I am studying leetcode questions for an upcoming interview and I dont understand a part of the backtracking algorithm. Specifically I don't understand how the code after the recursive call "backtracking(i+1) will be accessed by Python. This code is supposed to find all possible subsets of an array of numbers, for example the input [1,2,3] would result in [1], [2], [3], [1,2] ,[1,3], [2,] (not sure if I got all of them). the backtracking code that I'm looking at on a tutorial is this class Solution: def subsets (self, nums: List[int])-> List[List[int]]: n= len(nums) res, sol = [], [] def backtracking: if i == n res.append(sol[:]) return #dont pick nums backtrack(i+1) #pick nums sol.append(nums) backtrack(i+1) sol.pop(); backtrack(0) return res I dont understand how the compiler reaches the code after the first backtrack(i+1). Wont that call keep the code in a constant loop that either sends us back to the method if i<n or returns us to backtrack(0) if i ==n? Continue reading...
