#Permutation vs combination code

When this holds true, everything after input (the inversion point) cannot be rearranged to produce a larger permutation, since the elements are in descending order (elements right of inversion point). We know we have reached the inversion point when input > input. How do we find the inversion point? We iterate over the input, starting at the last element.The array will now be arranged such that it is ordered as the next permutation - meaning we can return the result.Because everything after the inversion point will inherently be sorted in descending order - we can just reverse this section of the array instead.Then we sort in ascending order, the elements after the inversion point. Once we find the inversion point, we swap the element at that position with the next element in the array that is larger than it.The inversion point is the position in the array after which the elements stop descending (remember we said that if it continually descends, no further permutations are possible).So, if we can find the inversion point of an input - we can calculate the next permutation.For example, no next permutation exists for For any given input that is in descending order, no next permutation is possible.A better way is to first recognize a few key traits that allow us to form a solution:.Time complexity would be O(n!) and space complexity would be O(n). This is obviously less than ideal: what if the given input is the last possible permutation? That's a lot of wasted effort.The brute force approach is to generate all possible permutations, and when we hit a permutation that is the same as the input, return the permutation that follows.For this problem we only need to find a single permutation, and it must be the next permutation that would come in lexicographical order.

This problem is slightly different to generating all possible permutations.Problem statement: Given an input array of integers, find the next possible permutation and return the result.Article explaining the algorithm and solution.See PrintPermutations() and FindPermutationsNoDupes().įinding the next permutation of an integer array

#Permutation vs combination code

As always, the code for these two variations is at the end of this post.

As there are n! permutations our time complexity is O(n! n), and our space complexity is O(n!).

The only difference is that we use a modified version of the input string (letters and their counts) and decrement the count of a letter when we have used it (when count = 0 we skip the letter).

Once this is done, the algorithm looks almost identical.

Sets are represented using the notation.

This means every set has an empty subset.

A subset is any combination of elements from a set.

A set with no elements is still a set and is known as an empty set.

Explanation of sets and subsets (slideshow).

I encourage you to stick with each problem even if it doesn't click the first time around. It just happens that the best way to really learn this material was to solve some problems after covering theory. I don't expect to get back into this style of post again until I have finished the remaining curriculum. This post is longer than usual & very heavy on problem solving, instead of the usual 'learning about a DS/Algo, then writing an example'.