LeetCode – Subsets II (Java)

 

Given a set of distinct integers, S, return all possible subsets.

Note:
Elements in a subset must be in non-descending order.
The solution set must not contain duplicate subsets.
For example,
If S = [1,2,3], a solution is:

[
  [3],
  [1],
  [2],
  [1,2,3],
  [1,3],
  [2,3],
  [1,2],
  []
]

Thoughts

Comparing this problem with Subsets can help better understand the problem.

Java Solution

public ArrayList<ArrayList<Integer>> subsetsWithDup(int[] num) {
	if (num == null)
		return null;
 
	Arrays.sort(num);
 
	ArrayList<ArrayList<Integer>> result = new ArrayList<ArrayList<Integer>>();
	ArrayList<ArrayList<Integer>> prev = new ArrayList<ArrayList<Integer>>();
 
	for (int i = num.length-1; i >= 0; i--) {
 
		//get existing sets
		if (i == num.length - 1 || num[i] != num[i + 1] || prev.size() == 0) {
			prev = new ArrayList<ArrayList<Integer>>();
			for (int j = 0; j < result.size(); j++) {
				prev.add(new ArrayList<Integer>(result.get(j)));
			}
		}
 
		//add current number to each element of the set
		for (ArrayList<Integer> temp : prev) {
			temp.add(0, num[i]);
		}
 
		//add each single number as a set, only if current element is different with previous
		if (i == num.length - 1 || num[i] != num[i + 1]) {
			ArrayList<Integer> temp = new ArrayList<Integer>();
			temp.add(num[i]);
			prev.add(temp);
		}
 
		//add all set created in this iteration
		for (ArrayList<Integer> temp : prev) {
			result.add(new ArrayList<Integer>(temp));
		}
	}
 
	//add empty set
	result.add(new ArrayList<Integer>());
 
	return result;
}

Feed the method [1,2,3] the following will be result at each iteration.

[2]
[2][2,2]
[2][2,2][1,2][1,2,2][1]
Get [] finally.

Related posts:

  1. LeetCode – Permutations (Java)
  2. LeetCode – Basic Calculator II (Java)
  3. Leetcode – Binary Tree Postorder Traversal (Java)
  4. LeetCode – Subsets (Java)
Category >> Algorithms  
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  • Govardhan

    I think this video has a very simple solution
    https://www.youtube.com/watch?v=XtJ-dpLmK4Y

  • KoKo

    This solution is for array contains duplicates numbers:
    “Given a collection of integers that might contain duplicates, nums, return all possible subsets.”

  • Dexter

    public static void subsets(int[] arr){

    for(int i=1; i <= arr.length; i++){

    ps(arr, i, 0);

    }

    }

    public static void print(int[] arr, int e){

    for(int i=0; i<=e; i++){

    System.out.print(" "+ arr[i]);

    }

    System.out.println("");

    }

    public static void ps(int[] arr, int length, int s){

    for(int j = length-1; j < arr.length; j++){

    swap(arr,length-1, j);

    print(arr, length-1);

    swap(arr,length-1, j);

    }

    }

    public static void swap(int[] arr, int i, int j){

    int temp = arr[i];

    arr[i] = arr[j];

    arr[j] = temp;

    }

  • Burhan COKCA

    Its kind of easier if ArrayList apis are allowed. Whenever you modify the result just check if the list exist in the result.

    public List<List> subsetsWithDup(int[] nums) {

    List<List> result = new ArrayList<List>();

    result.add(new ArrayList());

    Arrays.sort(nums);

    for (int i = 0; i < nums.length; i ++) {

    int curSize = result.size();

    for (int j = 0; j < curSize; j ++) {

    List cur = new ArrayList(result.get(j));

    cur.add(nums[i]);

    if(!result.contains(cur))

    result.add(cur);

    }

    }

    return result;

    }

  • LOts2Learn

    Hey What exactly is difference between the 2 questions subset and subset II ?? Both the questions are exactly the same.

  • Xing

    Hi,

    Why I can’t use ” result.addAll(prev); ” as what you did in Subsets?

    Thanks,