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validString.js
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/**
Sherlock considers a string to be valid if all characters of the string appear the same number of times. It is also valid if he can remove just 1 character at 1 index in the string, and the remaining characters will occur the same number of times. Given a string s, determine if it is valid. If so, return YES, otherwise return NO.
Output Format
Print YES if string
is valid, otherwise, print NO.
Sample Input 0
aabbcd
Sample Output 0
NO
*/
const isValid = (s) => {
let counter = {};
let freq = {};
// count of occurrences of characters
// for example aaabbcc => {a: 3, b: 2, c: 2}
Array.from(s).forEach((char) => {
counter[char] = counter[char] || 0;
counter[char]++;
});
// count the frequency of those occurence counts,
// for example {a: 3, b: 2, c: 2} => {3: 1, 2: 2}
Object.keys(counter).forEach((k) => {
freq[counter[k]] = freq[counter[k]] || 0;
freq[counter[k]]++;
});
// all our frequencies. e.g. [3, 2]
let freqArr = Object.keys(freq).map(Number);
// Simplest case: if there is only one frequency, it's valid
if (freqArr.length === 1) {
return "YES";
}
// If not a single frequency string, check next simplest case
// there are exactly two frequencies
// e.g. [3, 2]
let twoFrequencies = freqArr.length === 2;
let [a, b] = freqArr;
// frequency occurs only once
let oneOccurence = freq[a] === 1 || freq[b] === 1;
// the difference is a singleton character
// e.g. abbcc
let singleton = (freq[a] === 1 && a === 1) || (freq[b] === 1 && b === 1);
// differing frequency is exactly 1 more than others
// e.g. true if aaabbcc, false if abbcc
let diffOfOne = (freq[a] === 1 ? a - b : b - a) === 1;
// if single character or frequencies differ by one
// and there are exactly two frequencies
// and there is only one occurence of the singleton or differing frequency
if ((singleton || diffOfOne) && twoFrequencies && oneOccurence) {
return "YES";
}
return "NO";
};
console.log(isValid("aabbcd")); // NO
// console.log(isValid("aabbccddeefghi")); // NO
// console.log(isValid("abcdefghhgfedecba")); // YES