Permutations II LeetCode Programming Solutions | LeetCode Problem Solutions in C++, Java, & Python [💯Correct]

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In this post, you will find the solution for the Permutations II in C++, Java & Python-LeetCode problem. We are providing the correct and tested solutions to coding problems present on LeetCode. If you are not able to solve any problem, then you can take help from our Blog/website.

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Link for the Problem – Permutations II– LeetCode Problem

Permutations II– LeetCode Problem

Problem:

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

Permutations II– LeetCode Solutions

Permutations II in C++:

class Solution {
 public:
  vector<vector<int>> permuteUnique(vector<int>& nums) {
    vector<vector<int>> ans;

    sort(begin(nums), end(nums));
    dfs(nums, vector<bool>(nums.size()), {}, ans);

    return ans;
  }

 private:
  void dfs(const vector<int>& nums, vector<bool>&& used, vector<int>&& path,
           vector<vector<int>>& ans) {
    if (path.size() == nums.size()) {
      ans.push_back(path);
      return;
    }

    for (int i = 0; i < nums.size(); ++i) {
      if (used[i])
        continue;
      if (i > 0 && nums[i] == nums[i - 1] && !used[i - 1])
        continue;
      used[i] = true;
      path.push_back(nums[i]);
      dfs(nums, move(used), move(path), ans);
      path.pop_back();
      used[i] = false;
    }
  }
};

Permutations II in Java:

class Solution {
  public List<List<Integer>> permuteUnique(int[] nums) {
    List<List<Integer>> ans = new ArrayList<>();

    Arrays.sort(nums);
    dfs(nums, new boolean[nums.length], new ArrayList<>(), ans);

    return ans;
  }

  private void dfs(int[] nums, boolean[] used, List<Integer> path, List<List<Integer>> ans) {
    if (path.size() == nums.length) {
      ans.add(new ArrayList<>(path));
      return;
    }

    for (int i = 0; i < nums.length; ++i) {
      if (used[i])
        continue;
      if (i > 0 && nums[i] == nums[i - 1] && !used[i - 1])
        continue;
      used[i] = true;
      path.add(nums[i]);
      dfs(nums, used, path, ans);
      path.remove(path.size() - 1);
      used[i] = false;
    }
  }
}

Permutations II in Python:

class Solution:
  def permuteUnique(self, nums: List[int]) -> List[List[int]]:
    ans = []
    used = [False] * len(nums)

    def dfs(path: List[int]) -> None:
      if len(path) == len(nums):
        ans.append(path)
        return

      for i, num in enumerate(nums):
        if used[i]:
          continue
        if i > 0 and nums[i] == nums[i - 1] and not used[i - 1]:
          continue
        used[i] = True
        dfs(path + [num])
        used[i] = False

    nums.sort()
    dfs([])

    return ans