Avoid Flood in The City LeetCode solution in JavaScript

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var avoidFlood = function (rains) {
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  const n = rains.length;
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  const filledLakes = new Map();
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  const res = new Array(n).fill(-1);
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  const dryDays = [];
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  for (let i = 0; i < n; i++) {
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    const lake = rains[i]; // lake to rain on
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    if (lake === 0) {
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      // It is a dry day
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      dryDays.push(i);
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    } else if (!filledLakes.has(lake)) {
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      // The lake is not filled yet, so we let it be filled (we just don't want it to be rained on again and be flooded)
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      filledLakes.set(lake, i);
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    } else {
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      // The lake is already filled. We want to see if a dry day was available after the lake was previously rained on so that we can empty the lake
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      const lake_index = filledLakes.get(lake); //
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      const dry_index = binarySearch(lake_index);
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      if (dry_index === dryDays.length) return []; // there was no dry day after the lake was previouly filled
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      res[dryDays[dry_index]] = lake; // mark the earliest dry day that was used in our result array
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      filledLakes.set(lake, i); // we need to update the day that the lake is rained on again
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      dryDays.splice(dry_index, 1); // remove the dry day that was used (this is not very efficient, but it just makes our code cleaner)
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    }
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  }
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  dryDays.forEach((day) => (res[day] = 1));
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  return res;
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  function binarySearch(target) {
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    let left = 0;
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    let right = dryDays.length - 1;
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    while (left <= right) {
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      const mid = left + Math.floor((right - left) / 2);
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      if (dryDays[mid] < target) left = mid + 1;
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      else right = mid - 1;
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    }
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    return left;
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  }
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};