Permutations LeetCode Solution

Problem – Permutations LeetCode Solution

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]]


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

Permutations LeetCode Solution in C++

class Solution {
    vector<vector<int> > permute(vector<int> &num) {
	    vector<vector<int> > result;
	    permuteRecursive(num, 0, result);
	    return result;
    // permute num[begin..end]
    // invariant: num[0..begin-1] have been fixed/permuted
	void permuteRecursive(vector<int> &num, int begin, vector<vector<int> > &result)	{
		if (begin >= num.size()) {
		    // one permutation instance
		for (int i = begin; i < num.size(); i++) {
		    swap(num[begin], num[i]);
		    permuteRecursive(num, begin + 1, result);
		    // reset
		    swap(num[begin], num[i]);

Permutations LeetCode Solution in Python

def permute(self, nums):
    res = []
    self.dfs(nums, [], res)
    return res
def dfs(self, nums, path, res):
    if not nums:
        # return # backtracking
    for i in xrange(len(nums)):
        self.dfs(nums[:i]+nums[i+1:], path+[nums[i]], res)

Permutations LeetCode Solution in Java

    public List<List<Integer>> permute(int[] nums) {

        if (nums == null || nums.length == 0)
            return new ArrayList<>();

        List<List<Integer>> finalResult = new ArrayList<>();
        permuteRecur(nums, finalResult, new ArrayList<>(), new boolean[nums.length]);
        return finalResult;

    private void permuteRecur(int[] nums, List<List<Integer>> finalResult, List<Integer> currResult, boolean[] used) {

        if (currResult.size() == nums.length) {
            finalResult.add(new ArrayList<>(currResult));

        for (int i = 0; i < nums.length; i++) {
            if (used[i])
            used[i] = true;
            permuteRecur(nums, finalResult, currResult, used);
            used[i] = false;
            currResult.remove(currResult.size() - 1);

Permutations LeetCode Solution in JavaScript

var permute = function(letters) {
    let res = [];
    dfs(letters, [], Array(letters.length).fill(false), res);
    return res;

function dfs(letters, path, used, res) {
    if (path.length == letters.length) {
        // make a deep copy since otherwise we'd be append the same list over and over
    for (let i = 0; i < letters.length; i++) {
        // skip used letters
        if (used[i]) continue;
        // add letter to permutation, mark letter as used
        used[i] = true;
        dfs(letters, path, used, res);
        // remove letter from permutation, mark letter as unused
        used[i] = false;
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