Maximum Bags With Full Capacity of Rocks LeetCode solution in JavaScript

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var maximumBags = function (capacity, rocks, additionalRocks) {
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  var differences = [];
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  var count = 0;
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  for (let i in rocks) {
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    if (capacity[i] - rocks[i] > 0) {
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      // check if rocks[i] is less than capacity[i]
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      differences.push(capacity[i] - rocks[i]); // add the difference
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    } else {
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      count++; // rocks[i] is at full capacity, so the result goes up by 1
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    }
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  }
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  differences.sort((a, b) => a - b); // sort the differences
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  var r = additionalRocks;
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  while (r != 0 && differences.length > 0) {
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    // loop through differences until there is no more rocks or all of the bags of rocks are at full capacity
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    if (differences[0] <= r) {
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      r -= differences[0]; // minus the difference of the rocks
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      count++; // add to result because another bag of rocks is at full capacity
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    } else {
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      break; // rocks can not make any other bags at full capacity, so stop right here
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    }
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    differences.shift(); // remove the first difference
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  }
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  return count;
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};