03.Permutations.md

46. Permutations

Medium

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.

Approach

- we have n choices initially, after each selection we can't take that element again
- take a mark array to keep track of what elements fron the nums array which are already select in the current path
for(int i = 0; i < nums.length; i++) {
    if(mark[i] == false) {
        mark[i] = true;
        tmp.add(nums[i]);
        backtrack(nums, mark, tmp, ans);
        tmp.remove(tmp.size()-1);
        mark[i] = false;
    }
}

Solution

class Solution {
    public List<List<Integer>> permute(int[] nums) {
        boolean[] mark = new boolean[nums.length];
        List<List<Integer>> ans = new ArrayList<>();
        backtrack(nums, mark, new ArrayList<>(), ans);
        return ans;
    }
    
    private void backtrack(int[] nums, boolean[] mark, List<Integer> tmp, List<List<Integer>> ans) {
        if(tmp.size() == nums.length) {
            ans.add(new ArrayList<>(tmp));
            return;
        }
        
        for(int i = 0; i < nums.length; i++) {
            if(mark[i] == false) {
                mark[i] = true;
                tmp.add(nums[i]);
                backtrack(nums, mark, tmp, ans);
                tmp.remove(tmp.size()-1);
                mark[i] = false;
            }
        }
    }
}

Complexity Analysis

- Time Complexity: O(N^N)
- Space Complexity: O(N)