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  • + 0 comments

    Here is my one line Python solution!

    def twoStrings(s1, s2):
        return "NO" if len(set(s1).intersection(set(s2))) == 0 else "YES"
    
  • + 0 comments

    how to solve this proplem with kmp

  • + 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";
    }
    
  • + 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"
    
  • + 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"