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}

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Tags

hackerrank

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