Maximizing XOR HackerRank Solution


Maximizing XOR HackerRank Solution
Source : https://www.hackerrank.com/challenges/maximizing-xor



Source : https://www.hackerrank.com/challenges/maximizing-xor


Solution


// Karthikalapati.blogspot.com
// To maximize A xor B, we want A and B to differ as much as possible at every bit index.
// We first find the most significant bit that we can force to differ by looking at L and R.
// For all of the lesser significant bits in A and B, we can always ensure that they differ
// and still have L <= A <= B <= R. Our final answer will be the number represented by all
// 1s starting from the most significant bit that differs between A and B
// Example:
// L = 10111
// R = 11100
// _X___ <-- that's most significant differing bit
// 01111 <-- here's our final answer
//
// Notice that we never directly calculate the values of A and B
static int maximizingXor(int L, int R) {
int xored = L ^ R;
int significantBit = 31 - Integer.numberOfLeadingZeros(xored);
int result = (1 << (significantBit + 1)) - 1;
return result;
}
// Discuss on HackerRank: https://www.hackerrank.com/challenges/maximizing-xor/forum/comments/284317

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