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 For more about this, see A Project for 2021.Longest Substring Without Repeating Characters Leetcode Solution I’m posting some LeetCode model solutions this year. Permutations II on GitHub LeetCode Model Solutions Note that some swaps are skipped in this example, since the input contains a duplicate 1 element: i=0, j=0 Here’s what happens when the input is 1,1,2. So it seems reasonable that this would cover every possible swap.Īs with last week’s recursive problem, instrumentation can help illuminate the flow of control in a recursive algorithm. So (ignoring duplicates), we swap position 0 with positions 0, 1, …, n-1, position 1 with 1, 2, …, n-1, and so on until i=n. And j covers every position from i to the last element. Notice that i covers every position in nums from 0 to the last element. The key is to make sure we use all possible swap positions (except where the swap would have no effect because the source and destination elements are the same). Why does this process work? Since permutations are arrangements of the input values, it makes sense that we could generate these arrangements by swapping. If we sent a reference, we would overwrite the result of previous swaps. This is how we generate new permutations. One implementation detail: When we make the recursive call, we need to send a copy of nums, not just a reference to nums. To keep the code simple, we also do the swap in this case, though it has no effect since we’re swapping an element with itself. As a special case, we also make the recursive call when i=j. If they’re different, we swap them and recursively start the process again at the next starting position i+1. For each j (the current position), we’ll check the values at nums and nums. The iteration loop on j goes from i (the starting position) to the end of the current permutation, nums. We’ll use a combination of iteration and recursion. If it is, the current permutation is done, so we can add it to our list of results, and return: if i is past the end of the current permutationĪdd the current permutation to the result j: The current position in the current permutationįor the recursive base case, we check if our starting position is past the end of the current permutation.i: The starting position in the current permutation.nums: One unique permutation of the input list.In the backtrack function, we’ll need four variables, the first three of which are passed as parameters: The 0 argument in the backtrack call means start the process at position 0 in nums (the input list). Then we call the recursive function and return the result. This is a simple way to avoid duplicate permutations. The first step is to sort this input array. The main function accepts a list of integers. The solution uses two functions: the main function, which sets up the recursion and returns the result, and the recursive backtracking function. The canonical algorithm for finding permutations is recursive backtracking, so we’ll take that approach. For this problem, the input list can have duplicate elements (that’s the extra requirement compared to the previous problem, LeetCode 46: Permutations), but the answer must not have any duplicate permutations. For example, the three elements can be arranged into $3!=6$ unique permutations:, ,, ,, and. Problem Statement: Given a list of integers that may contain duplicates, return all possible unique permutations of those integers, in any order.Ī permutation is an arrangement of elements.
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