Xor profit problem

#1


This is my code. works fine for custom test cases but shows wrong answer upon final submission

#2

bro
u are implementing a quite brute force approach

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,

Examples 1:
L = 8 R = 20
L ^ R = (01000) ^ (10100) = (11100)
Now as L ^ R is of form (1xxxx) we
can get maximum XOR as (11111) by
choosing A and B as 15 and 16 (01111
and 10000)

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; 
}

understand this this is a very efficient solution

#3

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#4

https://ide.codingblocks.com/s/219762 in this code why did we do val=val>>1. supoose the value of val is 101 then it left shifting it will make it 1010 which we don not want . we only want to count the number of bits in the number so right shifting operator should be used but the code is still working so i am a bit confused

#5

@tanjuljain19 hey, val>>1 would right shift the value ,so if val is 101, val>>1 will be 10.
What you are saying is left shift operator, and it is <<.
If this resolves your doubt mark it as resolved.