Hackerrank: Sherlock and the Valid String Solution

Approach

Build 2 hash map of

• occurence of character
• occurence of occurence

Work on the hash map of occurence of occurence

• always YES if the hash map size is 1
• always NO if the hash map size is different from 2

With case of hash map size of 2

• always YES if one of the pair is [1, 1]
• always YES if one pair's value is one and subtract of its key and the other is 1
• the rest is NO

Implementation

1function isValid(s) {
2  const charOccMap = new Map()
3  const occOccMap = new Map()
4  s.split("").forEach(c => charOccMap.set(c, (charOccMap.get(c) || 0) + 1))
5  Array.from(charOccMap.values()).forEach(occ =>
6    occOccMap.set(occ, (occOccMap.get(occ) || 0) + 1)
7  )
8
9  if (occOccMap.size === 1) {
10    return "YES"
11  }
12  if (occOccMap.size !== 2) {
13    return "NO"
14  }
15  const validPairCase1 = (pair1, pair2) =>
16    pair1[1] === 1 && pair1[0] - pair2[0] === 1
17  const validPairCase2 = (pair1, pair2) =>
18    (pair1[0] === 1 && pair1[1] === 1) || (pair2[0] === 1 && pair2[1] === 1)
19  const [occX, occY] = Array.from(occOccMap)
20  return validPairCase1(occX, occY) ||
21    validPairCase1(occY, occX) ||
22    validPairCase2(occX, occY)
23    ? "YES"
24    : "NO"
25}