Sherlock and the Valid String - HackerRank - Easiest Solution

Hello everyone, today, while going through the problem of Sherlock & Valid String, I literally tried a lot and was not getting a simple and sober solution.

In fact, when I myself started implementing it, I found that the problem is not that complicated the way the solutions are available in other blogs, so I started the problem and here is the solution I came up with.

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function isValid(s) { const countMap = {}; const countList = []; const arr = s.split(''); arr.forEach((a) => countMap[a] = (countMap[a] || 0) + 1); Object.keys(countMap).forEach(k => countList.push(countMap[k])); countList.sort(); const countSet = new Set(countList); const len = countList.length; if (countSet.size > 2) { return 'NO'; } else if (countSet.size === 1) { return 'YES'; } else if (countSet.size === 2) { const firstNum = countList[0]; const lastNum = countList[len - 1]; const lastList = countList.slice(1); const firstList = countList.slice(0, len - 1); const lastListSet = new Set(lastList); const firstListSet = new Set(firstList); return ( (lastListSet.size === 1 && firstNum === 1) || (lastListSet.size === 1 && lastNum - firstNum === 1) || (firstListSet.size === 1 && lastNum - firstNum === 1) ) ? 'YES' : 'NO'; } }

Explanation

Suppose the input was aabbc:

1. Create a countMap with the number of occurrences of each character.
• countMap = { a: 2, b: 2, c:1 };
2. Make a list of the values of above maps( not keys ) and sort it.
• countList = [1, 2, 2]
3. Convert countList to countSet
1. countSet = [1, 2]
4. If (countSet.size > 2) the solution is not possible.
5. If (countSet.size === 1), we have the "YES"
6. If (countSet.size === 2), we have to process:
1. We will create 6 variables for our ease, those are:
1. firstNum  i.e first number of countList (sorted) = 1 (for our example)
2. lastList i.e sub array of countList from 1..end = [2, 2]
3. lastListSet i.e set over lastList, for removing duplicates = [2]
4. lastNum i.e last number of countList (sorted) = 2
5. firstList i.e sub array of countList from 0..end-1 = [1, 2]
6. firstListSet i.e set over firstList, for removing duplicates = [1, 2]
2. There are only 3 conditions which is "YES" those are:
1. [ 1, X, X, X, X]
2. [ X-1, X-1, X-1, X-1, X]
3. [X, X, X, X, X+1]
3. For our example 2.1 is true
4. Apart from them there is no other way, a String will give "YES" and thats what we did

Happy coding,

:)