Maximum XOR : DP Bitmask

#1

I am not able to access the editorial. Can you please share the tutorial of the questin ? Thank you.

#2

Maximum XOR

Given two integers l and r , your task is to find the maximum xor of two integers both of which lying between l and r inclusive.
Input Format

The first line contains l and r separated by a space.
Constraints

l, r<=10^18
Output Format

Output is a single number denoting the maximum xor
Sample Input

79242383109441603 533369389165030783

Sample Output

576460752303423487

#3

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#4

I didn’t understood the use of " if ((((l>>poz)^(r>>poz))&1LL)==1LL) ret += br;" as given in the editorial. What is this condition doing ? Can you please explain ?

#5

I mean what is the basic idea behind approaching this problem ?

#6

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. It is explained below with some examples,
int maxXORInRange( int L, int R)

{

// get xor of limits

int LXR = L ^ R;

// loop to get msb position of L^R

int msbPos = 0;

while (LXR)

{

msbPos++;

LXR >>= 1;

}

// construct result by adding 1,

// msbPos times

int maxXOR = 0; 

    int two  =1;

while (msbPos--)

{

maxXOR += two;

two <<= 1;

}

return maxXOR;

}

#7

so basically, get the position of the most significant bit of L^R
now, add 1 for each position of the position obtained

#8

why this actually works is quite simple, if a state has been reached of 1xxx in L^R there must definitely be an intermediate state of 1111

#9

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