Check Anagram Strings Using Hash Map

Given two strings s and t, return true if t is an anagram of s, otherwise return false. An anagram uses the same characters with the same frequency.

Difficulty:

JavaScript Solution

function isAnagram(s, t) {
  if (s.length !== t.length) return false;
  const map = {};

  for (let ch of s) map[ch] = (map[ch] || 0) + 1;
  for (let ch of t) {
    if (!map[ch]) return false;
    map[ch]--;
  }
  return true;
}

Explanation: We count character frequencies using a hash map. If both strings have the same frequency for every character, they are anagrams.

⏱ O(n) | 💾 O(1)

Java Solution

class Solution {
  public boolean isAnagram(String s, String t) {
    if (s.length() != t.length()) return false;
    int[] count = new int[26];

    for (char c : s.toCharArray()) count[c - 'a']++;
    for (char c : t.toCharArray()) {
      if (--count[c - 'a'] < 0) return false;
    }
    return true;
  }
}

Explanation: An integer array works as a hash table to track character counts efficiently.

⏱ O(n) | 💾 O(1)

Python Solution

def is_anagram(s, t):
    if len(s) != len(t):
        return False
    count = {}
    for ch in s:
        count[ch] = count.get(ch, 0) + 1
    for ch in t:
        if ch not in count or count[ch] == 0:
            return False
        count[ch] -= 1
    return True

Explanation: A dictionary stores character frequencies. Mismatches indicate non-anagrams.

⏱ O(n) | 💾 O(1)

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Check Anagram Strings Using Hash Map