Answer by Łukasz Zephyr for Bits counting algorithm (Brian Kernighan) in an...
This question is really about meaning of N in big O notation, not complexity of algorithm.N means size of the data. But in case where where "data" is a single number you need to define what you...
View ArticleAnswer by PengOne for Bits counting algorithm (Brian Kernighan) in an integer...
This algorithm goes through as many iterations as there are set bits. So if we have a 32-bit word with only the high bit set, then it will only go once through the loop. In the worst case, it will pass...
View ArticleAnswer by The_Sympathizer for Bits counting algorithm (Brian Kernighan) in an...
There are floor(lg(N)) + 1 significant bits in N -- that's a base-2 logarithm. The number of 1 bits in n is at most this. So the time will have asymptotic upper bound O(lg(N)) = O(log(N)).
View ArticleBits counting algorithm (Brian Kernighan) in an integer time complexity
Can someone explains why Brian Kernighan's algorithm takes O(log N) to count set bits (1s) in an integer. A simple implementation of this algorithm is below (in JAVA)int count_set_bits(int n){ int...
View ArticleAnswer by Uri London for Bits counting algorithm (Brian Kernighan) in an...
I would like to add to this discussion, a true O(Log N) algorithm, where N is the number of bits (32 or 64):int HammingWeight(DWORD dw) { dw = (dw & 0x55555555) + ((dw & 0xaaaaaaaa) >>...
View ArticleAnswer by wiesiu_p for Bits counting algorithm (Brian Kernighan) in an...
This is the matter of number representation only.I normal life we use decimal number system which allows us to write any number with digits 0-9. It allow us to write N with log(N) with base 10 number...
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