给定一组n个整数，返回总和为0的k个元素的所有子集

Question

given a unsorted set of n integers, return all subsets of size k (ie each set has k unique elements) that sum to 0.

So I gave the interviewer the following solution ( which I studied on GeekViewpoint ). No extra space used, everything is done in place, etc. But of course the cost is a high time complexity of O(n^k) where k=tuple in the solution.

public void zeroSumTripplets(int[] A, int tuple, int sum) {
  int[] index = new int[tuple];
  for (int i = 0; i < tuple; i++)
    index[i] = i;
  int total = combinationSize(A.length, tuple);
  for (int i = 0; i < total; i++) {
    if (0 != i)
      nextCombination(index, A.length, tuple);
    printMatch(A, Arrays.copyOf(index, tuple), sum);
  }// for
}// zeroSumTripplets(int[], int, int)

private void printMatch(int[] A, int[] ndx, int sum) {
  int calc = 0;
  for (int i = 0; i < ndx.length; i++)
    calc += A[ndx[i]];
  if (calc == sum) {
    Integer[] t = new Integer[ndx.length];
    for (int i = 0; i < ndx.length; i++)
      t[i] = A[ndx[i]];
    System.out.println(Arrays.toString(t));
  }// if
}// printMatch(int[], int[], int)

But then she imposed the following requirements:

• must use hashmap in answer so to reduce time complexity
• Must absolutely -- ABSOLUTELY -- provide time complexity for general case
• hint when k=6, O(n^3)

She was more interested in time-complexity more than anything else.

Does anyone know a solution that would satisfy the new constraints?

---

EDIT:

Supposedly, in the correct solution, the map is to store the elements of the input and the map is then to be used as a look up table just as in the case for k=2 .

When the size of the subset is 2 (ie k=2 ), the answer is trivial: loop through and load all the elements into a map. Then loop through the inputs again this time searching the map for sum - input[i] where i is the index from 0 to n-1 , which would then be the answers. Supposedly this trivial case can be extended to where k is anything.

Answer 1

Since no-one else has made an attempt, I might as well throw in at least a partial solution. As I pointed out in an earlier comment, this problem is a variant of the subset sum problem and I have relied heavily on documented approaches to that problem in developing this solution.

We're trying to write a function subsetsWithSum(A, k, s) that computes all the k-length subsets of A that sum to s. This problem lends itself to a recursive solution in two ways:

1. The solution of subsetsWithSum(x ₁ ... x _n , k, s) can be found by computing subsetsWithSum(x ₂ ... x _n , k, s) and adding all the valid subsets (if any) that include x ₁ ; and
2. All the valid subsets that include element x _i can be found by computing subsetsWithSum(A - x _i , k-1, sx _i ) and adding x _i to each subset (if any) that results.

The base case for the recursion occurs when k is 1, in which case the solution to subsetsWithSum(A, 1, s) is the set of all single element subsets where that element is equal to s.

So a first stab at a solution would be

/**
 * Return all k-length subsets of A starting at offset o that sum to s.
 * @param A - an unordered list of integers.
 * @param k - the length of the subsets to find.
 * @param s - the sum of the subsets to find.
 * @param o - the offset in A at which to search.
 * @return A list of k-length subsets of A that sum to s.
 */
public static List<List<Integer>> subsetsWithSum(
        List<Integer> A,
        int k,
        int s,
        int o)
{
    List<List<Integer>> results = new LinkedList<List<Integer>>();

    if (k == 1)
    {
        if (A.get(o) == s)
            results.add(Arrays.asList(o));
    }
    else
    {
        for (List<Integer> sub : subsetsWithSum(A, k-1, s-A.get(o), o+1))
        {
            List<Integer> newSub = new LinkedList<Integer>(sub);
            newSub.add(0, o);
            results.add(0, newSub);
        }
    }

    if (o < A.size() - k)
        results.addAll(subsetsWithSum(A, k, s, o+1));

    return results;
}

Now, notice that this solution will often call subsetsWithSum(...) with the same set of arguments that it has been called with before. Hence, subsetsWithSum is just begging to be memoized .

To memoize the function, I've put the arguments k, s and o into a three element list which will be the key to a map from these arguments to a result computed earlier (if there is one):

public static List<List<Integer>> subsetsWithSum(
        List<Integer> A,
        List<Integer> args,
        Map<List<Integer>, List<List<Integer>>> cache)
{
    if (cache.containsKey(args))
        return cache.get(args);

    int k = args.get(0), s = args.get(1), o = args.get(2);
    List<List<Integer>> results = new LinkedList<List<Integer>>();

    if (k == 1)
    {
        if (A.get(o) == s)
            results.add(Arrays.asList(o));
    }
    else
    {
        List<Integer> newArgs = Arrays.asList(k-1, s-A.get(o), o+1);

        for (List<Integer> sub : subsetsWithSum(A, newArgs, cache))
        {
            List<Integer> newSub = new LinkedList<Integer>(sub);
            newSub.add(0, o);
            results.add(0, newSub);
        }
    }

    if (o < A.size() - k)
        results.addAll(subsetsWithSum(A, Arrays.asList(k, s, o+1), cache));

    cache.put(args, results);
    return results;
}

To use the subsetsWithSum function to compute all the k-length subsets that sum to zero, one can use the following function:

public static List<List<Integer>> subsetsWithZeroSum(List<Integer> A, int k)
{
    Map<List<Integer>, List<List<Integer>>> cache =
            new HashMap<List<Integer>, List<List<Integer>>> ();
    return subsetsWithSum(A, Arrays.asList(k, 0, 0), cache);
}

Regrettably my complexity calculating skills are a bit (read: very) rusty, so hopefully someone else can help us compute the time complexity of this solution, but it should be an improvement on the brute-force approach.

Edit: Just for clarity, note that the first solution above should be equivalent in time complexity to a brute-force approach. Memoizing the function should help in many cases, but in the worst case the cache will never contain a useful result and the time complexity will then be the same as the first solution. Note also that the subset-sum problem is NP-complete meaning that any solution has an exponential time complexity. End Edit.

Just for completeness, I tested this with:

public static void main(String[] args) {
    List<Integer> data = Arrays.asList(9, 1, -3, -7, 5, -11);

    for (List<Integer> sub : subsetsWithZeroSum(data, 4))
    {
        for (int i : sub)
        {
            System.out.print(data.get(i));
            System.out.print(" ");
        }

        System.out.println();
    }
}

and it printed:

9 -3 5 -11
9 1 -3 -7

Answer 2

I think your answer was very close to what they were looking for, but you can improve the complexity by noticing that any subset of size k can be thought of as two subsets of size k/2 . So instead of finding all subsets of size k (which takes O(n^k) assuming k is small), use your code to find all subsets of size k/2 , and put each subset in a hashtable, with its sum as the key.

Then iterate through each subset of size k/2 with a positive sum (call the sum S ) and check the hashtable for a subset whose sum is -S . If there is one then the combination of the two subsets of size k/2 is a subset of size k whose sum is zero.

So in the case of k=6 that they gave, you would find all subsets of size 3 and compute their sums (this will take O(n^3) time). Then checking the hashtable will take O(1) time for each subset, so the total time is O(n^3) . In general this approach will take O(n^(k/2)) assuming k is small, and you can generalize it for odd values of k by taking subsets of size floor(k/2) and floor(k/2)+1 .

Answer 3

@kasavbere -

Recently a friend had one of those harrowing all-day interviews for a C++ programming job with Google. His experience was similar to yours.

It inspired him to write this article - I think you might enjoy it:

The Pragmatic Defense