Hackerrank: Frequency Queries Solution

Approach

Two hash tables

• one mapping element with its occurence
• one mapping occurrence with elements having that occurence

Implementation

1function freqQuery(queries) {
2  const res = []
3  const elOccMap = {}
4  const occElMap = {}
5  for (const [op, val] of queries) {
6    switch (op) {
7      case 1: {
8        const curOcc = elOccMap[val]
9        const newOcc = (curOcc || 0) + 1
10
11        if (!occElMap[curOcc]) {
12          occElMap[curOcc] = new Map()
13        }
14        occElMap[curOcc].delete(val)
15        if (!occElMap[newOcc]) {
16          occElMap[newOcc] = new Map()
17        }
18        occElMap[newOcc].set(val, true)
19
20        elOccMap[val] = newOcc
21        break
22      }
23      case 2: {
24        const curOcc = elOccMap[val]
25        if (!curOcc) {
26          break
27        }
28        const newOcc = curOcc - 1
29
30        if (!occElMap[curOcc]) {
31          occElMap[curOcc] = new Map()
32        }
33        occElMap[curOcc].delete(val)
34        if (!occElMap[newOcc]) {
35          occElMap[newOcc] = new Map()
36        }
37        occElMap[newOcc].set(val, true)
38
39        elOccMap[val] = newOcc
40        break
41      }
42      case 3: {
43        if (occElMap[val] && occElMap[val].size > 0) {
44          res.push(1)
45        } else {
46          res.push(0)
47        }
48        break
49      }
50    }
51  }
52  return res
53}