Download A Collection of Bit Programming Interview Questions solved by Dr Antonio Gulli PDF

By Dr Antonio Gulli

ISBN-10: 1495330729

ISBN-13: 9781495330728

Bits is the second one of a chain of 25 Chapters dedicated to algorithms, challenge fixing, and C++ programming. This ebook is ready low point bit programming

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Read or Download A Collection of Bit Programming Interview Questions solved in C++ PDF

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Represent an integer with variable length encoding using delta encoding Solution Code 28. Compute the average with no division Solution 1. Given an unsigned int, swap the bits in odd and even positions Solution Assume that an unsigned int is 32bits and that we swap the bits in position with those in position,. In order to select all the even bits we can AND with bitmask 0xAAAAAAAAAA, which is a 32 bit number with even bits set (0xA is decimal 10, 1010 binary). For selecting all the odd bits we can AND with bitmask 0×55555555, which is a number with all even bits sets (0x5 is decimal5, 0101 in binary).

Code int complement2(int n) { n = ~n; n = n + 1; return n; } 14. Isolate the rightmost bit set to 1 Solution Let’s build some example. Suppose that , if we compute in two’s complement we get . Therefore selects the rightmost bit set to 1. 15. Create a mask for trailing zeros Solution If we are able to identify the rightmost bit set to 1, then we can create the mask for trailing zeros just subtracting 1. The solution is therefore 16. Compute parity for a 32 bit number Solution Given a number, we can drop the last bit with the expression.

Result : result; } 21. Convert a number from base b1 to base b2 Solution We can generalize the solution presented in the previous exercise. empty()) return s; bool negative = (s[0] == '-'); int n = 0, reminder; for (unsigned int i = (negative ? size(); ++i) n = n * b1 + (isdigit(s[i]) ? push_back(reminder >= 10 ? end()); return result; } 22. Given a set S, compute the powerset of S Solution The powerset of is the set of all the subsets of S. For instance given the set ,the powerset is . If we represent the presence (absence) of element in with a bit set to 1 (respectively, 0), then we can build the powerset by generating all the bitmasks from 0 to where is the size of.

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