90, Subsets II

I Problem

Given an integer array nums that may contain duplicates, return all possible subsets (the power set).

The solution set must not contain duplicate subsets. Return the solution in any order.

Example 1
Input: nums = [1, 2, 2]
Output: [[], [1], [1, 2], [1, 2, 2], [2], [2, 2]]

Example 2
Input: nums = [0]
Output: [[], [0]]

Constraints

• 1 <= nums.length <= 10
• -10 <= nums[i] <= 10

Related Topics

• Array
• Backtracking
• Bit Manipulation

II Solution

Approach 1: Backtracking

pub fn subsets_with_dup(mut nums: Vec<i32>) -> Vec<Vec<i32>> {
    nums.sort_unstable();

    const DFS: fn(usize, &[i32], &mut Vec<i32>, &mut Vec<Vec<i32>>) = 
        |i, nums, subset, res| {
            res.push(subset.clone());
            if i == nums.len() {
                return;
            }

            for j in i..nums.len() {
                if j > i && nums[j] == nums[j - 1] {
                    continue;
                }

                subset.push(nums[j]);
                DFS(j + 1, nums, subset, res);
                subset.pop();
            }
        };
    let mut res = vec![];

    DFS(0, &nums, &mut vec![], &mut res);

    res
}

interface QuadrConsumer<A, B, C, D> {
    void accept(A a, B b, C c, D d);
}

QuadrConsumer<Integer, int[], List<Integer>, List<List<Integer>>> dfs =
        (i, nums, subset, res) -> {
            res.add(new ArrayList<>(subset));
            if (i == nums.length) {
                return;
            }

            for (int j = i; j < nums.length; j++) {
                if (j > i && nums[j] == nums[j - 1]) {
                    continue;
                }

                subset.addLast(nums[j]);
                this.dfs.accept(j + 1, nums, subset, res);
                subset.removeLast();
            }
        };

public List<List<Integer>> subsetsWithDup(int[] nums) {
    Arrays.sort(nums);
    List<List<Integer>> res = new ArrayList<>();

    this.dfs.accept(0, nums, new ArrayList<>(), res);

    return res;
}

Approach 2: Iteration

pub fn subsets_with_dup(mut nums: Vec<i32>) -> Vec<Vec<i32>> {
    nums.sort_unstable();
    let mut res = vec![vec![]];
    let (mut pre_len, mut len) = (0, 0);

    for i in 0..nums.len() {
        (pre_len, len) = (len, res.len());

        if i > 0 && nums[i] != nums[i - 1] {
            pre_len = 0;
        }

        for j in pre_len..len {
            let mut subset = res[j].clone();
            subset.push(nums[i]);
            res.push(subset);
        }
    }

    res
}

public List<List<Integer>> subsetsWithDup(int[] nums) {
    Arrays.sort(nums);
    List<List<Integer>> res = new ArrayList<>() {{
        this.add(new ArrayList<>());
    }};
    int preSize = 0;
    int size = 0;

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

        if (i > 0 && nums[i] != nums[i - 1]) {
            preSize = 0;
        }

        for (int j = preSize; j < size; j++) {
            ArrayList<Integer> subset = new ArrayList<>(res.get(j));
            subset.add(nums[i]);
            res.add(subset);
        }
    }

    return res;
}