Permutations

Backtracking/Permutations

Problem

Given an array nums of distinct integers, return all the possible permutations. You can return the answer in any order.

For example:

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]]

Input: nums = [1]
Output: [[1]]

Thought Process

• We are essentially picking and placing the number in a seperate list and recursing on the new list without that number present

Solution

class Solution:
    def permute(self, nums: List[int]) -> List[List[int]]:
        soln = []
        self.helper(nums, [], soln)
        return soln
    
    def helper(self, nums, permList, soln):
        
        if len(nums) == 0:
            soln.append(list(permList))
            return
        for i in range(len(nums)):
            permList.append(nums[i])
            remaining = nums[:i] + nums[i+1:]
            self.helper(remaining,permList,soln)
            permList.pop()

Key Points

• In our for-loop, we separate the specific number from the rest of the numbers, append the seperated number to our new list, and recurse on the new seperated list

Time Complexity

• Time: $O(n*n!)$
• Space: $O(n * n!)$