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  1. Prepare
  2. Algorithms
  3. Strings
  4. Two Strings
  5. Discussions

Two Strings

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2281 Discussions

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  • alban_tyrex
     + 0 comments

    Here is my simple c++ solution, video explanation here : https://youtu.be/xwLiYvExM6I

    string twoStrings(string s1, string s2) {
    map<char, int>mp;
    for(int i = 0; i < s1.size(); i++) mp[s1[i]] = 1;
    for(int i = 0; i < s2.size(); i++){
        if(mp[s2[i]]) return "YES";
    }
    return "NO";
}

  • psaumet_dvp
     + 0 comments

    Version 1:

    def twoStrings(s1, s2):
    for char in s1:
        if char in s2:
            return "YES"
    return "NO"

    Version 2:

    def twoStrings(s1, s2):
    return "YES" if any(char in s2 for char in s1) else "NO"

    Version 3:

    def twoStrings(s1, s2):
    return "YES" if set(s1) & set(s2) else "NO"

  • illogicalsleepi1
     + 0 comments
    def twoStrings(s1, s2):
    x=list(set(s1))
    y=list(set(s2))
    for i in x:
        for j in y:
            if i==j:
                return 'YES'
    return 'NO'

    or

    def twoStrings(s1, s2):
    set1 = set(s1)
    set2 = set(s2)
    return "YES" if set1 & set2 else "NO"

    or

    def twoStrings(s1, s2):
    def bitmask(s):
        bitmask = 0
        for char in s:
            bitmask |= (1 << (ord(char) - ord('a')))
        return bitmask
    bm1 = bitmask(s1)
    bm2 = bitmask(s2)
    return "YES" if bm1 & bm2 else "NO"

  • vuhuutuan0
     + 0 comments

    My answer with Typescript, I am a bit surprised by this question, is it too simple?

    function twoStrings(s1: string, s2: string): string {
    for (let c of s1) if (s2.includes(c)) return 'YES'
    return 'NO'
}

  • highsoft_soi
     + 0 comments

    Perl:

    sub twoStrings {
    my @str1 = split("", shift);
    my @str2 = split("", shift);
    my %h;
    for (my $k = 0; $k <= scalar(@str1) - 1; $k++) {
        $h{$str1[$k]} = 0;
    }
    for (my $i = 0; $i <= scalar(@str2) - 1; $i++) {
        if (exists($h{$str2[$i]})) {
            return "YES";
        }
        
    }
    return "NO";
}