LeetCode #3090 — EASY

Maximum Length Substring With Two Occurrences

Build confidence with an intuition-first walkthrough focused on hash map fundamentals.

The Problem

Problem Statement

Given a string s, return the maximum length of a substring such that it contains at most two occurrences of each character.

Example 1:

Input: s = "bcbbbcba"

Output: 4

Explanation:

The following substring has a length of 4 and contains at most two occurrences of each character: "bcbbbcba".

Example 2:

Input: s = "aaaa"

Output: 2

Explanation:

The following substring has a length of 2 and contains at most two occurrences of each character: "aaaa".

Constraints:

2 <= s.length <= 100
s consists only of lowercase English letters.

Step 01

Brute Force Baseline

Problem summary: Given a string s, return the maximum length of a substring such that it contains at most two occurrences of each character.

Baseline thinking

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

Pattern signal: Hash Map · Sliding Window

Example 1

"bcbbbcba"

Example 2

"aaaa"

Step 02

Core Insight

What unlocks the optimal approach

We can try all substrings by brute-force since the constraints are very small.

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

Upper-end input sizes

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 #3090: Maximum Length Substring With Two Occurrences
class Solution {
    public int maximumLengthSubstring(String s) {
        int[] cnt = new int[26];
        int ans = 0;
        for (int i = 0, j = 0; j < s.length(); ++j) {
            int idx = s.charAt(j) - 'a';
            ++cnt[idx];
            while (cnt[idx] > 2) {
                --cnt[s.charAt(i++) - 'a'];
            }
            ans = Math.max(ans, j - i + 1);
        }
        return ans;
    }
}

// Accepted solution for LeetCode #3090: Maximum Length Substring With Two Occurrences
func maximumLengthSubstring(s string) (ans int) {
	cnt := [26]int{}
	i := 0
	for j, c := range s {
		idx := c - 'a'
		cnt[idx]++
		for cnt[idx] > 2 {
			cnt[s[i]-'a']--
			i++
		}
		ans = max(ans, j-i+1)
	}
	return
}

# Accepted solution for LeetCode #3090: Maximum Length Substring With Two Occurrences
class Solution:
    def maximumLengthSubstring(self, s: str) -> int:
        cnt = Counter()
        ans = i = 0
        for j, c in enumerate(s):
            cnt[c] += 1
            while cnt[c] > 2:
                cnt[s[i]] -= 1
                i += 1
            ans = max(ans, j - i + 1)
        return ans

// Accepted solution for LeetCode #3090: Maximum Length Substring With Two Occurrences
fn maximum_length_substring(s: String) -> i32 {
    let cs: Vec<_> = s.bytes().collect();
    let mut left = 0;
    let mut right = 0;

    let mut ret = 0;
    let mut table = vec![0; 26];
    while left < s.len() && right < s.len() {
        let idx = (cs[right] - b'a') as usize;
        table[idx] += 1;
        while table[idx] > 2 && left < s.len() {
            let idx_left = (cs[left] - b'a') as usize;
            table[idx_left] -= 1;
            left += 1;
        }

        ret = std::cmp::max(ret, right - left + 1);
        right += 1;
    }

    ret as i32
}

fn main() {
    let ret = maximum_length_substring("bcbbbcba".to_string());
    println!("ret={ret}");
}

#[test]
fn test() {
    assert_eq!(maximum_length_substring("bcbbbcba".to_string()), 4);
    assert_eq!(maximum_length_substring("aaaa".to_string()), 2);
}

// Accepted solution for LeetCode #3090: Maximum Length Substring With Two Occurrences
function maximumLengthSubstring(s: string): number {
    let ans = 0;
    const cnt: number[] = Array(26).fill(0);
    for (let i = 0, j = 0; j < s.length; ++j) {
        const idx = s[j].charCodeAt(0) - 'a'.charCodeAt(0);
        ++cnt[idx];
        while (cnt[idx] > 2) {
            --cnt[s[i++].charCodeAt(0) - 'a'.charCodeAt(0)];
        }
        ans = Math.max(ans, j - i + 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.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.

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.