Transform to Chessboard


  • 0

    Click here to see the full article post


  • 0
    P

    Thanks a lot!

    int ones = Integer.bitCount(k1 & Nones);

    Is "k1 & Nones" necessary here?


  • 0
    L

    if (!(count.get(k1) == N/2 && count.get(k2) == (N+1)/2) &&
    !(count.get(k2) == N/2 && count.get(k1) == (N+1)/2))
    return -1;

    can be simplified to :
    if (Math.abs(count.get(k1)-count.get(k2))>1){
    return -1;
    }


  • 0
    D

    nice solution.

    i think if each row and each column is half ones or half zeros (or with an extra 1/0 when odd length) then it is impossible for there to be more or less than 2 distinct rows/columns. therefore those checks are redundant


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