Longest Substring with Even Occurrences

Write a function:
def solution(S)
that, given a string S consisting of N lowercase English letters, returns the length of the longest substring in which every letter occurs an even number of times. A substring is defined as a contiguous segment of a string. If no such substring exists, return 0.
Write an efficient algorithm for the following assumptions:
N is an integer within the range [1..100,000];
string S consists only of lowercase letters ('a'-'z').

Function
solution(S: String) → int

Examples
Example 1
S = "bdaaadadb"
return = 6
Substrings in which every letter occurs an even number of times are "aa", "adad", "daaadad" and "aaadad". The length of the longest of them is 6.
Example 2
S = "abacb"
return = 0
There is no non-empty substring in which every letter occurs an even number of times.
Example 3
S = "zthtzh"
return = 6
Every letter in the whole string occurs an even number of times.

Track a 26-bit mask where bit i flips each time letter i appears. A substring has all-even counts exactly when the mask at its start equals the mask at its end. So store the first index where each mask appears, with mask 0 at index -1. Walk the string, flip the bit, and if the mask was seen before, the candidate length is i minus the first index. Otherwise record it. Answer is the max, or 0. The pitfall is forgetting to seed mask 0 at -1, which breaks cases like "zthtzh" where the whole string qualifies. Another trap is updating the map on repeat sightings, which shrinks your lengths. Only store the first occurrence. Time is O(N), space is at most 2^26 keys but really bounded by N. StealthCoder is the hedge if the bitmask idea won't come to you under the clock.

If this hits your live OA and you blank, StealthCoder solves it in seconds, invisible to the proctor.