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Solution in O(g) time, where g is the MSF of (l xor r). For a 32 bit unsigned integer, worst-case g would be 32 operations.
int maxXor(int l, int r) { int a = l ^ r; int max = 0; while (a != 0) { max |= a; a = a >> 1; } return max; }
Maximizing XOR
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Solution in O(g) time, where g is the MSF of (l xor r). For a 32 bit unsigned integer, worst-case g would be 32 operations.