# Sherlock and the Valid String HackerRank Solution

In this Sherlock and the Valid String HackerRank solution, 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  character at  index in the string, and the remaining characters will occur the same number of times. Given a string , determine if it is valid. If so, return YES, otherwise return NO.

Example

This is a valid string because frequencies are .

This is a valid string because we can remove one  and have  of each character in the remaining string.

This string is not valid as we can only remove  occurrence of . That leaves character frequencies of .

Function Description

Complete the isValid function in the editor below.

isValid has the following parameter(s):

• string s: a string

Returns

• string: either YES or NO

Input Format

A single string .

Constraints

• Each character

Sample Input

aabbcd


Sample Output

NO


Explanation

is the minimum number of removals required to make it a valid string. It can be done in following two ways:

Remove c and d to get aabb.
Or remove a and b to get abcd.

## Sherlock and the Valid String HackerRank solution

I will Provide solution in Multiple programming languages for you. If you are not able to find the code in required language then please share in comments so that our team can help you.

### Problem Solution in C#

public static string isValid(string s)
{
var charFreq = s.Distinct()
.ToDictionary(c => c, c => s.Count(t => t == c));
var timesFreq = charFreq.Values.Distinct().
ToDictionary(f => f, f => charFreq.Count(cf => cf.Value == f))
.OrderBy(x => x.Value).ThenBy(x => x.Key).ToList();
return timesFreq.Count switch
{
1 => "YES",
2 => timesFreq[0] is { Key: 1, Value: 1 } || timesFreq[1].Key == timesFreq[0].Key - 1 ? "YES" : "NO",
_ => "NO"
};
}

### Problem Solution in JavaScript

function isValid(s) {
let hastMap = {};
for(let i = 0; i<s.length;i++){
hastMap[s[i]] = (hastMap[s[i]] || 0) + 1;
}

const arr = Object.values(hastMap)
const check = arr.filter(e => e != arr[0]).length;
if(check > 1) {
return 'NO'
}else {
return 'YES'
}

}

### Problem Solution in Python

def isValid(s):
v = {}
for char in s:
if char in v.keys():
v[char] += 1
else:
v[char] = 1
a = list(v.values())
for i in range(len(set(a))):
if len(a) - a.count(a[i]) > 1:
return "NO"
return "YES"

### Problem Solution in Java

public static String isValid(String s) {
Map<Character, Integer> frequancy = new HashMap<>();
for(int i = 0; i < s.length(); i++){
frequancy.put(s.charAt(i), frequancy.getOrDefault(s.charAt(i), 0) + 1);
}
int n = frequancy.get(s.charAt(0));
boolean isRemoved = false;
for (Map.Entry<Character, Integer> e : frequancy.entrySet()) {
if(Math.abs(e.getValue() - n) == 1 ||
(Math.abs(e.getValue() - n) > 1 && e.getValue() == 1)){
if(!isRemoved){
isRemoved = true;
}else{
return "NO";
}
}else if(Math.abs(e.getValue() - n) > 1){
return "NO";
}
}
return "YES";

}

### Problem Solution in C++

string isValid(string s)
{
map<char,int> occurences;
for(int i = 0; i < s.size(); i++)
{
occurences[s[i]]++;
}

map<int, int> frequency;
for(const auto& el : occurences)
{
frequency[el.second]++;
}

if(frequency.size() < 2)
{
return "YES";
}
else if(frequency.size() > 2)
{
return "NO";
}
else if(frequency.size() == 2)
{
auto it = frequency.begin();
auto it2 = next(frequency.begin());

if(abs((*it).first - (*it2).first) == 1 &&
((*it).second == 1  || (*it2).second == 1))
{
return "YES";
}

if(abs((*it).first - (*it2).first) != 1)
{
if(((*it).second == 1 && (*it).first == 1) ||
((*it2).second == 1 && (*it2).first == 1))
{
return "YES";
}

}
}

return "NO";
}
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