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Valid Anagram

Character Frequency Count — Fixed-Size Array

LeetCode 242 • Easy • Arrays & Hashing

Input: s = "anagram", t = "nagaram" → Output: true (same letters, rearranged)

TimeO(n)single pass

SpaceO(1)fixed 26 slots

Processed: 0/7 Non-zero: 0 Verified: 0/26

Current s[i]

Current t[i]

Count updated

Balanced (0)

Imbalanced

Index —

Count[s[i]] —

Count[t[i]] —

Result —

Speed:

Ready

Press Play to watch the algorithm, or Step to advance one operation at a time.
For each index i, increment count[s[i]] and decrement count[t[i]]. If all counts end at zero, it's an anagram.

✎ Whiteboard

Size: 3

⌨ Type It — Build Muscle Memory

Practice until you don't need to look. Use the guide comments below as scaffolding. Type the full implementation in the editor. The green highlights are the nuances to burn into memory.

# ─── VALID ANAGRAM (LeetCode 242) ───

# Pattern: Character frequency count

# Time: O(n) Space: O(1) ← fixed 26 slots

# 1. Define: isAnagram(self, s, t)

# 2. Early exit — lengths must match:

# if len(s) != len(t):

# return False

# 3. Init count array:

# count = [0] * 26 ← one slot per letter

# 4. Single loop — process both strings:

# for i in range(len(s)):

# 5. Map char → index with ord():

# count[ord(s[i]) - ord('a')] += 1 ← s increments

# count[ord(t[i]) - ord('a')] -= 1 ← t decrements

# 6. Verify all zeros:

# for val in count:

# if val != 0:

# return False

# 7. All balanced: return True

# Key insight: ord(c) - ord('a') maps a-z → 0-25

# Anagram ⇔ every count ends at exactly 0

# Vars: count (list[26]), i, val

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