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# 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 = 
Output: []
``````

Constraints:

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

### Permutations LeetCode Solution in C++

``````class Solution {
public:
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
result.push_back(num);
return;
}

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

``````# DFS
def permute(self, nums):
res = []
self.dfs(nums, [], res)
return res

def dfs(self, nums, path, res):
if not nums:
res.append(path)
# 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) {
return;
}

for (int i = 0; i < nums.length; i++) {
if (used[i])
continue;
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
res.push(Array.from(path));
return;
}
for (let i = 0; i < letters.length; i++) {
// skip used letters
if (used[i]) continue;
// add letter to permutation, mark letter as used
path.push(letters[i]);
used[i] = true;
dfs(letters, path, used, res);
// remove letter from permutation, mark letter as unused
path.pop();
used[i] = false;
}
}
``````
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