Although i have solved the question. But for optimization i was reading solution of this problem. But it didn’t get what explanation and code want to explain. Can you please explain me?
Problem on XOR profit problem
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,
so u are basically finding the last set bit for the xor of the L and the R value
the last set bit will give u the position for a number which occurs b/w the range and on setting all the bits uptill the position of MSB will give u the maximum profit
so we find the position of leftmost set bit of XOR of a ,b
and then set all the bits upto the msb which provides the required answer
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