valid anagram

Link : https://neetcode.io/problems/is-anagram

solution 1

function isAnagram_01(s, t) {
  const x = {};
  const y = {};
  if (s.length !== t.length) {
    return false;
  }
  for (let i = 0; i < s.length; i++) {
    // if (x.hasOwnProperty(s[i])) {
    //   x[s[i]] = ++x[s[i]];
    // } else {
    //   x[s[i]] = 1;
    // }

    // if (y.hasOwnProperty(t[i])) {
    //   y[t[i]] = ++y[t[i]];
    // } else {
    //   y[t[i]] = 1;
    // }
    x[s[i]] = 1 + (x[s[i]] ?? 0);
    y[t[i]] = 1 + (y[t[i]] ?? 0);
  }

  for (const property in x) {
    if (x[property] !== y[property]) {
      return false;
    }
  }
  return true;
}

solution 2

var isAnagram_02 = function (s, t) {
  if (s.length !== t.length) {
    return false;
  }
  const lookup = {};

  for (let i = 0; i < s.length; i++) {
    if (lookup.hasOwnProperty(s[i])) {
      lookup[s[i]] = ++lookup[s[i]];
    } else {
      lookup[s[i]] = 1;
    }
  }

  for (let i = 0; i < t.length; i++) {
    if (lookup.hasOwnProperty(t[i])) {
      lookup[t[i]] = --lookup[t[i]];
    } else {
      lookup[t[i]] = 1;
    }
  }

  for (const property in lookup) {
    if (lookup[property] > 0) {
      return false;
    }
  }
  return true;
};