Minimum Window Substring
Reported by candidates from Character.AI's online assessment. Pattern, common pitfall, and the honest play if you blank under the timer.
The Character.AI OA reported in June 2026 hands you Minimum Window Substring, and the trap is duplicates. If t is "aa" and s has one "a", a naive set-based check says you're done and you return the wrong answer. This is a sliding-window problem with character counts, and with strings up to 10^5 long, brute force dies fast. You have 24 to 72 hours, so learn the one pattern and the one edge case. If you blank mid-assessment, StealthCoder runs invisibly as a safety net and reads the problem on screen, but you should still know the shape of the answer.
The problem
Given strings s and t, find the shortest contiguous part of s that contains every character of t with its required multiplicity. Return that part of s. If no part qualifies, return an empty string. For these test cases, a shortest qualifying window, when one exists, is unique. Treat uppercase and lowercase letters as different characters. Function minWindow(s: String, t: String) → String Examples Example 1 s = "ADOBECODEBANC" t = "ABC" return = "BANC" The final four characters contain each required letter; no shorter window does. Example 2 s = "a" t = "a" return = "a" The only character supplies the target. Example 3 s = "a" t = "aa" return = "" There is only one copy of a, so no valid window exists. Constraints 1 ≤ s.length, t.length ≤ 10^5. Both strings contain only uppercase and lowercase English letters. The total input fits in memory.
Reported by candidates. Source: FastPrep
Pattern and pitfall
Build a count map of t. Keep a second window map, or one shared map that goes negative. Expand the right pointer, adding each character. Track a 'formed' counter that rises only when a character's window count reaches its required count, not when it first appears. That's the multiplicity fix. Once formed equals the number of distinct characters in t, shrink from the left while still valid, recording the best start and length each time. Then drop the left character and keep going. Pitfalls: comparing the full maps on every step, which gets slow, and treating presence as enough. Uppercase and lowercase are different, so use a 128-size array or a hash map. Return an empty string if no window was ever recorded. Runtime is O(n + m). StealthCoder is the hedge if the pointer logic slips under pressure, but write this one by hand once tonight.
Drill it cold or hedge it with StealthCoder. Either way, don't walk into the OA hoping you remember the trick.
You can drill Minimum Window Substring cold, or you can hedge it. StealthCoder runs invisibly during screen share and surfaces a working solution in under 2 seconds. The proctor sees the IDE. They don't see what's behind it. Made for the candidate who got the OA invite this morning and has 72 hours, not six months.
Get StealthCoderRelated leaked OAs
This OA pattern shows up on LeetCode as minimum window substring. If you have time before the OA, drill that.
You've seen the question.
Make sure you actually pass Character.AI's OA.
Character.AI reuses patterns across OAs. Made for the candidate who got the OA invite this morning and has 72 hours, not six months. Works on HackerRank, CodeSignal, CoderPad, and Karat.
Minimum Window Substring FAQ
What's the trick to Minimum Window Substring?+
Use two pointers and a count of how many distinct required characters are fully satisfied. Expand right until all are satisfied, then shrink left as far as the window stays valid. Record the smallest window at each valid state. It runs in linear time.
How hard is this really for the Character.AI OA?+
It's a well-known hard-tagged problem, but the pattern is mechanical once you've seen it. The difficulty is bookkeeping, not insight. If you can write a clean sliding window with a count map, you can finish it.
Why does t = "aa" and s = "a" matter?+
It tests multiplicity. A solution that only checks whether each character appears will wrongly accept it. You need the window count of each character to be at least its count in t, so example 3 returns an empty string.
Is sliding window still the expected approach in 2026?+
Yes. This problem was reported in June 2026 and the standard linear solution is still what's expected. Brute force over all substrings is O(n^2) or worse and won't pass at lengths up to 10^5.
How do I prepare in 48 hours?+
Write this solution from scratch twice without looking. Then trace example 1 by hand to see the window shrink to BANC. Test edge cases: single characters, t longer than s, repeated letters, and mixed case. That covers most of what can go wrong.