Cycles of length n in an undirected and connected graph

Given an undirected and connected graph and a number n, count total number of cycles of length n in the graph. A cycle of length n simply means that the cycle contains n vertices and n edges. And we have to count all such cycles that exist. 

Example : 

Input :  n = 4

Output : Total cycles = 3
Explanation : Following 3 unique cycles 
   0 -> 1 -> 2 -> 3 -> 0
   0 -> 1 -> 4 -> 3 -> 0
   1 -> 2 -> 3 -> 4 -> 1
Note* : There are more cycles but
these 3 are unique as 0 -> 3 -> 2 -> 1
-> 0 and 0 -> 1 -> 2 -> 3 -> 0 are 
same cycles and hence will be counted as 1.

To solve this Problem, DFS(Depth First Search) can be effectively used. Using DFS we find every possible path of length (n-1) for a particular source (or starting point). Then we check if this path ends with the vertex it started with, if yes then we count this as the cycle of length n. Notice that we looked for path of length (n-1) because the n^th edge will be the closing edge of cycle.

Every possible path of length (n-1) can be searched using only V – (n – 1) vertices (where V is the total number of vertices). 
For above example, all the cycles of length 4 can be searched using only 5-(4-1) = 2 vertices. The reason behind this is quite simple, because we search for all possible path of length (n-1) = 3 using these 2 vertices which include the remaining 3 vertices. So, these 2 vertices cover the cycles of remaining 3 vertices as well, and using only 3 vertices we can’t form a cycle of length 4 anyways. 

One more thing to notice is that, every vertex finds 2 duplicate cycles for every cycle that it forms. For above example 0^th vertex finds two duplicate cycle namely 0 -> 3 -> 2 -> 1 -> 0 and 0 -> 1 -> 2 -> 3 -> 0. Hence the total count must be divided by 2 because every cycle is counted twice.

Implementation:

Total cycles of length 4 are 3

Time Complexity: O(V*V)
Space Complexity: O(V)

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