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]] Explanation: All 3! = 6 permutations are returned.

Example 2

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

Solution Approach

We use a visited array approach. At each position in the current permutation we try every element that has not been used yet. When the current list reaches size n, we have a complete permutation.

Maintain a boolean used[] array to track which elements are in the current path.
At each recursive call, iterate over all elements; skip those already used.
Add the element, mark it used, recurse, then unmark and remove (backtrack).
When current.size() == nums.length, record the permutation.

Java

import java.util.*;

class Solution {
    public List<List<Integer>> permute(int[] nums) {
        List<List<Integer>> result = new ArrayList<>();
        boolean[] used = new boolean[nums.length];
        backtrack(nums, used, new ArrayList<>(), result);
        return result;
    }

    private void backtrack(int[] nums, boolean[] used,
                           List<Integer> current,
                           List<List<Integer>> result) {
        if (current.size() == nums.length) {
            result.add(new ArrayList<>(current));
            return;
        }
        for (int i = 0; i < nums.length; i++) {
            if (used[i]) continue;
            used[i] = true;
            current.add(nums[i]);
            backtrack(nums, used, current, result);
            current.remove(current.size() - 1);
            used[i] = false;
        }
    }
}

Complexity Analysis:

Time complexity: O(n * n!). n! permutations, each takes O(n) to copy.
Space complexity: O(n) recursion depth plus O(n * n!) for the output.