XOR Profit problem

#1

How to do this problem in O(n) time?

#2

hello @rahul_bansal
An efficient solution is to consider pattern of binary values from L to R. We can see that first bit from L to R either changes from 0 to 1 or it stays 1 i.e. if we take the XOR of any two numbers for maximum value their first bit will be fixed which will be same as first bit of XOR of L and R itself. After observing the technique to get first bit, we can see that if we XOR L and R, the most significant bit of this XOR will tell us the maximum value we can achieve i.e. let XOR of L and R is 1xxx where x can be 0 or 1 then maximum XOR value we can get is 1111 because from L to R we have all possible combination of xxx and it is always possible to choose these bits in such a way from two numbers such that their XOR becomes all 1.

code

#include <iostream>
using namespace std;
int main()
{
    int x,y;
    cin>>x>>y;
    int num = x^y;
    int msb=0;
    while(num!=0) {
        num=num>>1;
        msb++;
    }
    int result = 1;
    while(msb--) {
        result=result<<1;
    }
    cout<<result-1;

    return 0;
}

time complexiy-> O(maximum no of bits in x or y)

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#3

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