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;
}

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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

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

: 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://)