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  1. Prepare
  2. Algorithms
  3. Strings
  4. Sherlock and the Valid String
  5. Discussions

Sherlock and the Valid String

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  • biren_yadav114
     + 0 comments

    python -

    def isValid(s):

    # Write your code here
s_dict = {}
count=0
for i in s:
    if i in s_dict:
        s_dict[i]+=1
    else:
        s_dict[i]=1
dict_val_list = list(s_dict.values())

for i in range(1,len(s_dict)):
    if dict_val_list[0] == dict_val_list[i]:
        continue          
    else:
        if count==0:
            count+=1
        else:
            return 'NO'
return 'YES'

  • Peace_Ranger
     + 0 comments

    The input "aabbbccc" is not a valid one cz at least 2 characters has to be removed to make this valid. But the editorial solutions consider this input as "valid". How's that possible? Am I missing something obvious here?

  • priyanshugevra
     + 0 comments
     public static String isValid(String s) {
        Map<Character,Integer> map = new HashMap<>();
        int n = s.length();
        for(int i=0 ; i<n ; i++){
            map.put(s.charAt(i), map.getOrDefault(s.charAt(i), 0)+1);
        } 
        Map<Integer,Integer> freqMap = new HashMap<>();
        for(int fq : map.values())
        {
            freqMap.put(fq, freqMap.getOrDefault(fq, 0)+1);
        }
          if (freqMap.size() == 1) {
            return "YES"; 
        }
          if (freqMap.size() == 2) {
            for(int i : freqMap.values()){
                if(i==1) return "YES";
            }
        }

                can any one explain why it is paasing only 18/20 test cases? also please give the failed test case.

            `        
    return "NO";

}

  • highsoft_soi
     + 0 comments

    Perl:

    sub isValid {
    my @s = split("", shift);
    
    my %h;
    my %c;

    foreach (@s) {
        $h{$_} +=1;
    }
    foreach (values(%h)) {
        $c{$_} +=1;
    }

    return "YES" if (keys(%c) == 1);    
    
    if (keys %c == 2) {
        my @keys = keys(%c);
        my ($freq1, $freq2) = @keys;
        my ($count1, $count2) = @c{@keys};

        if (($freq1 == 1 && $count1 == 1) || ($freq2 == 1 && $count2 == 1)) {
            return "YES";
        }

        if ((abs($freq1 - $freq2) == 1) && ($count1 == 1 || $count2 == 1)) {
            return "YES";
        }
    }

    return "NO";
}

  • vuhuutuan0
     + 0 comments

    My answer with Typescript, can be increase [point] to increase number of character can be remove

    function isValid(s: string): string {
    // 0. count every char in [s]
    let hash = new Map<string, number>()
    for (let c of s) hash.set(c, (hash.get(c) || 0) + 1)

    // 1. define a [value_average] and [point] that can be decreate
    let values = Array.from(hash.values())
    let value_average = values[0]
    let point = 1

    // 2. loop from 1 to end, check value gap and decreate point
    //  -> if [point] is out, return 'NO'
    //  -> orelse return 'YES'
    for (let i = 1; i < values.length; i++) {
        let value = values[i]

        if (value != value_average) {
            if (point == 0) return 'NO'
            point--
        }
    }

    return 'YES'
}