Discussion on Maximizing XOR Challenge

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496 Discussions

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2 weeks ago+ 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;
}

3 weeks ago+ 0 comments

def maximizingXor(l, r):
    m=0
    for i in range(l,r+1):
        for j in range(l,r+1):
            if(m>(i^j)):
                pass
            else:
                m=i^j
    return m

3 weeks ago+ 0 comments

I optimized XOR maximization algorithm reduces complexity from O(log r) to O(log l) by iterating only over the smaller number L. It uses bitwise comparison to maximize the XOR value, flipping bits of l or r as needed while respecting the range [L,R] . The final XOR result is calculated using a fast power function, ensuring minimal operations and efficiency. This method ensures the maximum XOR value is computed while preserving the range constraints.

code with more detalied explain: https://github.com/MohamedHussienMansour/Maximum-XOR-problem-with-O-log-L-

1 month ago+ 0 comments

Lots of naive solutions in the comments. Here is a Python solution:

def maximizingXor(l, r):
    count = 0
    while r != l:
        r >>= 1
        l >>= 1
        count += 1
    return (1 << count) - 1

2 months ago+ 0 comments

Here is my Python solution!

def binary(num):
    remainders = []
    while num != 0 :
        remainders.insert(0, str(num % 2))
        num //= 2
    return "".join(remainders)

def xor(num1, num2):
    newnum = []
    if len(num1) > len(num2):
        num2 = (len(num1) - len(num2)) * "0" + num2
    elif len(num2) > len(num1):
        num1 = (len(num2) - len(num1)) * "0" + num1
    for i in range(len(num1)):
        if num1[i] == num2[i]:
            newnum.append("0")
        else:
            newnum.append("1")
    return "".join(newnum)

def decimal(num):
    newnum = 0
    for i in range(len(num)):
        newnum += int(num[-i - 1]) * 2 ** i
    return newnum

def maximizingXor(l, r):
    values = []
    for num1 in range(l, r + 1):
        for num2 in range(num1 + 1, r + 1):
            values.append(decimal(xor(binary(num1), binary(num2))))
    return max(values)

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