You have an undirected graph with N nodes and M edges. The degree of the connected trio in the graph is the number of edges connected to the nodes of the trio where one node belongs to the node in the trio, while the other node doesn’t belong to the trio. Your task is to print the minimum degree of the connected trio.
Note:
A connected trio means a set of three nodes, where all three are interconnected.
The first line of the input contains ‘T’ denoting the number of test cases.
The first line of each test case contains ‘N’ and ‘M’ denoting the number of nodes and number of edges.
Each of the next M lines contains two space-separated integers u and v, denoting node u and node v are connected by an edge.
Output Format:
For each test case, return an integer denoting the minimum degree of the connected trio.
Print the output for each test case in a separate line.
Note:
You don't need to print anything, it has already been taken care of. Just implement the given function.
1 <= T <= 3
2 <= N <= 100
0 <= M <= min(500, N*(N-1)/2)
0 <= u[i], v[i] <= N-1
where, ‘N’ and ‘M’ denoting the number of nodes and number of edges and u[i], v[i] are nodes to be connected by an edge.
Time Limit: 1 sec.
2
6 7
0 1
0 3
1 2
1 3
2 3
2 4
1 5
4 0
3
0
In test case 1:

There are two trios here in this graph [0, 1, 3] and [1, 2, 3].
In trio [0, 1, 3] : we have 3 edges where one node is connected to trio and other isn’t namely { (1-- 5) , (1 -- 2), (3 -- 2) }
In trio [ 1, 2, 3] : we have 4 edges where one node is connected to trio and other isn’t namely { (1-- 5) , (1 -- 0), (3 -- 0) , (2 -- 4) }
The minimum of the two is 3, hence the answer.
In test case 2:
There are no edges thus answer is 0.
2
5 6
0 1
1 2
2 3
3 1
1 4
4 2
6 6
0 3
0 1
4 1
1 5
5 3
3 4
3
0
Think to iterate over all trios of nodes and check if there is an edge between all pairs. Can you think of a solution now?
Algorithm:
ans= min(ans, deg[ i ] + deg[ j ] + deg[ k ] - 6 )
O(N ^ 3), where 'N' is the number of nodes in the graph
We are iterating over all the trio of nodes, which can reach up to N * (N-1) * (N-2) / 6 i.e. O(N^3).
O(N ^ 2), where ‘N’ is the number of nodes in the graph
We will be making an adjacency matrix in the problem. Thus space complexity is O(N^2).