Understand what the interviewer is asking for by using test cases and questions about the problem.
- Established a set (2-3) of test cases to verify their own solution later.
- Established a set (1-2) of edge cases to verify their solution handles complexities.
- Have fully understood the problem and have no clarifying questions.
- Have you verified any Time/Space Constraints for this problem?
Be sure that you clarify the input and output parameters of the problem:
Run through a set of example cases:
HAPPY CASE Input: nums = [1,2,3] Output: [[1,2,3],[1,3,2],[2,1,3],[2,3,1],[3,1,2],[3,2,1]] Input: nums = [0,1] Output: [[0,1],[1,0]] EDGE CASE Input: nums =  Output: []
Match what this problem looks like to known categories of problems, e.g. Linked List or Dynamic Programming, and strategies or patterns in those categories.
Plan the solution with appropriate visualizations and pseudocode.
General Idea: The concept is to resolve the problem recursively by thinking of finding permutations as mainly swapping values in a list. Look at it as set theory that there are N! permutations of a list of size N. We also know that the permutations are going to be the leaves of the tree, which means we will have N! leaves. In order to get to each one of those leaves, we had to go through N calls. That's O(N*N!). Again a little more than the total number of nodes because some nodes are shared among more than one path.
1. Pick one element in the list, and remove it from the list of available integers 2. Generate all the permutations for the remaining list and append them to the first element 3. To do that we keep track of current which is the currently available integers, and permutation, which is the currently generated permutation.
⚠️ Common Mistakes
Implement the code to solve the algorithm.
class Solution: def permute(self, nums: List[int]) -> List[List[int]]: res =  n = len(nums) def solve(n, perm): if n==0: res.append(perm.copy()) return None for i in range(len(nums)): # To do that we keep track of current which is the currently available integers, and permutation, which is the currently generated permutation. if nums[i] in perm: continue # Generate all the permutations for the remaining list and append them to the first element perm.append(nums[i]) solve(n-1, perm) # Pick one element in the list, and remove it from the list of available integers perm.pop() solve(n, ) return res
Review the code by running specific example(s) and recording values (watchlist) of your code's variables along the way.
Evaluate the performance of your algorithm and state any strong/weak or future potential work.
1 x 2 + 2 x 3 + 6 x 4 + ... + (n-1)! x n = 1! x 2 + 2! x 3 + 3! x 4 + ... + (n-1)! x n = 2! + 3! + 4! + ... + n!