Quick Overview

This question evaluates proficiency in string manipulation, algorithmic efficiency, and time/space complexity analysis within the Coding & Algorithms domain for data scientist roles.

Determine If Two Strings Are Anagrams Efficiently

Company: Google

Role: Data Scientist

Category: Coding & Algorithms

Difficulty: medium

Interview Round: Onsite

##### Scenario Backend service needs to verify whether two user-provided strings are anagrams for text-matching features. ##### Question Implement a Python function is_anagram(s1, s 2) that returns True if the two input strings are anagrams of each other, otherwise False. Explain time and space complexity. ##### Hints Normalize case; compare sorted strings or use a character-count hash map.

Overview: This question evaluates proficiency in string manipulation, algorithmic efficiency, and time/space complexity analysis within the Coding & Algorithms domain for data scientist roles.

Given two strings s1 and s2, return True if they are anagrams of each other when compared case-insensitively, otherwise return False. Treat every character, including spaces and punctuation, as significant after converting both strings to lowercase.

Constraints

  • 0 <= len(s1), len(s2) <= 200000
  • Characters are printable ASCII (code points 32 to 126).
  • Comparison is case-insensitive; all characters (including spaces and punctuation) are significant.
  • Expected time complexity: O(n).

Hints

  1. Normalize both strings to lowercase.
  2. If lengths differ after normalization, return False immediately.
  3. Use a hash map (dictionary) to count characters in s1 and decrement with s2.
  4. Alternatively, compare sorted lowercase strings (O(n log n)).
  5. With ASCII, an array of size 128 can store counts.

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Approach

Convert both strings to lowercase, then compare their character frequency distributions using a dictionary. First increment counts for characters in s1, then decrement for characters in s2. If any character is missing or over-decremented, return False. If all counts balance to zero, the strings are anagrams.

Time complexity:
O(n)
Space complexity:
O(1) with fixed ASCII alphabet; O(k) otherwise