1110MediumHash TableStringBit ManipulationPrefix Sum

The Longest Stretch With Every Tally Even

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Trains the technique from

LeetCode 1371Find the Longest Substring Containing Vowels in Even Counts

This is an original problem, written from a brief that listed the technique, the difficulty, the topics, the function shape and the input bounds — none of that problem's wording, examples, hints or editorials. The link is there so you can map your practice onto the standard set.

Same function shape, different story and different numbers.

A line line holds only lowercase letters. The five letters a, e, i, o and u are the tallied letters, and the other twenty-one are ignored.

A stretch of neighbouring letters is balanced when every tallied letter appears an even number of times inside it, and appearing no times at all counts as even.

Return the length of the longest balanced stretch. The empty stretch is balanced, so the answer is never below zero.

Examples

Example 1

Input: line = "abcabc"
Output: 6

The only tallied letter here is a and it turns up twice, so the whole line is balanced.

Example 2

Input: line = "ae"
Output: 0

One a and one e, each an odd number of times, and dropping either letter leaves the other odd, so nothing but the empty stretch is balanced.

Example 3

Input: line = "azbzc"
Output: 4

The single a is tallied once, so a balanced stretch has to leave it out. The four letters after it hold no tallied letter at all.

Constraints

1 <= line.length <= 5 * 10^5
line holds only lowercase English letters

The signature

The editor is preloaded with this. It matches the parent problem's shape, so a solution that works here transfers to a judge unchanged.

Python

def find_the_longest_substring(line: str) -> int:

Java

public int findTheLongestSubstring(String line)

The intended approach

Time O(n) over the line.Space O(1), since only thirty-two states exist.

What makes a stretch balanced is not how many tallied letters it holds but only whether each count is odd or even, and that is five yes-or-no facts, so thirty-two states in all. Walk the line keeping the odd-or-even state of the five counts across everything read so far. A tallied letter flips its own fact and leaves the other four alone, and an ignored letter changes nothing. A stretch between two positions is balanced exactly when the state before it and the state at its end agree, because every count that flipped an odd number of times inside would have left them different. So hold, for each of the thirty-two states, the earliest position it was ever seen at. Meeting a state already recorded gives a balanced stretch reaching back to that earliest position, and keeping the earliest rather than the latest is what makes it the longest. The starting state counts as seen just before the line begins, which is what lets a stretch from the very first letter be found.

Pythonentry point: find_the_longest_substring

def find_the_longest_substring(line):
    seat = {"a": 1, "e": 2, "i": 4, "o": 8, "u": 16}
    first = {0: -1}
    state = 0
    best = 0
    for i in range(len(line)):
        state ^= seat.get(line[i], 0)
        if state in first:
            span = i - first[state]
            if span > best:
                best = span
        else:
            first[state] = i
    return best

The obvious version

Take every stretch outright. Fix a starting letter, walk forward tallying the five letters as they turn up, and after each step check all five tallies for evenness, keeping the longest stretch that passes. Costs the line length squared.

This is not filler. It was written independently of the fast solution and the two were run against thousands of random inputs to confirm they agree — which is how the answer above is known to be right. Read it when the intended approach feels like a rabbit pulled out of a hat.

Python

def bf_find_the_longest_substring(line):
    n = len(line)
    marked = "aeiou"
    best = 0
    for start in range(n):
        tally = [0, 0, 0, 0, 0]
        for end in range(start, n):
            spot = marked.find(line[end])
            if spot >= 0:
                tally[spot] += 1
            ok = True
            for value in tally:
                if value % 2:
                    ok = False
                    break
            if ok and end - start + 1 > best:
                best = end - start + 1
    return best

How this problem was checked

Reference solution executed against 15 test cases — 15 passed
3 worked examples confirmed by running the code
0 eight-word sequences shared with the parent problem's text
Agrees with an independent brute-force implementation on random inputs
Test set proven to catch deliberately broken solutions
Reviewed by a separate quality pass
Function arity matches the parent problem's

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