Best Anagram Match
Reported by candidates from Google's online assessment. Pattern, common pitfall, and the honest play if you blank under the timer.
The trap in Google's Best Anagram Match, reported in September 2026, is the input size, not the idea. Checking every candidate by sorting it looks fine on the examples and falls apart when the target and the list both hit 10^5. You need to return the lexicographically smallest exact anagram, or an empty string if nothing matches. It's a hash-table and counting problem wearing an easy costume. If you blank mid-assessment, StealthCoder runs invisibly on your desktop and hands you the approach in real time, but the pattern below is short enough to carry in your head.
The problem
Given a lowercase target string and a list of lowercase candidate words, return the lexicographically smallest candidate that is an exact anagram of the target. An exact anagram has the same length and the same character frequencies. Return the empty string when no candidate qualifies. Function findBestAnagram(target: String, candidates: String[]) → String Examples Example 1 target = "listen" candidates = ["silent","enlist","google"] return = "enlist" Both silent and enlist are anagrams; enlist is lexicographically smaller. Example 2 target = "abc" candidates = ["ab","abd"] return = "" No candidate has the same multiset. Constraints 1 <= target.length <= 10^5. 0 <= candidates.length <= 10^5. All strings contain lowercase English letters.
Reported by candidates. Source: FastPrep
Pattern and pitfall
Build a frequency count of the target, 26 slots. For each candidate, skip it immediately if its length differs from the target's. Otherwise compare its counts to the target's, and if they match, keep it if it's smaller than your current best. Use a plain string comparison for the minimum. The pitfall is sorting every candidate, which costs O(L log L) each and can blow up when strings are long. Counting keeps each check linear in its length. Watch the edge cases: an empty candidates list returns an empty string, and a candidate with the same letters but a different length (like "ab" against "abc") must fail. Also don't forget that identical strings count as anagrams. Total work is bounded by the sum of candidate lengths. If the live OA freezes your head, StealthCoder is the hedge that surfaces this exact approach while you type.
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Best Anagram Match FAQ
How hard is Best Anagram Match really?+
Easy to medium. The logic is a frequency comparison plus a running minimum. What makes it feel harder is the 10^5 bounds, which punish sorting every candidate. If you count letters and filter by length first, it's a short function.
What's the trick to get the lexicographically smallest answer?+
Don't sort the candidates up front. Scan once, and whenever a candidate is a valid anagram, compare it to your stored best and replace it if it's smaller. Start with a null or empty best so the no-match case returns an empty string naturally.
Should I sort each word or use a count array?+
Use a count array of 26. Sorting each word costs L log L and adds up across 10^5 candidates. Counting is linear per word. Sorting works on small inputs, but the constraints here are built to make it risky.
What edge cases should I test before submitting?+
Test an empty candidates list, a candidate shorter or longer than the target, a candidate identical to the target, duplicates in the list, and a single-character target. Example 2 shows a length mismatch must return an empty string, so check that explicitly.
How do I prepare for this in 48 hours?+
Write the counting solution from scratch twice, once with a 26-slot array and once with a hash map. Then practice the group-anagrams style of problem so frequency keys feel automatic. Time yourself on edge cases, since that's where Google-style OAs trip people up.