How do you solve Permutations on LeetCode?

Permutations (LeetCode #46) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n!) time complexity and O(n) space complexity. Key insight: Understanding backtracking is crucial for generating permutations.

What is the brute force solution for Permutations?

The brute force approach for Permutations is Brute Force, with O(n!) time complexity and O(n) space complexity. The brute-force approach generates all possible arrangements of the input array by swapping elements. This method is straightforward but can be inefficient as it explores all combinations. The optimal Optimal Solution approach improves this to O(n!).

What algorithm or data structure is used to solve Permutations?

Permutations uses the following concepts: Array, Backtracking. The recommended patterns to study are: Backtracking, Recursion.

What are the interview tips for Permutations?

Explain your thought process clearly while coding. Discuss time and space complexity as you write your solution. Practice writing test cases to validate your solution.

What are common mistakes when solving Permutations?

Not handling the base case correctly in recursion. Forgetting to backtrack properly, leading to incorrect results.

#46

Permutations

Medium

ArrayBacktrackingBacktrackingRecursion

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n!)	O(n!)
Space	O(n)	O(n)

💡

Intuition

Time O(n!)Space O(n)

The optimal solution leverages backtracking to efficiently explore all permutations without unnecessary swaps. This method is similar to the brute force but avoids redundant operations.

⚙️

Algorithm

4 steps

1Step 1: Initialize a result list to store permutations.
2Step 2: Create a backtracking function that takes the current permutation and a set to track used elements.
3Step 3: If the current permutation length equals the input array length, add it to the result.
4Step 4: Iterate through the input array, adding unused elements to the current permutation and marking them as used.

solution.py17 lines

1def permute(nums):
2    def backtrack(path, used):
3        if len(path) == len(nums):
4            result.append(path[:])
5            return
6        for i in range(len(nums)):
7            if used[i]: continue
8            used[i] = True
9            path.append(nums[i])
10            backtrack(path, used)
11            path.pop()
12            used[i] = False
13
14    result = []
15    used = [False] * len(nums)
16    backtrack([], used)
17    return result

ℹ

Complexity note: The time complexity remains O(n!) due to the number of permutations. The space complexity is O(n) for the recursion stack and the path storage.

1Understanding backtracking is crucial for generating permutations.
2Recognizing when to use a used array can optimize the solution.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.