Added 9 solutions. Updated Readme

2017-04-09 19:12:53 -05:00
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+//     Author: Rodney Shaghoulian
+//     Github: github.com/rshaghoulian
+// HackerRank: hackerrank.com/rshaghoulian
+
+import java.util.Scanner;
+
+// 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
+
+public class Solution {
+    public static void main(String [] args) {
+        Scanner scan = new Scanner(System.in);
+        int L = scan.nextInt();
+        int R = scan.nextInt();
+        scan.close();
+        
+        int xored  = L ^ R;
+        int significantBit = 31 - Integer.numberOfLeadingZeros(xored);
+        int result = (1 << (significantBit + 1)) - 1;
+
+        System.out.println(result);
+    }
+}