Cracking the Coding Interview, 4 Edition - 150 Programming Interview Questions and Solutions

Solutions to Chapter 8 | Recursion
in the set (the “yes” state) or (2) the element is not in the set (the “no” state). This means
that each subset is a sequence of yesses / nos—e.g., “yes, yes, no, no, yes, no”
»»
This gives us 2^n possible subsets. How can we iterate through all possible sequences
of “yes” / “no” states for all elements? If each “yes” can be treated as a 1 and each “no” can
be treated as a 0, then each subset can be represented as a binary string.
»»
Generating all subsets then really just comes down to generating all binary numbers
(that is, all integers). Easy!
1 ArrayList getSubsets2(ArrayList set) {
2 ArrayList allsubsets =
3 new ArrayList();
4 int max = 1 0) {
10 if ((k & 1) > 0) {
11 subset.add(set.get(index));
12 }
13 k >>= 1;
14 index++;
15 }
16 allsubsets.add(subset);
17 }
18 return allsubsets;
19 }
CareerCup.com
1 7 2