Valid Anagram 15 XP Easy

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LeetCode #242: Valid Anagram — a hash map frequency counting problem.

Two strings are anagrams if they contain the exact same characters with the exact same frequencies.

Approach: Count character frequencies in both strings. If the counts match, they're anagrams.

💭 Time to code! Use the step-by-step approach described above. Don't peek at the solution — try first! 🚀

Complexity: Time O(n), Space O(1) — at most 26 letters in the hash map.

📋 Instructions

Implement `isAnagram(s, t)` that returns True if t is an anagram of s. Use a dictionary to count character frequencies. Don't use `sorted()` — practice the hash map approach!

First check len(s) != len(t) → return False. Create count = {}. For c in s: count[c] = count.get(c, 0) + 1. For c in t: count[c] = count.get(c, 0) - 1, if count[c] < 0: return False. Return True.

⚠️ Try solving it yourself first — you'll learn more!

def isAnagram(s, t):
    if len(s) != len(t):
        return False
    count = {}
    for c in s:
        count[c] = count.get(c, 0) + 1
    for c in t:
        count[c] = count.get(c, 0) - 1
        if count[c] < 0:
            return False
    return True

🧪 Test Cases

Input

isAnagram("anagram", "nagaram")

Expected

True

Boolean check

Input

isAnagram("rat", "car")

Expected

False

Boolean check

Input

isAnagram("listen", "silent")

Expected

True

Boolean check

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main.py

Hi! I'm Rex 👋

Output

Ready. Press ▶ Run or Ctrl+Enter.

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