# Mastering LeetCode: Generating All Permutations of an Array

## A Step-by-Step Guide to Solving the Permutations Problem Using Backtracking in Python

## Introduction

Generating permutations is a fundamental problem in computer science, often appearing in coding interviews and algorithm challenges. Understanding how to efficiently generate permutations can provide insights into various combinatorial problems and improve your problem-solving skills.

In this article, we'll explore the "Permutations" problem on LeetCode and solve it using a backtracking approach.

### LeetCode Problem Statement:

Given an array

`nums`

of distinct integers, returnall the possible permutations. You can return the answer inany 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`

areunique.

## Understanding Permutations

Permutations refer to all possible arrangements of a set of elements. For an array of `n`

distinct elements, there are `n!`

(n factorial) possible permutations. Generating these permutations is useful in various applications, including solving puzzles, optimizing routes, and more.

## Solution Approach

To solve the permutations problem, we'll use a backtracking approach. Backtracking is an algorithmic technique for solving recursive problems by trying to build a solution incrementally and removing solutions that fail to satisfy the problem constraints.

## Detailed Solution Breakdown

### Step 1: Function Definition

We'll start by defining the main function `permute`

that will take the list of integers `nums`

as input and return a list of lists containing all possible permutations.

### Step 2: Helper Function

We'll define a helper function `backtrack`

that will be used to generate permutations by swapping elements. This function will take the current position `start`

as an argument.

### Step 3: Swapping Elements

In the `backtrack`

function, we'll loop through the elements starting from the `start`

index to the end of the list. For each index `i`

, we'll swap the elements at indices `start`

and `i`

, recursively call `backtrack`

with the next position `start + 1`

, and then swap the elements back to their original positions.

### Code Implementation

Here's the complete Python code for generating permutations:

```
from typing import List
def permute(nums: List[int]) -> List[List[int]]:
def backtrack(start: int):
if start == len(nums):
result.append(nums[:])
for i in range(start, len(nums)):
nums[start], nums[i] = nums[i], nums[start] # swap
backtrack(start + 1)
nums[start], nums[i] = nums[i], nums[start] # swap back
result = []
backtrack(0)
return result
# Test cases
print(permute([1, 2, 3])) # Output: [[1,2,3],[1,3,2],[2,1,3],[2,3,1],[3,1,2],[3,2,1]]
print(permute([0, 1])) # Output: [[0,1],[1,0]]
print(permute([1])) # Output: [[1]]
```

### Explanation of the Code

**Function Definition**:- The
`permute`

function takes a list`nums`

and returns a list of lists containing all permutations.

- The
**Backtracking Function**:We define a nested function

`backtrack`

that takes the current position`start`

as an argument.If

`start`

equals the length of`nums`

, we've reached a complete permutation, so we append a copy of`nums`

to the`result`

list.

**Swapping Elements**:We loop through the elements starting from the

`start`

index.For each index

`i`

, we swap the elements at indices`start`

and`i`

.We recursively call

`backtrack`

with`start + 1`

.After the recursive call, we swap the elements back to their original positions.

**Result Storage**:We initialize an empty list

`result`

to store all permutations.We call

`backtrack`

starting from index 0.Finally, we return the

`result`

list containing all permutations.

## Time Complexity Analysis

The time complexity of this solution is O(n * n!), where `n`

is the length of the input list. This is because there are `n!`

permutations, and we need to copy each permutation of length `n`

to the result list. Given the constraint (1 <= nums.length <= 6), this complexity is acceptable.

## Common Questions and Pitfalls

### Q1: What is the maximum length of the input list?

- A1: The length of the input list is between 1 and 6.

### Q2: Are duplicate elements allowed in the input list?

- A2: No, all integers in the input list are unique.

### Q3: Why use backtracking for this problem?

- A3: Backtracking efficiently explores all possible permutations by making choices and undoing them when necessary.

## FAQ Section

### What is the maximum length of the input list?

The input list can have a length between 1 and 6.

### Are duplicate elements allowed in the input list?

No, all integers in the input list must be unique.

### Why use backtracking for this problem?

Backtracking is efficient for generating all permutations by making choices and backtracking to previous states when necessary.

## Conclusion

Generating permutations is a classic problem that can be efficiently solved using a backtracking approach. By understanding and implementing the solution step-by-step, you can enhance your problem-solving skills and tackle similar combinatorial problems with confidence.

Practice more permutation problems to deepen your understanding and improve your coding abilities.