![]() ![]() See the image above for clarification.ĭo this for all the cases and it will generate all possible permutations of the given array. 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. ![]() Backtrack and fix another element at index l and recur for index l+1 to r.To generate all the permutations of an array from index l to r, fix an element at index l and recur for the index l+1 to r.Complexity Analysis for Leetcode problem Permutations ExamplesĤ 1 2 3 Algorithm for Leetcode problem PermutationsĪll the permutations can be generated using backtracking.Explanation for Leetcode problem Permutations.Algorithm for Leetcode problem Permutations.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. 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
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