Alert Using Same Key-Card Three or More Times in a One Hour Period LeetCode solution in JavaScript

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/**
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 * @param {string[]} keyName
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 * @param {string[]} keyTime
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 * @return {string[]}
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 */
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var alertNames = function (keyName, keyTime) {
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  // so we don't keep duplicates
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  const abusers = new Set();
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  // map: name->times[] (sorted)
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  const times = {};
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  for (let i = 0; i < keyName.length; i++) {
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    const name = keyName[i];
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    const time = keyTime[i];
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    if (times[name] == null) times[name] = [];
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    times[name].push(time);
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    if (times[name].length > 2) {
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      times[name].sort();
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      const len = times[name].length;
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      // we check all triples for a time difference below 1 hour.
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      // as times are sorted, we need to check all i and i+2
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      for (let i = 0; i < len - 2; i++) {
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        if (belowHour(times[name][i], times[name][i + 2])) {
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          abusers.add(name);
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        }
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      }
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    }
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  }
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  const ar = Array.from(abusers);
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  // sort them lexicographically
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  ar.sort();
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  return ar;
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  /*
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    same hour: true
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    hour diff is above 1: false
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    hour diff is 1: m2 must be <= m1
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    */
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  function belowHour(t1, t2) {
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    const [h1, m1] = t1.split(":").map((num) => parseInt(num));
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    const [h2, m2] = t2.split(":").map((num) => parseInt(num));
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    if (h1 === h2) return true;
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    if (h2 - h1 > 1) return false;
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    if (m2 <= m1) return true;
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  }
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};