Valid Anagram

EASY✓ CHIP-TIMEDLC #242 — FULL STATEMENT ↗

The drill: Two lowercase words, one question: are they the same multiset of letters? Shuffling is free — what matters is that every letter shows up the same number of times on both sides.

THE BRIEFING — THE FULL DRILL, IN MY OWN WORDS

Two lowercase strings arrive side by side, and the question is whether one is simply a rearrangement of the other's letters — same letters, same counts, order doesn't matter.

Every character in the first string must be accounted for somewhere in the second, with matching frequency; nothing may be added, dropped, or substituted along the way.

Strings of different lengths can never be anagrams of each other, which settles the answer immediately without inspecting a single letter.

EX 01

s = "silent" · t = "listen"

→ true

THE CLASSIC PAIR

EX 02

s = "race" · t = "care"

→ true

EX 03

s = "race" · t = "racer"

→ false

LENGTH MISMATCH, QUICK EXIT

strings hold lowercase letters only, on the order of a few thousand characters at most
differing lengths mean an automatic no
letter order carries no meaning, only counts
the answer is a single true/false verdict

THE HINTS — TAKE ONLY WHAT YOU NEED

HINT 1 THE NUDGE

Order is the only thing an anagram scrambles. What property of a word survives any shuffle completely untouched?

HINT 2 THE STRUCTURE

Force both words into one agreed-upon order and true rearrangements become identical strings — or skip ordering entirely and compare how often each letter appears.

HINT 3 ONE STEP FROM THE ANSWER

Lowercase means at most 26 letters. One 26-slot tally — add along s, subtract along t — and any nonzero slot ends the debate. A length check up front makes the quick exit free.

COACH'S BOARD — THE PATTERN, STEP BY STEP

ONE TALLY, TWO WORDSPATTERN · 26-SLOT COUNTERs = "race" · t = "care"

LETTER TALLY (+s, −t)

— empty —

STEP 1

s = "race", t = "care" — same length 4, so a mismatch can't be ruled out by length alone.

STEP 1 / 8 · ← → WORK TOO

THE SPLITS — TWO PACES, TWO LANGUAGES

grind/valid-anagram.py — RACE PACE

LANG ▸

PACE ▸

class Solution:
    def isAnagram(self, s: str, t: str) -> bool:
        if len(s) != len(t):
            return False
        count = [0] * 26
        for ch in s:
            count[ord(ch) - ord("a")] += 1
        for ch in t:
            count[ord(ch) - ord("a")] -= 1
        return all(c == 0 for c in count)

TIME O(N)SPACE O(1)PYTHON · RACE PACE · 10 LN

class Solution:
    def isAnagram(self, s: str, t: str) -> bool:
        if len(s) != len(t):
            return False
        return sorted(s) == sorted(t)

TIME O(N LOG N)SPACE O(N)PYTHON · TRAINING PACE · 5 LN

class Solution {
    public boolean isAnagram(String s, String t) {
        if (s.length() != t.length()) {
            return false;
        }
        int[] count = new int[26];
        for (int i = 0; i < s.length(); i++) {
            count[s.charAt(i) - 'a']++;
            count[t.charAt(i) - 'a']--;
        }
        for (int c : count) {
            if (c != 0) {
                return false;
            }
        }
        return true;
    }
}

TIME O(N)SPACE O(1)JAVA · RACE PACE · 18 LN

class Solution {
    public boolean isAnagram(String s, String t) {
        if (s.length() != t.length()) {
            return false;
        }
        char[] a = s.toCharArray();
        char[] b = t.toCharArray();
        Arrays.sort(a);
        Arrays.sort(b);
        return Arrays.equals(a, b);
    }
}

TIME O(N LOG N)SPACE O(N)JAVA · TRAINING PACE · 12 LN

✓ CHIP-TIMED — ALL 4 SOLUTIONS RAN GREEN AGAINST SELF-AUTHORED CASES IN CI · JDK 21 · CPYTHON 3.12 · NOTHING PUBLISHES RED