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.

Approach: Backtracking with Used Array

Algorithm

1. Use a recursive backtracking function that builds a permutation one element at a time
2. Track which elements have already been used with a boolean[] used array
3. When the current permutation has the same length as nums, add a copy to the result
4. Otherwise, iterate through every element, skip used ones, add it, recurse, then backtrack

Time & Space Complexity

• Time Complexity: O(n · n!) - there are n! permutations, each costing O(n) to build
• Space Complexity: O(n) - recursion depth plus the used array (excluding the output)

Java Implementation

import java.util.ArrayList;
import java.util.List;

public class Permutations {

    /**
     * Return all possible permutations of the array.
     * @param nums Array of distinct integers
     * @return List of all permutations
     */
    public static List<List<Integer>> permute(int[] nums) {
        List<List<Integer>> result = new ArrayList<>();
        backtrack(nums, new boolean[nums.length], new ArrayList<>(), result);
        return result;
    }

    private static 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);

            used[i] = false;
            current.remove(current.size() - 1);
        }
    }

    // Test method
    public static void main(String[] args) {
        int[] nums1 = {1, 2, 3};
        System.out.println("Input: nums = [1,2,3]");
        System.out.println("Output: " + permute(nums1));
        // Expected: [[1,2,3],[1,3,2],[2,1,3],[2,3,1],[3,1,2],[3,2,1]]

        int[] nums2 = {0, 1};
        System.out.println("Input: nums = [0,1]");
        System.out.println("Output: " + permute(nums2));
        // Expected: [[0,1],[1,0]]

        int[] nums3 = {1};
        System.out.println("Input: nums = [1]");
        System.out.println("Output: " + permute(nums3));
        // Expected: [[1]]
    }
}

Key Points

1. Distinct Elements: No duplicate handling is needed since all integers are unique
2. Used Tracking: The used array prevents reusing an element within one permutation
3. Copy on Complete: Add new ArrayList<>(current) to avoid sharing mutable references
4. Backtracking Pattern: Mark used -> add -> recurse -> unmark -> remove
5. Small Constraint: nums.length <= 6, so at most 720 permutations