LeetCode - 242. Valid Anagram

Thoughts:

When anything is about frequency, count, or occurrence, I always think of hash map. I can use an object for storing the frequency of each character in the first String. Then, I can iterate over the second string and decrement the frequency of each character in the object. The time complexity will be O(n).

Code:

/**
 * @param {string} s
 * @param {string} t
 * @return {boolean}
 */
var isAnagram = function (s, t) {
  if (s.length !== t.length) {
    return false;
  }
  const map = {};
  for (let char of s) {
    map[char] = (map[char] || 0) + 1;
  }
  for (let char of t) {
    if (!map[char]) {
      return false;
    }
    map[char]--;
  }
  return true;
};

Reference

Original link