The number of 1's in the range 0..X (X is positive) is easy to calculate (Can you get a simple recurrence which does this in O(log X) ?) Another observation is that the number of 1's in -X is equal to the number of 0's in ~(-X) = X - 1. Using this, it is easy to calculate the answer for negative ranges as well.
Aish aishraj
aishraj
/ Hint.md
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November 13, 2011 09:57
— forked from yuvipanda/Hint.md
2's Compliment Solution (InterviewStreet CodeSprint Fall 2011)
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