Maximum Equal Frequency LeetCode solution in JavaScript

1
// *NOTE: ternary operator is to handle if key is undefined since javascript doesn't have a default object value
2

3
var maxEqualFreq = function (nums) {
4
  // create cache to count the frequency of each number
5
  const numCnt = {};
6

7
  // create cache to count the number of 'number of frequencies'
8
  // e.g. [1,2,3,2,4,4,3,1,5]
9
  //        {  1: 1           1 number appeared with a frequency of 1
10
  //           2: 3   }      3 numbers appeared with a frequency of 2
11
  const cntFreq = {};
12

13
  // holds the maximum frequency a number has appeared
14
  let maxFreq = 0;
15

16
  // holds solution
17
  let result = 0;
18

19
  // iterate through numbers
20
  for (let i = 0; i < nums.length; i++) {
21
    // get current number;
22
    const num = nums[i];
23

24
    // subtract one from the current frequency of the current number
25
    cntFreq[numCnt[num]] = cntFreq[numCnt[num]] ? cntFreq[numCnt[num]] - 1 : -1;
26

27
    // increase the count of the current number
28
    numCnt[num] = numCnt[num] ? numCnt[num] + 1 : 1;
29

30
    // add one to the current frequency of the current number
31
    cntFreq[numCnt[num]] = cntFreq[numCnt[num]] ? cntFreq[numCnt[num]] + 1 : 1;
32

33
    // if the current frequency is more than the max frequency update the max frequency
34
    if (maxFreq < numCnt[num]) maxFreq = numCnt[num];
35

36
    // if max frequency is 1 or
37
    // if the max frequency multiplied by the number of values with that frequency equals the current index or
38
    // if the result of the 1 less than the max frequency equals to the current index
39
    // update result
40
    if (
41
      maxFreq === 1 ||
42
      maxFreq * cntFreq[maxFreq] === i ||
43
      (maxFreq - 1) * (cntFreq[maxFreq - 1] + 1) === i
44
    ) {
45
      result = i + 1;
46
    }
47
  }
48
  return result;
49
};