), where |Σ| is the size of the character set, which is 26 for lowercase English letters. The space is used to store the count of each character.

Learn more about how to find time and space complexity quickly.