LeetCode #3445 — HARD

Maximum Difference Between Even and Odd Frequency II

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

The Problem

Problem Statement

You are given a string s and an integer k. Your task is to find the maximum difference between the frequency of two characters, freq[a] - freq[b], in a substring subs of s, such that:

subs has a size of at least k.
Character a has an odd frequency in subs.
Character b has a non-zero even frequency in subs.

Return the maximum difference.

Note that subs can contain more than 2 distinct characters.

Example 1:

Input: s = "12233", k = 4

Output: -1

Explanation:

For the substring "12233", the frequency of '1' is 1 and the frequency of '3' is 2. The difference is 1 - 2 = -1.

Example 2:

Input: s = "1122211", k = 3

Output: 1

Explanation:

For the substring "11222", the frequency of '2' is 3 and the frequency of '1' is 2. The difference is 3 - 2 = 1.

Example 3:

Input: s = "110", k = 3

Output: -1

Constraints:

3 <= s.length <= 3 * 10⁴
s consists only of digits '0' to '4'.
The input is generated that at least one substring has a character with an even frequency and a character with an odd frequency.
1 <= k <= s.length

Step 01

Brute Force Baseline

Problem summary: You are given a string s and an integer k. Your task is to find the maximum difference between the frequency of two characters, freq[a] - freq[b], in a substring subs of s, such that: subs has a size of at least k. Character a has an odd frequency in subs. Character b has a non-zero even frequency in subs. Return the maximum difference. Note that subs can contain more than 2 distinct characters.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Sliding Window

Example 1

"12233"
4

Example 2

"1122211"
3

Example 3

"110"
3

Core Insight

What unlocks the optimal approach

Fix the two characters.
Use prefix sum (maintain 2 characters' parities as status).

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Largest constraint values

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #3445: Maximum Difference Between Even and Odd Frequency II
class Solution {
    public int maxDifference(String S, int k) {
        char[] s = S.toCharArray();
        int n = s.length;
        final int inf = Integer.MAX_VALUE / 2;
        int ans = -inf;
        for (int a = 0; a < 5; ++a) {
            for (int b = 0; b < 5; ++b) {
                if (a == b) {
                    continue;
                }
                int curA = 0, curB = 0;
                int preA = 0, preB = 0;
                int[][] t = {{inf, inf}, {inf, inf}};
                for (int l = -1, r = 0; r < n; ++r) {
                    curA += s[r] == '0' + a ? 1 : 0;
                    curB += s[r] == '0' + b ? 1 : 0;
                    while (r - l >= k && curB - preB >= 2) {
                        t[preA & 1][preB & 1] = Math.min(t[preA & 1][preB & 1], preA - preB);
                        ++l;
                        preA += s[l] == '0' + a ? 1 : 0;
                        preB += s[l] == '0' + b ? 1 : 0;
                    }
                    ans = Math.max(ans, curA - curB - t[curA & 1 ^ 1][curB & 1]);
                }
            }
        }
        return ans;
    }
}

// Accepted solution for LeetCode #3445: Maximum Difference Between Even and Odd Frequency II
func maxDifference(s string, k int) int {
	n := len(s)
	inf := math.MaxInt32 / 2
	ans := -inf

	for a := 0; a < 5; a++ {
		for b := 0; b < 5; b++ {
			if a == b {
				continue
			}
			curA, curB := 0, 0
			preA, preB := 0, 0
			t := [2][2]int{{inf, inf}, {inf, inf}}
			l := -1

			for r := 0; r < n; r++ {
				if s[r] == byte('0'+a) {
					curA++
				}
				if s[r] == byte('0'+b) {
					curB++
				}

				for r-l >= k && curB-preB >= 2 {
					t[preA&1][preB&1] = min(t[preA&1][preB&1], preA-preB)
					l++
					if s[l] == byte('0'+a) {
						preA++
					}
					if s[l] == byte('0'+b) {
						preB++
					}
				}

