Permutations II

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Given a collection of numbers, nums, that might contain duplicates, return all possible unique permutations in any order.

Example 1:

Input: nums = [1,1,2]

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

Example 2:

Input: nums = [1,2,3]

Output: [[1,2,3],[1,3,2],[2,1,3],[2,3,1],[3,1,2],[3,2,1]]

Constraints:

1 <= nums.length <= 8
-10 <= nums[i] <= 10

Solution

Let's fill in the logic for backtracking1:

path_length: indicates the number of elements in path
used: an array storing the status of whether an element is used.
is_leaf: path_length == len(nums), when path_length is the same as nums's length.
get_edges and is_valid: all elements that are unused (used[i]==Flase).

In the implementation, we use an array used to store the whether an element has been used or not. Everytime we use a number, we add it into the path and we mark it used (used[idx] = True). Since there are duplicate values, let's apply methods in Deduplication to deduplicate the results. we want to sort the nums array so that duplicate elements are consecutive, and only proceed when the current element is the first of its occurrance.

Implementation

def permuteUnique(self, nums: List[int]) -> List[List[int]]:
    used = [False] * len(nums)
    ans = []
    nums.sort()
    
    def dfs(path_length, path):
        if path_length == len(nums):
            ans.append(path[:])
            return
        
        for i in range(0, len(nums)):
            if used[i]: continue
            if(i > 0 and nums[i] == nums[i-1] and not used[i-1]): continue
            path.append(nums[i])
            used[i] = True
            dfs(path_length + 1, path)
            used[i] = False
            path.pop()
    
    dfs(0, [])
    return ans

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Discover Your Strengths and Weaknesses: Take Our 2-Minute Quiz to Tailor Your Study Plan:

What's the output of running the following function using the following tree as input?

1def serialize(root):
2    res = []
3    def dfs(root):
4        if not root:
5            res.append('x')
6            return
7        res.append(root.val)
8        dfs(root.left)
9        dfs(root.right)
10    dfs(root)
11    return ' '.join(res)
12

1import java.util.StringJoiner;
2
3public static String serialize(Node root) {
4    StringJoiner res = new StringJoiner(" ");
5    serializeDFS(root, res);
6    return res.toString();
7}
8
9private static void serializeDFS(Node root, StringJoiner result) {
10    if (root == null) {
11        result.add("x");
12        return;
13    }
14    result.add(Integer.toString(root.val));
15    serializeDFS(root.left, result);
16    serializeDFS(root.right, result);
17}
18

1function serialize(root) {
2    let res = [];
3    serialize_dfs(root, res);
4    return res.join(" ");
5}
6
7function serialize_dfs(root, res) {
8    if (!root) {
9        res.push("x");
10        return;
11    }
12    res.push(root.val);
13    serialize_dfs(root.left, res);
14    serialize_dfs(root.right, res);
15}
16

1 2 3 6 x x

1 2 3 x x x 6 x x

1 2 6 x x x 3 x x

I don't know

Permutations II

Solution

Implementation

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