35 lines
1.2 KiB
Java
35 lines
1.2 KiB
Java
// Author: Rodney Shaghoulian
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// Github: github.com/RodneyShag
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// HackerRank: hackerrank.com/RodneyShag
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import java.util.Scanner;
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// To maximize A xor B, we want A and B to differ as much as possible at every bit index.
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// We first find the most significant bit that we can force to differ by looking at L and R.
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// For all of the lesser significant bits in A and B, we can always ensure that they differ
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// and still have L <= A <= B <= R. Our final answer will be the number represented by all
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// 1s starting from the most significant bit that differs between A and B
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// Example:
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// L = 10111
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// R = 11100
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// _X___ <-- that's most significant differing bit
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// 01111 <-- here's our final answer
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//
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// Notice that we never directly calculate the values of A and B
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public class Solution {
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public static void main(String[] args) {
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Scanner scan = new Scanner(System.in);
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int L = scan.nextInt();
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int R = scan.nextInt();
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scan.close();
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int xored = L ^ R;
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int significantBit = 31 - Integer.numberOfLeadingZeros(xored);
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int result = (1 << (significantBit + 1)) - 1;
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System.out.println(result);
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}
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}
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