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  1. Prepare
  2. Algorithms
  3. Strings
  4. String Construction
  5. Discussions

String Construction

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983 Discussions

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  • alban_tyrex
     + 0 comments

    The solution here is to count the number of unique occurrences of the characters, here is the explanation : https://youtu.be/UqqNcIUnsv8

    solution 1 : using map

    int stringConstruction(string s) {
    map<char, int> mp;
    int r = 0;
    for(int i = 0; i < s.size(); i++) if(!mp[s[i]]){
        mp[s[i]] = 1; r++;
    }
    return r;
}

    solution 2 : using set

    int stringConstruction(string s) {
    set<char> se(s.begin(), s.end());
    return se.size();
}

  • illogicalsleepi1
     + 0 comments
    def stringConstruction(s):
     return len(list(set(s)))

    or

    def stringConstruction(s):
    bitmask = 0
    for char in s:
        bitmask |= (1 << (ord(char) - ord('a')))
    return bin(bitmask).count('1')

  • vuhuutuan0
     + 0 comments

    My answer with Typescript,

    function stringConstruction(s: string): number {
    // count number of unique character in [s]
    return new Set(s.split('')).size
}

  • highsoft_soi
     + 0 comments

    Perl:

    sub existElement {
    my $element = shift;
    my $arr = shift;
    for (my $i = 0; $i <= scalar(@$arr) - 1; $i++){
        if ($arr->[$i] eq $element) {
            return ($arr->[$i] eq $element) ? 1 : 0;
            last;
        }        
    }
}

sub stringConstruction {
    my @s = split("", shift);
    my $res = [0];
    my $cnt = 0;
    foreach my $el (@s) {
        if (!existElement($el, $res)) {
            push(@$res, $el);
            $cnt++;
        }        
    }
    return $cnt;
}

  • dorirgob
     + 0 comments

    Java:

    public static int stringConstruction(String s) {
  boolean[] freqs = new boolean[26]; // a to z
  int cost = 0;
  for (char c : s.toCharArray()) {
    // If the character hasn't been added before, increase the cost
    if (!freqs[c - 'a']) {
      freqs[c - 'a'] = true;
      cost++;
    }
  }
  return cost;
}