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

Solutions to Chapter 8 | Recursion
8.3 Write a method that returns all subsets of a set.
pg 64
SOLUTION
We should first have some reasonable expectations of our time and space complexity. How
many subsets of a set are there? We can compute this by realizing that when we generate a
subset, each element has the “choice” of either being in there or not. That is, for the first element,
there are 2 choices. For the second, there are two, etc. So, doing 2 * 2 * ... * 2 n times
gives us 2^n subsets. We will not be able to do better than this in time or space complexity.
Approach #1: Recursion
This is a great problem to implement with recursion since we can build all subsets of a set using
all subsets of a smaller set. Specifically, given a set S, we can do the following recursively:
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Let first = S[0]. Let smallerSet = S[1, ..., n].
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Compute all subsets of smallerSet and put them in allsubsets.
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For each subset in allsubsets, clone it and add first to the subset.
The following code implements this algorithm:
1 ArrayList getSubsets(ArrayList set,
2 int index) {
3 ArrayList allsubsets;
4 if (set.size() == index) {
5 allsubsets = new ArrayList();
6 allsubsets.add(new ArrayList()); // Empty set
7 } else {
8 allsubsets = getSubsets(set, index + 1);
9 int item = set.get(index);
10 ArrayList moresubsets =
11 new ArrayList();
12 for (ArrayList subset : allsubsets) {
13 ArrayList newsubset = new ArrayList();
14 newsubset.addAll(subset); //
15 newsubset.add(item);
16 moresubsets.add(newsubset);
17 }
18 allsubsets.addAll(moresubsets);
19 }
20 return allsubsets;
21 }
Approach #2: Combinatorics
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When we’re generating a set, we have two choices for each element: (1) the element is
1 7 1
Cracking the Coding Interview | Concepts and Algorithms