![]() ![]() Time complexity will be O(3n), which came from O(3+3²+3³++3n). ![]() You can customize the available time you have, difficulty, topics, etc. Solution: this is not exactly backtracking problem, however, we recursively add the next digit to the previous combinations. Length of num_calls = 64, which != n * n! = 4 * (4*3*2*1) = 96. Grind 75 is a better version of Blind 75 which goes beyond 75 questions. The key to solve the problem is still the backtracking algorithm. The problem is different from the previous permutation problem on the condition that the input array can contain duplicates. ![]() Some people say its worst case O(n * n!), but looking at the len of num_calls doesn't verify this claim. Given a zero-based permutation nums ( 0-indexed ), build an array ans of the same length where ans i nums nums i for each 0 < i < nums.length and return it. Permutations II - LeetCode 4.12 (155 votes) Solution Overview As the name of the problem suggests, this problem is an extension of the Permutation problem. I can't make sense of any of the answers that I have seen thus far for the time and space complexity of this solution. class Solution:īacktrack(combo + ], rem + rem) I used both iteration and recursion to solve this particular problem. Here is my backtracking solution for the problem, where I added the num_calls variable to keep track of the number of times that the backtrack function is called recursively. The following code is what I used to solve the permutation problem. The question is as follows: Given a collection of distinct integers, return all possible permutations. ![]()
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