How to efficiently count the number of pairs having a xor b equal to m where 1<=a<=n , 1<=b<=n and n varies 2 to 2e5 and m also varies 1 to n
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Auto comment: topic has been updated by nkb-xyz (previous revision, new revision, compare).
Hint: If a and m are fixed, can you find a value of b, such that a ^ b = m?
yeah b = m^a and complexity will be o(n) right?? got it ... it this final like cant we go beyond that like logarithmic or like lesser than this
My bad, you still need to check if the corresponding b is <= n, so n — 1 is not correct, in general it is less than that. So I failed in my attempt to make it better than O(n)
i think its not possible... i gave up on my approach
Good job! How many possible values of a are there?
(Don't overthink the complexity — what you are shooting for is O(n) )
For linear time we can just iterate a from 1 to n and check if the corresponding b is in 1 to n or nah. I think OP is asking for a sublinear algorithm.