Valid Anagram

EASY✓ CHIP-TIMEDLC #242 — FULL STATEMENT

The briefingthe full drill, in my own words

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.

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.

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

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

The hintstake 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 boardthe 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 splitstwo 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-TIMEDALL 4 SOLUTIONS RAN GREEN AGAINST SELF-AUTHORED CASES IN CI · JDK 21 · CPYTHON 3.12 · NOTHING PUBLISHES RED