#2 Easy Time O(n) Space O(1) Animated

Valid Anagram

Given two strings s and t, return true if t is an anagram of s.

LeetCode ↗

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1 Predict
2 Trace
3 Explain
4 Implement

Pattern Hashing for constant-time memory

Trade one linear scan for a small lookup structure so each value is handled once.

Approach: Character Frequency Map

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Core Insight

Trade one linear scan for a small lookup structure so each value is handled once.

Common Pitfall

Do not overwrite information you still need before checking the current value.

Description

Given two strings s and t, return true if t is an anagram of s.

Constraints

Time Complexity: O(n)
Space Complexity: O(1)

Solution — Character Frequency Map

TimeO(n) SpaceO(1)

Implementation Plan

Translate the invariant into these small, testable moves.

Create the lookup structure.
Scan each value once.
Check the needed relationship before updating state.
Return as soon as the answer is determined.

def isAnagram(s, t):
    if len(s) != len(t):
        return False

    count = {}
    for char in s:
        count[char] = count.get(char, 0) + 1
    for char in t:
        if char not in count:
            return False
        count[char] -= 1
        if count[char] < 0:
            return False
    return True

Contains Duplicate

Two Sum

Valid Anagram

Steps

Description

Constraints

Tags

Solution — Character Frequency Map