Permutations LeetCode Solution | Easy Approach

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Permutations | Given an array nums of distinct integers, return all the possible permutations. You can return the answer in any order.

Example 1:

Input: nums = [1,2,3]
Output: [[1,2,3],[1,3,2],[2,1,3],[2,3,1],[3,1,2],[3,2,1]]

Example 2:

Input: nums = [0,1]
Output: [[0,1],[1,0]]

Example 3:

Input: nums = [1]
Output: [[1]]

Constraints:

  • 1 <= nums.length <= 6
  • -10 <= nums[i] <= 10
  • All the integers of nums are unique.

Permutations Solutions

Time: O(nn!)
Space: O(nn!)

C++

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

    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;
      used[i] = true;
      path.push_back(nums[i]);
      dfs(nums, move(used), move(path), ans);
      path.pop_back();
      used[i] = false;
    }
  }
};

Java

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

    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;
      used[i] = true;
      path.add(nums[i]);
      dfs(nums, used, path, ans);
      path.remove(path.size() - 1);
      used[i] = false;
    }
  }
}

Python

class Solution:
  def permute(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.copy())
        return

      for i, num in enumerate(nums):
        if used[i]:
          continue
        used[i] = True
        path.append(num)
        dfs(path)
        path.pop()
        used[i] = False

    dfs([])
    return ans

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