Finding Pairs With a Certain Sum LeetCode solution in JavaScript

1
function Counter(arr) {
2
  // input: array
3
  // output: corresponding occurrence-based dictionary
4

5
  let keyOccDict = {};
6

7
  arr.forEach((val) => (keyOccDict[val] = (keyOccDict[val] || 0) + 1));
8

9
  return keyOccDict;
10
}
11

12
var FindSumPairs = function (nums1, nums2) {
13
  this.arrA = nums1;
14
  this.arrB = nums2;
15

16
  // maintain mapping between distinct number and occurrence for array A
17
  this.dictA = Counter(this.arrA);
18

19
  // maintain mapping between distinct number and occurrence for array B
20
  this.dictB = Counter(this.arrB);
21

22
  return;
23
};
24

25
FindSumPairs.prototype.add = function (index, val) {
26
  // update mapping for arrB
27

28
  let old_value = this.arrB[index];
29
  this.arrB[index] += val;
30
  let new_value = this.arrB[index];
31

32
  this.dictB[old_value] = this.dictB[old_value] - 1;
33
  this.dictB[new_value] = (this.dictB[new_value] || 0) + 1;
34

35
  return;
36
};
37

38
FindSumPairs.prototype.count = function (tot) {
39
  /*
40
      Goal:
41

42
      Find the method count of a + b = total,
43
      where a comes from array A, and b comes from array B
44

45
      apply the concept learned from Leetcode #1 Two sum
46
      a + b = total <=> b = total - a
47

48
      speed-up by dictionary
49
    */
50

51
  let counter = 0;
52

53
  for (const [a, occ_a] of Object.entries(this.dictA)) {
54
    let b = tot - a;
55

56
    counter += occ_a * (this.dictB[b] || 0);
57
  }
58

59
  return counter;
60
};