				ans = max(ans, curA-curB-t[curA&1^1][curB&1])
			}
		}
	}

	return ans
}

# Accepted solution for LeetCode #3445: Maximum Difference Between Even and Odd Frequency II
class Solution:
    def maxDifference(self, S: str, k: int) -> int:
        s = list(map(int, S))
        ans = -inf
        for a in range(5):
            for b in range(5):
                if a == b:
                    continue
                curA = curB = 0
                preA = preB = 0
                t = [[inf, inf], [inf, inf]]
                l = -1
                for r, x in enumerate(s):
                    curA += x == a
                    curB += x == b
                    while r - l >= k and curB - preB >= 2:
                        t[preA & 1][preB & 1] = min(t[preA & 1][preB & 1], preA - preB)
                        l += 1
                        preA += s[l] == a
                        preB += s[l] == b
                    ans = max(ans, curA - curB - t[curA & 1 ^ 1][curB & 1])
        return ans

// Accepted solution for LeetCode #3445: Maximum Difference Between Even and Odd Frequency II
use std::cmp::{max, min};
use std::i32::{MAX, MIN};

impl Solution {
    pub fn max_difference(S: String, k: i32) -> i32 {
        let s: Vec<usize> = S
            .chars()
            .map(|c| c.to_digit(10).unwrap() as usize)
            .collect();
        let k = k as usize;
        let mut ans = MIN;

        for a in 0..5 {
            for b in 0..5 {
                if a == b {
                    continue;
                }

                let mut curA = 0;
                let mut curB = 0;
                let mut preA = 0;
                let mut preB = 0;
                let mut t = [[MAX; 2]; 2];
                let mut l: isize = -1;

                for (r, &x) in s.iter().enumerate() {
                    curA += (x == a) as i32;
                    curB += (x == b) as i32;

                    while (r as isize - l) as usize >= k && curB - preB >= 2 {
                        let i = (preA & 1) as usize;
                        let j = (preB & 1) as usize;
                        t[i][j] = min(t[i][j], preA - preB);
                        l += 1;
                        if l >= 0 {
                            preA += (s[l as usize] == a) as i32;
                            preB += (s[l as usize] == b) as i32;
                        }
                    }

                    let i = (curA & 1 ^ 1) as usize;
                    let j = (curB & 1) as usize;
                    if t[i][j] != MAX {
                        ans = max(ans, curA - curB - t[i][j]);
                    }
                }
            }
        }

        ans
    }
}

// Accepted solution for LeetCode #3445: Maximum Difference Between Even and Odd Frequency II
function maxDifference(S: string, k: number): number {
    const s = S.split('').map(Number);
    let ans = -Infinity;
    for (let a = 0; a < 5; a++) {
        for (let b = 0; b < 5; b++) {
            if (a === b) {
                continue;
            }
            let [curA, curB, preA, preB] = [0, 0, 0, 0];
            const t: number[][] = [
                [Infinity, Infinity],
                [Infinity, Infinity],
            ];
            let l = -1;
            for (let r = 0; r < s.length; r++) {
                const x = s[r];
                curA += x === a ? 1 : 0;
                curB += x === b ? 1 : 0;
                while (r - l >= k && curB - preB >= 2) {
                    t[preA & 1][preB & 1] = Math.min(t[preA & 1][preB & 1], preA - preB);
                    l++;
                    preA += s[l] === a ? 1 : 0;
                    preB += s[l] === b ? 1 : 0;
                }
                ans = Math.max(ans, curA - curB - t[(curA & 1) ^ 1][curB & 1]);
            }
        }
    }
    return ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(k)

Approach Breakdown

BRUTE FORCE

O(n × k) time

O(1) space

For each starting index, scan the next k elements to compute the window aggregate. There are n−k+1 starting positions, each requiring O(k) work, giving O(n × k) total. No extra space since we recompute from scratch each time.

SLIDING WINDOW

O(n) time

O(k) space

The window expands and contracts as we scan left to right. Each element enters the window at most once and leaves at most once, giving 2n total operations = O(n). Space depends on what we track inside the window (a hash map of at most k distinct elements, or O(1) for a fixed-size window).

Shortcut: Each element enters and exits the window once → O(n) amortized, regardless of window size.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Shrinking the window only once

Wrong move: Using `if` instead of `while` leaves the window invalid for multiple iterations.

Usually fails on: Over-limit windows stay invalid and produce wrong lengths/counts.

Fix: Shrink in a `while` loop until the invariant is valid again.