Minimum Number of Taps to Open to Water a Garden LeetCode solution in JavaScript

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var minTaps = function (n, ranges) {
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  let intervals = [];
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  for (let i = 0; i < ranges.length; i++) {
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    let l = i - ranges[i];
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    let r = i + ranges[i];
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    intervals.push([l, r]);
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  }
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  intervals.sort((a, b) => {
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    if (a[0] === b[0]) return b[1] - a[1];
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    return a[0] - b[0];
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  });
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  // Find the starting idx
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  let startIdx;
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  for (let i = 0; i < intervals.length; i++) {
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    let [s, e] = intervals[i];
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    if (s <= 0) {
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      if (startIdx === undefined) startIdx = i;
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      else if (intervals[startIdx][1] < e) startIdx = i;
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    } else break;
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  }
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  if (startIdx === undefined) return -1;
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  let q = [startIdx],
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    openedTaps = 1;
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  while (q.length) {
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    let max;
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    while (q.length) {
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      let idx = q.pop();
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      let [start, end] = intervals[idx];
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      if (end >= n) return openedTaps;
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      for (let i = idx + 1; i < intervals.length; i++) {
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        let [nextStart, nextEnd] = intervals[i];
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        // If next interval's start is less than the current interval's end
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        if (nextStart <= end) {
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          if (!max && nextEnd > end) max = { i, end: nextEnd };
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          // If the next interval's end is greater than the current interval's end
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          else if (max && nextEnd > max.end) max = { i, end: nextEnd };
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        } else break;
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      }
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    }
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    if (max) {
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      q.push(max.i);
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      openedTaps++;
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    }
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  }
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  return -1;
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};