Determine if the given directed graph is a tournament. A **tournament** is a directed graph in which every pair of distinct vertices is connected by a single directed edge.

**Example**

For

n = 5,

fromV = [2, 1, 3, 4, 4, 4, 1, 2, 3, 4] and

toV = [3, 2, 1, 3, 2, 1, 5, 5, 5, 5],

the output should be

isTournament(n, fromV, toV) = true.

Here's how the given graph looks like:

**Input/Output**

**[time limit] 4000ms (js)**

**[input] integer n**A positive integer n representing the number of vertices in the given graph.

*Guaranteed constraints:*

1 ≤ n ≤ 10.**[input] array.integer fromV**An array of integers containing integers less than or equal to n.

*Guaranteed constraints:*

0 ≤ fromV.length ≤ 50,

1 ≤ fromV[i] ≤ n.**[input] array.integer toV**For each i in range [0, fromV.length)there is an edge from the vertex number fromV[i] to the vertex toV[i] in the given directed graph.

*Guaranteed constraints:*

toV.length = fromV.length,

1 ≤ toV[i] ≤ n.**[output] boolean**true if the given graph is a tournament, false otherwise.

## Solution

If we list all possible pairs of nodes for \(n=5\) and mark the ones for the given graph it may look like this:

`(1, 1), <strong>(1, 2)</strong>, (1, 3), (1, 4), <strong>(1, 5)</strong><br />(2, 1), (2, 2), <strong>(2, 3)</strong>, (2, 4), <strong>(2, 5)</strong><br /><strong>(3, 1)</strong>, (3, 2), (3, 3), (3, 4), <strong>(3, 5)</strong><br /><strong>(4, 1)</strong>, <strong>(4, 2)</strong>, <strong>(4, 3)</strong>, (4, 4), <strong>(4, 5)</strong><br />(5, 1), (5, 2), (5, 3), (5, 4), (5, 5)`

We now mirror the connections and see this nice pattern, where only the diagonal remains unmarked:

`(1, 1), <strong>(1, 2)</strong>, <strong>(1, 3)</strong>, <strong>(1, 4)</strong>, <strong>(1, 5)</strong><br /><strong>(2, 1)</strong>, (2, 2), <strong>(2, 3)</strong>, <strong>(2, 4)</strong>, <strong>(2, 5)</strong><br /><strong>(3, 1)</strong>, <strong>(3, 2)</strong>, (3, 3), <strong>(3, 4)</strong>, <strong>(3, 5)</strong><br /><strong>(4, 1)</strong>, <strong>(4, 2)</strong>, <strong>(4, 3)</strong>, (4, 4), <strong>(4, 5)</strong><br /><strong>(5, 1)</strong>, <strong>(5, 2)</strong>, <strong>(5, 3)</strong>, <strong>(5, 4)</strong>, (5, 5)`

Creating a set of unique connection tuples, is pretty straightforward using python3, by concatenating the two arrays \(f\) and \(t\). Duplicates get kicked out by the set creation: {*zip(f + t, f + t)}. Since the given constraints \(1 \leq f[i] \leq n\) and \(1 \leq t[i] \leq n\) all numbers are valid and we only need to work with the count. How many distinct tuples do we have? n-squared minus the diagonal. This allows us to implement the following routine:

`isTournament = lambda n, f, t: len({*zip(f + t, t + f)}) == n * n - n`

The routine would be perfectly small, but fails if the number is correct, but a duplicate connection gets removed. That's why we also need to check if the original length is okay to form the actual matrix:

`isTournament = lambda n, f, t: len({*zip(f + t, t + f)}) == n * n - n and n * n - n == 2 * len(f)`

Merging the two conditions to form a smaller solution yields the following final form:

`isTournament = lambda n, f, t: n * n - n == 4 * len(f) + len({*zip(f + t, t + f)})`

Using an adjacency matrix \(M\) we built before, we can solve the problem more elegantly by using an environment like Julia or Matlab:

`isequal(M + M', 1 - eye(size(M)...))`