LeetCode #47: Permutations II (C/C++).

medium

source: https://leetcode.com/problems/permutations-ii/
C/C++ Solution to LeetCode problem 47. Permutations II.

Problem

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

Examples

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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

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

Solution

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This can be solved using Backtracking (see more backtracking problems)

For this problem, to avoid having duplicated permutations (because we have repeated numbers):

• First we sort the array.
• We need to keep track of the numbers we add to a solution/permutation.
• If the number to add to a permutation is the same number than the previous one, and the previous one has not been added, then we skip it.

class Solution {
private:
  vector<vector<int>> result;

  void permutations(vector<int>& nums, vector<int>& tmpSol, vector<bool>& proc) {
    if (tmpSol.size() == nums.size()) {
      result.push_back(tmpSol);
      return;
    }

    for (int i=0; i<nums.size(); i++) {
      if (proc[i] || (i>0 && nums[i] == nums[i-1] && !proc[i-1])) 
        continue;
      proc[i] = true;
      tmpSol.push_back(nums[i]);
      permutations(nums, tmpSol, proc);
      tmpSol.pop_back();
      proc[i] = false;
    }
  }
public:
  vector<vector<int>> permuteUnique(vector<int>& nums) {
    vector<bool> proccessed(nums.size(), false);
    vector<int> tmpSol;
    sort(nums.begin(), nums.end());
    permutations(nums, tmpSol, proccessed);

    return result;
  }
};