LeetCode #1371 — MEDIUM

Find the Longest Substring Containing Vowels in Even Counts

Move from brute-force thinking to an efficient approach using hash map strategy.

The Problem

Problem Statement

Given the string s, return the size of the longest substring containing each vowel an even number of times. That is, 'a', 'e', 'i', 'o', and 'u' must appear an even number of times.

Example 1:

Input: s = "eleetminicoworoep"
Output: 13
Explanation: The longest substring is "leetminicowor" which contains two each of the vowels: e, i and o and zero of the vowels: a and u.

Example 2:

Input: s = "leetcodeisgreat"
Output: 5
Explanation: The longest substring is "leetc" which contains two e's.

Example 3:

Input: s = "bcbcbc"
Output: 6
Explanation: In this case, the given string "bcbcbc" is the longest because all vowels: a, e, i, o and u appear zero times.

Constraints:

1 <= s.length <= 5 x 10^5
s contains only lowercase English letters.

Step 01

Brute Force Baseline

Problem summary: Given the string s, return the size of the longest substring containing each vowel an even number of times. That is, 'a', 'e', 'i', 'o', and 'u' must appear an even number of times.

Baseline thinking

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

Pattern signal: Hash Map · Bit Manipulation

Example 1

"eleetminicoworoep"

Example 2

"leetcodeisgreat"

Example 3

"bcbcbc"

Step 02

Core Insight

What unlocks the optimal approach

Represent the counts (odd or even) of vowels with a bitmask.
Precompute the prefix xor for the bitmask of vowels and then get the longest valid substring.

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 #1371: Find the Longest Substring Containing Vowels in Even Counts
class Solution {
    public int findTheLongestSubstring(String s) {
        String vowels = "aeiou";
        int[] d = new int[32];
        Arrays.fill(d, 1 << 29);
        d[0] = 0;
        int ans = 0, mask = 0;
        for (int i = 1; i <= s.length(); ++i) {
            char c = s.charAt(i - 1);
            for (int j = 0; j < 5; ++j) {
                if (c == vowels.charAt(j)) {
                    mask ^= 1 << j;
                    break;
                }
            }
            ans = Math.max(ans, i - d[mask]);
            d[mask] = Math.min(d[mask], i);
        }
        return ans;
    }
}

// Accepted solution for LeetCode #1371: Find the Longest Substring Containing Vowels in Even Counts
func findTheLongestSubstring(s string) (ans int) {
    vowels := "aeiou"
    d := [32]int{}
    for i := range d {
        d[i] = 1 << 29
    }
    d[0] = 0
    mask := 0
    for i := 1; i <= len(s); i++ {
        c := s[i-1]
        for j := 0; j < 5; j++ {
            if c == vowels[j] {
                mask ^= 1 << j
                break
            }
        }
        ans = max(ans, i-d[mask])
        d[mask] = min(d[mask], i)
    }
    return
}

# Accepted solution for LeetCode #1371: Find the Longest Substring Containing Vowels in Even Counts
class Solution:
    def findTheLongestSubstring(self, s: str) -> int:
        d = {0: -1}
        ans = mask = 0
        for i, c in enumerate(s):
            if c in "aeiou":
                mask ^= 1 << (ord(c) - ord("a"))
            if mask in d:
                j = d[mask]
                ans = max(ans, i - j)
            else:
                d[mask] = i
        return ans

// Accepted solution for LeetCode #1371: Find the Longest Substring Containing Vowels in Even Counts
struct Solution;
use std::collections::HashMap;

impl Solution {
    fn find_the_longest_substring(s: String) -> i32 {
        let mut mask = 0;
        let mut hm: HashMap<u8, usize> = HashMap::new();
        hm.insert(0, 0);
        let mut res = 0;
        for (i, c) in s.char_indices() {
            if Self::vowel(c) != 0 {
                mask ^= 1 << Self::vowel(c);
            }
            hm.entry(mask).or_insert(i + 1);
            res = res.max(i + 1 - hm[&mask]);
        }
        res as i32
    }

    fn vowel(c: char) -> usize {
        match c {
            'a' => 1,
            'e' => 2,
            'i' => 3,
            'o' => 4,
            'u' => 5,
            _ => 0,
        }
    }
}

#[test]
fn test() {
    let s = "eleetminicoworoep".to_string();
    let res = 13;
    assert_eq!(Solution::find_the_longest_substring(s), res);
    let s = "leetcodeisgreat".to_string();
    let res = 5;
    assert_eq!(Solution::find_the_longest_substring(s), res);
    let s = "bcbcbc".to_string();
    let res = 6;
    assert_eq!(Solution::find_the_longest_substring(s), res);
}

// Accepted solution for LeetCode #1371: Find the Longest Substring Containing Vowels in Even Counts
function findTheLongestSubstring(s: string): number {
    const vowels = 'aeiou';
    const d: number[] = Array(32).fill(1 << 29);
    d[0] = 0;
    let [ans, mask] = [0, 0];
    for (let i = 1; i <= s.length; i++) {
        const c = s[i - 1];
        for (let j = 0; j < 5; j++) {
            if (c === vowels[j]) {
                mask ^= 1 << j;
                break;
            }
        }
        ans = Math.max(ans, i - d[mask]);
        d[mask] = Math.min(d[mask], i);
    }
    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(1)

Approach Breakdown

SORT + SCAN

O(n log n) time

O(n) space

Sort the array in O(n log n), then scan for the missing or unique element by comparing adjacent pairs. Sorting requires O(n) auxiliary space (or O(1) with in-place sort but O(n log n) time remains). The sort step dominates.

BIT MANIPULATION

O(n) time

O(1) space

Bitwise operations (AND, OR, XOR, shifts) are O(1) per operation on fixed-width integers. A single pass through the input with bit operations gives O(n) time. The key insight: XOR of a number with itself is 0, which eliminates duplicates without extra space.

Shortcut: Bit operations are O(1). XOR cancels duplicates. Single pass → O(n) time, O(1) space.

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.