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16. Valid Anagram

easyAsked at Databricks

Determine whether two strings are anagrams — Databricks surfaces this in early screens to test whether you reach for a frequency map, the same mental model behind deduplication passes in Delta Lake compaction jobs.

By Alex Chen, Founder, InterviewChamp.AI · Last verified

Problem

Given two strings s and t, return true if t is an anagram of s and false otherwise. An anagram uses all the original letters of s exactly once, rearranged.

Constraints

  • 1 <= s.length, t.length <= 5 * 10^4
  • s and t consist of lowercase English letters

Examples

Example 1

Input
s = "anagram", t = "nagaram"
Output
true

Explanation: Both strings contain exactly one 'a','n','g','r','m','a' — same frequency profile.

Example 2

Input
s = "rat", t = "car"
Output
false

Approaches

1. Brute force — sort and compare

Sort both strings and check equality. Simple but wastes O(n log n) on the sort.

Time
O(n log n)
Space
O(n)
function isAnagram(s, t) {
  if (s.length !== t.length) return false;
  return s.split('').sort().join('') === t.split('').sort().join('');
}

Tradeoff:

2. Frequency map

Count character frequencies in s, decrement for t, then verify all counts are zero. Single pass, O(1) space over a fixed alphabet.

Time
O(n)
Space
O(1)
function isAnagram(s, t) {
  if (s.length !== t.length) return false;
  const freq = new Array(26).fill(0);
  const a = 'a'.charCodeAt(0);
  for (let i = 0; i < s.length; i++) {
    freq[s.charCodeAt(i) - a]++;
    freq[t.charCodeAt(i) - a]--;
  }
  return freq.every(c => c === 0);
}

Tradeoff:

Databricks-specific tips

Databricks interviewers want you to connect the frequency-map pattern to real data workloads — for instance, how a shuffle-hash join partitions records by key frequency. Call out O(1) space explicitly; they penalize candidates who accept O(n) without noticing the fixed-alphabet shortcut.

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