90. Subsets II

Given a collection of integers that might contain duplicates, nums, return all possible subsets (the power set).

Note: The solution set must not contain duplicate subsets.

Example:

Input: [1,2,2]
Output:
[
  [2],
  [1],
  [1,2,2],
  [2,2],
  [1,2],
  []
]

	class Solution {
	public List<List<Integer>> subsetsWithDup(int[] nums) {
	List<List<Integer>> ans = new ArrayList<List<Integer>>();
	// Sort to track duplicates
	Arrays.sort(nums);

	List<Integer> prefix = new ArrayList<Integer>();
	ans.add(prefix);

	int start = 0, existing = 0;
	for (int i = 0; i < nums.length; i++) {
	// If current number is duplicate of prev, start from previous start, instead of 0
	start = i > 0 && nums[i] == nums[i - 1] ? existing : 0;
	// Iterate upto end of ans, add current element to each existing entry
	existing = ans.size();
	for (int j = start; j < existing; j++) {
	prefix = new ArrayList<Integer>(ans.get(j));
	prefix.add(nums[i]);
	ans.add(prefix);
	}
	}

	return ans;
	}
	}

	class Solution {
	public List<List<Integer>> subsetsWithDup(int[] nums) {
	List<List<Integer>> ans = new ArrayList<List<Integer>>();
	// Sort to track duplicates
	Arrays.sort(nums);

	List<Integer> prefix = new ArrayList<Integer>();

	dfs(nums, 0, prefix, ans);

	return ans;
	}

	private void dfs(int[] nums, int current, List<Integer> prefix, List<List<Integer>> ans) {
	ans.add(new ArrayList<Integer>(prefix));

	for (int i = current; i < nums.length; i++) {
	// Skip duplicates.. i > current to avoid out of bounds
	if (i > current && nums[i] == nums[i - 1]) {
	continue;
	}

	prefix.add(nums[i]);
	dfs(nums, i + 1, prefix, ans);
	prefix.remove(prefix.size() - 1);
	}
	}
	}

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