Maximizing XOR Discussions | Algorithms | HackerRank
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  2. Algorithms
  3. Bit Manipulation
  4. Maximizing XOR
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Maximizing XOR

  • snaran
     + 0 comments

    My way had the same concept as your picture: to calculate the highest order bit that is different and subtract one from it, though my way has a loop that you avoided by doing an ^.

    static int maxXor(int l, int r) {
        int lBit = Integer.highestOneBit(l);
        int rBit = Integer.highestOneBit(r);
        if (lBit == rBit) {
            if (lBit == 0) {
                return 0;
            } else {
                int newL = l & (~lBit);
                int newR = r & (~lBit);
                return maxXor(newL, newR);
            }
        } else {
            int max = Math.max(lBit, rBit);
            return (max << 1) -1;
        }
    }