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

Maximizing XOR

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489 Discussions

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  • alban_tyrex
     + 0 comments

    Here is my easy and straitghforward solution, you can watch video here : https://youtu.be/pCU8W_ofvDg

    int maximizingXor(int l, int r) {
    int result = 0;
    for(int a = l; a <= r; a++){
        for(int b = a; b<= r; b++){
            result = max(result, a ^ b);
        }
    }
    return result;
}

  • jmcparland
     + 0 comments

    Haskell

    module Main where

import Data.Bits (xor)

solve :: Int -> Int -> Int
solve l r = maximum [a `xor` b | a <- [l .. r - 1], b <- [a .. r]]

main :: IO ()
main = do
    l <- readLn :: IO Int
    r <- readLn :: IO Int
    print $ solve l r

  • Umesh94
     + 0 comments

    ![]( def maximizingXor(l, r): new_results = 0 for i in range(l, r + 1): for j in range(l,r + 1): old_result = i ^ j if old_result >= new_results: new_results = old_result return new_resultshttps://)

  • leoisaac_lameda
     + 0 comments
    from operator import xor
def maximizingXor(l, r):
    # Write your code here
    if l == r:
        return 0
    maxVal = 0
    a = l
    b = a + 1
    while b <= r:
        maxVal = max(maxVal, xor(a, b))
        b += 1
        a += 1
    return maxVal

  • birot
     + 0 comments

    Python, one-liner:

    return 2**(1+int(math.floor(math.log(l^r, 2.0))))-1