Leetcode permutation12/10/2023 ![]() ![]() Step3 then we set a variable k to the nums and declare a list p which is equal to nums except for k. Step2: make an empty list, then loop for in a range of length of nums. if the length of nums is 1 then we return nums as a sublist. ![]() Step1: We will solve this problem recursively, Firstly, we will add base case i.e. Output: ,]ĭef permute(self, nums: List) -> List]: Given an array nums of distinct integers, return all the possible permutations. In this tutorial, we will solve a leetcode problem Permutations in python. Reverse words in a String Leetcode Solution Middle of the Linked List Leetcode Solution ![]() Remove Nth Node From End Leetcode SolutionĬontainer With Most Water Leetcode Solutionĭetermine if String Halves Are Alike Leetcode Solution Remove Duplicates from Sorted Array Leetcode Solution Letter Combinations of a Phone Number Leetcode Solution Temp = S.upper() if S.islower() else S.Longest Substring Without Repeating Characters Leetcode Solution Return letterCasePermutation(self, S, i=0, results=None, call_self=None): Return f(*args,**kwargs, **params_items, call_self=f) Your solution (taken as a function instead of as a method) with this method applied to it: def lazy_curry_by_ref(results=list,**other_params): If want to keep the format of your solution (the recursive function is side effect only), then we need to have the decorator specifically know that parameter and return it as a result. The one to use here will be one lazily constructing parameters. If the structure of the outer function is simple (or you are through enough in how you write your decorator) then we can write a simple decorator for such functions to use. The idea is that each time we change a character, we can construct a new result for each old result, by simply slicing in the changed character on all those new results. Results +=, new_char, result]) for result in results] ![]() New_char = char.upper() if char.islower() else char.lower() It might not extend to all the similar problems you want to, but it could extend to some of them: def letterCasePermutation(S): Here is a very simple way to solve this specific problem without recursion and with only a single function. Here's an elegant solution using generators to create the permutations lazily: class Solution: # the list returned by the recursive call. # if it's not a letter, just prepend it to all the answers in First as a lowercase letterįor tail in self.letterCasePermutation(s): # if s is a letter, prepend it to all the answers in the list # base case the answer is a list with an empty string Then process the list returned by the recursive call to add back in the first element. If the sequence isn't empty, take off the first element of the sequence and recursively call the function with the rest of the list. If you don't like having the extra function or method "cluttering" the namespace, you can nest it inside the driver like this: class Solution:Ī common way to recursively process a string, list, or other sequence is to define the base case as an empty sequence which returns a list with with the base case answer. It's fairly common to use an auxiliary parameter as an index or accumulator in recursive functions. My question is, is there a nicer way I can go about accomplishing this task? One which can perform recursion only using one function? Self.letterCasePermutationAux("".join(temp), i+1) Temp = S.upper() if S.islower() else S.lower() class Solution(object):ĭef letterCasePermutationAux(self, S, i=0): I choose this question as an example but really any recursive question could act as a stand in here. LetterCasePermutation and one which acts as the worker ( letterCasePermutationAux) where the only difference is the arguments (I would usually have something like a counter and a reference to a vector to hold my answers). One which acts as the "interface" is this case I find that with alot of these leetcode question that I want to use recursion for I find myself making two versions of a function. My question pertains mostly to the pattern that I am following. Lowercase or uppercase to create another string. Given a string S, we can transform every letter individually to be ![]()
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