🧮Beginner-Friendly Guide "Maximum Difference Between Even and Odd Frequency I" LeetCode 3442 (C++ | JavaScript | Python)

#programming #cpp #javascript #python

Hey tech enthusiasts! 👋
Today, we’re diving into a fun and insightful string frequency problem — LeetCode 3442: Maximum Difference Between Even and Odd Frequency I.

This one’s all about counting how often characters appear and performing a bit of clever subtraction. Let’s break it down and solve it cleanly in C++, JavaScript, and Python.

📜 Problem Overview

You're given a string s made of lowercase English letters.

Your task is to compute the maximum difference diff = a₁ - a₂ such that:

a₁ is the frequency of a character with odd frequency.
a₂ is the frequency of a character with even frequency.

📌 Constraints:

3 <= s.length <= 100
s contains at least one character with an odd frequency and one with an even frequency.

✏️ Example Walkthroughs

🧾 Example 1:

Input: s = "aaaaabbc"
Frequencies:
- a: 5 (odd)
- b: 2 (even)
- c: 1 (odd)

maxOdd = 5
minEven = 2
Output: 5 - 2 = 3

🧾 Example 2:

Input: s = "abcabcab"
Frequencies:
- a: 3 (odd)
- b: 3 (odd)
- c: 2 (even)

maxOdd = 3
minEven = 2
Output: 3 - 2 = 1

🧠 Intuition & Strategy

The idea is very simple:

Count the frequency of each character.
Find the maximum odd frequency (maxOdd).
Find the minimum even frequency (minEven).
Return maxOdd - minEven.

📌 We ignore characters that don’t appear at all (frequency = 0).

⚙️ C++ Code

class Solution {
 public:
  int maxDifference(string s) {
    vector<int> count(26);
    int maxOdd = 0;
    int minEven = s.length();

    for (const char c : s)
      ++count[c - 'a'];

    for (const int freq : count) {
      if (freq == 0)
        continue;
      if (freq % 2 == 0)
        minEven = min(minEven, freq);
      else
        maxOdd = max(maxOdd, freq);
    }

    return maxOdd - minEven;
  }
};

🌐 JavaScript Version

var maxDifference = function(s) {
    let count = new Array(26).fill(0);
    for (let char of s) {
        count[char.charCodeAt(0) - 97]++;
    }

    let maxOdd = 0;
    let minEven = s.length;

    for (let freq of count) {
        if (freq === 0) continue;
        if (freq % 2 === 0) {
            minEven = Math.min(minEven, freq);
        } else {
            maxOdd = Math.max(maxOdd, freq);
        }
    }

    return maxOdd - minEven;
};

🐍 Python Version

class Solution:
    def maxDifference(self, s: str) -> int:
        from collections import Counter
        freq = Counter(s)

        max_odd = 0
        min_even = len(s)

        for count in freq.values():
            if count % 2 == 0:
                min_even = min(min_even, count)
            else:
                max_odd = max(max_odd, count)

        return max_odd - min_even

🧪 Test Case Scenarios

Input: "aaaaabbc"
Output: 3  → 'a' (5), 'b' (2), maxOdd = 5, minEven = 2

Input: "abcabcab"
Output: 1  → 'a', 'b' (3 each), 'c' (2), maxOdd = 3, minEven = 2

Input: "aabbccdd"
Output: 2  → All have frequency 2 (even), so add 1 'a' to make it 3 → Now, maxOdd = 3, minEven = 2

Edge Case: "aabccc"
Output: 1 → 'a' (2), 'b' (1), 'c' (3), maxOdd = 3, minEven = 2

🧮 Time & Space Complexity

Time Complexity: O(n)
- One pass to count characters (O(n))
- Second pass through 26 letters (constant)

Space Complexity: O(1)
- Fixed-size array or map for 26 lowercase letters

✨ Wrap Up

This problem is a great example of frequency-based logic in string analysis.
It’s simple, efficient, and perfect for early technical interviews. 🧠

📈 If you're brushing up on data structure patterns like frequency maps or prefix logic, this one's a great addition to your toolkit.

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Happy coding! 🧑‍💻🧠