Union-Disjoint Sets and Number theory (Knowledge of graphs and Connected components will be helpful).
Given an array A of length N, We create graph consisting of N vertices, each corresponding to an element in array A. Undirected edge (a,b) is added if and only if gcd(A_a, A_b) is one. That is, A[a] and A** are co-prime.
We need to check whether this graph is connected or not, and if it isn’t connected, Change the minimum number of elements in the array to make the graph connected, if the graph is always constructed in similar fashion.
Note, 2 \leq A* \leq 50 should hold at all times even after modifications.
SUPER QUICK EXPLANATION
- Use union disjoint sets to check whether the graph is already connected. Union operations can be performed naively in O(N^2).
- If the graph is connected, print the array. Otherwise, It can be proved that changing exactly one element in the array can make the graph connected.
- Choose one position in the array, replace it with a prime not used in the array. (Prime not used can be found by the sieve, and are guaranteed to exist since the graph is connected.)
So, another problem with primes.
This editorial has two parts, one focusing on checking whether the graph is connected, and other making the graph connected if it is not yet connected.
Initially discussing problem from graph view.
First of all, Iterate over every pair of number, and check if there are co-prime, if yes, add an edge between two vertices. This way, we have added all edges in O(N^2). Now, using DFS, we can see the number of connected components in this resultant graph. If it is already connected (Number of connected components is 1), our graph is already valid, so we can just print the given array as output. (We can also use Union-Disjoint sets for checking if the graph is connected.)
Now, Let’s consider that the graph is not connected. Think about a number which will be co-prime to all numbers (So that we will connect this vertex to all other vertices in the graph, getting connected graph irrespective of all other values in the array, in just one step.
I know you all are good enough to guess I’m talking about 1, but problem setter intentionally choose to keep 2 \leq A* \leq 50 to make this task a bit harder
Let’s us suppose we choose a prime since it will not be co-prime only to its multiples. Choose a prime above N/2+1 makes sure that only multiple of this number in the array is the element itself.
So, just choose a prime and replace any element in array with this prime. I choose 47.
Suppose evil setter already knows of this solution and set all values in the array as 47. Isn’t our solution trapped?
So, how do we handle this, Simple!!
Just take another prime. It can be proven that the problem will be solved in one step since we can choose one element in the array, replace it with any such prime number in the array.
But, we shall make his effort useless, by noticing that we can always make the graph connected, in just one step. For proof, see primes above or equal to N/2+1. Let’s call them good primes.
We can prove that if the graph has at least two good primes, then the graph is connected.
**Reason: ** Suppose we have two good primes in the array, so, It’s guaranteed that all other elements in the array will be connected to at least one of the good prime. Thus, Graph is divided into at most two components, one containing first good prime while other containing the second good prime. These two good primes will also be connected to each other since two distinct primes are always Co-prime to each other, Resulting in the graph being connected.
So, I took primes 41 and 47 for this purpose, If the array doesn’t consist of all 47, replace the first element with 47 and that’s it. Otherwise, replace the first element with 41.
The reason of not choosing a prime smaller than or equal to N/2 is that we will need to be careful about choosing which element to replace, and also handle multiples of this prime present in the array. It can be done easily, but choosing a good prime makes things simple enough.
To Problem setter, Thanks a lot for proofreading editorial as well as suggesting important points to be kept in the editorial.
I guess those all those who participated in October long challenge already know about one such problem. For those who didn’t, Coprime Components. Please suggest more such problems.
O(N^2) for initial graph construction dominate overall time complexity. Apart from that, all other operations take time of order O(N).
AUTHOR’S AND TESTER’S SOLUTIONS:
Feel free to Share your approach, If it differs. Suggestions are always welcomed.