A decreasing array is the last permutation of the array, and the next one is the reverse of it.Some facts regarding lexicographic permutations: This is unfair question to ask on interviews as it requires a fair amount of pre hand knowledge regarding lexicographic permutations to be able to solve this in a short amount of time.īut the question is interesting in that it explores flow of numbers in permutations, which is good to understand and hence I have included it here. See solution to the problem on github here: * The replacement must be in place and use only constant extra memory. * If such an arrangement is not possible, it must rearrange it as the lowest possible order (i.e., sorted in ascending order). % or returns when the argument is already the last possible permutation.* Implement next permutation, which rearranges numbers into the lexicographically next greater permutation of numbers. % Computes and returns the next lexicographical permutation of the given vector, (Spoilers at the bottom.) Annotated code (Java) boolean nextPermutation( int array) *) MATLAB function result = nextperm(arr) Now if you truly understand the algorithm, here’s an extension exercise for you: Design the algorithm for stepping backward to the previous lexicographical permutation. Thus, computing every permutation requires Θ( n! × n) run time. Overall, this algorithm to compute the next lexicographical permutation has Θ( n) worst-case time complexity, and Θ(1) space complexity. Thus we obtain the sequence (0, 1, 3, 0, 2, 3, 5), which is the next permutation that we wanted to compute.įind largest index i such that array array. In fact, we can avoid sorting and simply reverse the suffix, because the replaced element respects the weakly decreasing order. weakly increasing) order because we increased the prefix, so we want to make the new suffix as low as possible. (Note that if the suffix has multiple copies of the new pivot, we should take the rightmost copy – this plays into the next step.)įinally, we sort the suffix in non-decreasing (i.e. (The prefix is everything in the sequence except the suffix.) In the example, we end up with the new prefix (0, 1, 3) and new suffix (5, 3, 2, 0). If we swap the pivot with the smallest element in the suffix that is greater than the pivot, then the prefix is minimally increased. So some element in the suffix is greater than the pivot. the entire sequence is non-increasing – then this is already the last permutation.) The pivot is necessarily less than the head of the suffix (in the example it’s 5). Secondly, look at the element immediately to the left of the suffix (in the example it’s 2) and call it the pivot. Also note that such a suffix has at least one element, because a single element substring is trivially non-increasing.) (Note that we can identify this suffix in Θ( n) time by scanning the sequence from right to left. This suffix is already the highest permutation, so we can’t make a next permutation just by modifying it – we need to modify some element(s) to the left of it. In our example, the suffix with this property is (5, 3, 3, 0). In fact, there is no need to change the second element either, which brings us to the next point.įirstly, identify the longest suffix that is non-increasing (i.e. For example, there is no need to change the first element from 0 to 1, because by changing the prefix from (0, 1) to (0, 2) we get an even closer next permutation. Just like when we count up using numbers, we try to modify the rightmost elements and leave the left side unchanged. The key observation in this algorithm is that when we want to compute the next permutation, we must “increase” the sequence as little as possible. We will use the sequence (0, 1, 2, 5, 3, 3, 0) as a running example. We will use concrete examples to illustrate the reasoning behind each step of the algorithm. The simple and fast algorithm for performing this is what will be described on this page. It turns out that the best approach to generating all the permutations is to start at the lowest permutation, and repeatedly compute the next permutation in place. Moreover, if we insist on manipulating the sequence in place (without producing temporary arrays), then it’s difficult to generate the permutations in lexicographical order. But this method is tricky because it involves recursion, stack storage, and skipping over duplicate values. We could pick the first element, then recurse and pick the second element from the remaining ones, and so on. The naive way would be to take a top-down, recursive approach. Suppose we have a finite sequence of numbers like (0, 3, 3, 5, 8), and want to generate all its permutations. Next lexicographical permutation algorithm Introduction
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