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  1. Prepare
  2. Tutorials
  3. 30 Days of Code
  4. Day 29: Bitwise AND
  5. Discussions

Day 29: Bitwise AND

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427 Discussions

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  • jaykipkerich
     + 0 comments

    PYTHON 3 SOLUTION

    python:

        def bitwiseAnd(N, K):
        if K - 1 | K <= N:
                return K -1
        return K-2

  • jonlle19
     + 0 comments

    JavaScript/TypeScript

    function bitwiseAnd(N: number, K: number): number {
    let max = 0;
    let aANDb;

    for (let a = 1; a < N; a++) {
        for (let b = a + 1; b <= N; b++) {
            aANDb = a & b;
            if ((aANDb) < K && (aANDb) > max) {
                max = aANDb;
            }
        }
    }

    return max;
}

  • illogicalsleepi1
     + 1 comment
    def bitwiseAnd(N, K):
    if (K - 1) | K <= N:
        return K - 1
    return K - 2

    the max value before k will be k-1 is it exists print k-1 else k-2 the next largest value before k-1

    • borisvdb09
       + 0 comments
      public static int bitwiseAnd(int N, int K)
{
    if (((K - 1) | K) <= N)
    {
        return K - 1;
    }
    return K - 2;
}

          Corrected for C#

  • shyamjee969678
     + 0 comments

    Java 8

    public static int bitwiseAnd(int N, int K) {
    // Write your code here
        int ret=0;
        for(int A=1;A<N;A++)
        {
            for(int B=A+1;B<=N;B++)
            {
                if((A&B)<K)
                {
                    ret=Math.max(ret, A&B);
                }
            }

        }
    return ret;
}

  • n4m774r_d
     + 0 comments

    one liner preventing exceeding time limit:

    def bitwiseAnd(N, K):
    return max((a+1)&(K-1) for a in range(N) if K-1 != a+1)

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