Strings: Valid Anagram — Kotlin Solution

Problem Info

LeetCode #	242 — Valid Anagram
Difficulty	Easy
Topic	Strings, HashMap

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

An anagram uses all the original characters exactly once — just rearranged.

Example:

Input:  s = "anagram", t = "nagaram"
Output: true

Input:  s = "rat", t = "car"
Output: false

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Approach

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

Approach 1 — Sort both strings and compare: If they’re anagrams, sorting both gives the same string.

"anagram" → sorted → "aaagmnr"
"nagaram" → sorted → "aaagmnr"  ✓ match

Simple but O(n log n).

Approach 2 — Frequency count (O(n)): Count character frequencies in s, then decrement for each character in t. If any count goes non-zero — not an anagram.

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Kotlin Solution

Approach 1 — Sort (simple)

fun isAnagram(s: String, t: String): Boolean {
    if (s.length != t.length) return false
    return s.toCharArray().sorted() == t.toCharArray().sorted()
}

Approach 2 — Frequency Map (optimal)

fun isAnagram(s: String, t: String): Boolean {
    if (s.length != t.length) return false

    val count = HashMap<Char, Int>()

    for (c in s) count[c] = count.getOrDefault(c, 0) + 1
    for (c in t) count[c] = count.getOrDefault(c, 0) - 1

    return count.values.all { it == 0 }
}

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Why Kotlin Shines Here

toCharArray().sorted() — sort-and-compare in one chain:

s.toCharArray().sorted() == t.toCharArray().sorted()
// vs Java: Arrays.sort(s.toCharArray()) — can't compare arrays with ==
// Kotlin's sorted() returns a List, so == works correctly

getOrDefault shorthand via Kotlin’s own extension:

count.getOrDefault(c, 0) + 1
// or even cleaner:
count[c] = (count[c] ?: 0) + 1

all { it == 0 } — readable validation over all values:

return count.values.all { it == 0 }
// vs Java: count.values().stream().allMatch(v -> v == 0)

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Two Approaches Compared

Approach	Time	Space	Notes
Sort	O(n log n)	O(n)	Simple, readable
Frequency map	O(n)	O(n)	Optimal, interview preferred

Use the frequency map in interviews — it shows you understand the O(n) solution exists. Use sort when readability matters more than performance.

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Complexity (Frequency Map)

Time	O(n) — two passes through strings
Space	O(1) — at most 26 characters in the map

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Key Takeaway

Anagram = same characters, same frequency. Count up for one string, count down for the other. If everything cancels to zero — it’s an anagram.

🔗 Solve it on LeetCode →

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📚 Kotlin DSA Series

This post is part of the Kotlin DSA series — solving LeetCode problems using idiomatic Kotlin.

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