![]() ![]() For more about this, see A Project for 2021. 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. Can you solve this real interview question Validate Stack Sequences - Given two integer arrays pushed and popped each with distinct values, return true if this could have been the result of a sequence of push and pop operations on an initially empty stack, or false otherwise. 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. See the image above for clarification.ĭo this for all the cases and it will generate all possible permutations of the given array.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. Fixing the second position automatically fixes the third position. In the first column of second-level 1 is fixed at the first position, in the second column 2 is fixed at the first position and in the third column 3 is fixed at the first position.Īfter fixing an element at the first position, fix an element at the second position, consider the case in the second level and the first column, that is,, 1 is fixed at the first position, so we have 2 choices for the second position that is either 2 or 3. The image below the second level represents this situation. Explanation for Leetcode problem Permutationsįix an element in the first position, we have three choices 1, or 2, or 3. Repeat the above steps to generate all the permutations. ![]()
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