Minimum Number of Frogs Croaking LeetCode solution in JavaScript

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var minNumberOfFrogs = function (croakOfFrogs) {
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  const croakArr = new Array(5).fill(0); //Array to store occurence of each char
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  let overlap = 0; //Store the number of overlaps
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  for (let i = 0; i < croakOfFrogs.length; i++) {
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    switch (croakOfFrogs[i]) {
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      case "c":
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        croakArr[0] += 1;
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        //Check if new start, is overlapping with others
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        if (croakArr[0] - croakArr[4] - overlap > 1) {
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          ++overlap;
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        }
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        break;
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      case "r":
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        //Condition to check if r comes before c
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        if (croakArr[0] <= croakArr[1]) return -1;
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        croakArr[1] += 1;
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        break;
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      case "o":
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        //Condition to check if o comes before r
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        if (croakArr[1] <= croakArr[2]) return -1;
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        croakArr[2] += 1;
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        break;
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      case "a":
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        //Condition to check if a comes before o
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        if (croakArr[2] <= croakArr[3]) return -1;
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        croakArr[3] += 1;
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        break;
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      case "k":
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        //Condition to check if k comes before a
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        if (croakArr[3] <= croakArr[4]) return -1;
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        croakArr[4] += 1;
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        break;
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    }
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  }
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  //Check if all items of array have same count else return -1
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  //If all items have same count return overlap + 1
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  return Math.max(...croakArr) === Math.min(...croakArr) ? overlap + 1 : -1;
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};