Maximizing XOR Discussions | Algorithms | HackerRank
We use cookies to ensure you have the best browsing experience on our website. Please read our cookie policy for more information about how we use cookies.
  1. Prepare
  2. Algorithms
  3. Bit Manipulation
  4. Maximizing XOR
  5. Discussions

Maximizing XOR

Sort by

recency

|

490 Discussions

|

  • 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;
}

  • abhayad268207
     + 0 comments

    Simple and Easy Solution in Python If helps you please upvote to help others

    maximum=0
    for i in range(l,r+1):
        for j in range(i,r+1):
            print(i,j)
            val=i^j
            maximum= max(maximum,val)
    return maximum

  • 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