If x$,$ y in $\{0,1\}^$n are two binary strings of length n$,$ then we say x is a neighbour of y if they differ in exactly one bit position. For example, the strings x $=00100$ and y $=10100$ are neighbours, since they are the same everywhere except for the first position. On the other hand, x$=00100$ and y$=00011$ are not neighbours, since they differ in three bit positions.

Consider the following algorithmic problem. The input is a set X $\{0,1\}^$n of binary strings of length n$.$ The output is to decide if X can be partitioned into neighbouring strings. Formally, it should output "Yes" if there is a partition of X into a collection of pairs of neighbouring strings such that $($a$)$ every string appears in at least one pair of the partition and $($b$)$ no string appears in two distinct pairs of the partition. It should output $$No$$ if no such partition is possible. Give an efficient algorithm solving this problem. For full marks, your algorithm should run in O$(($X$^2)$n$)$ time. $($Prove corrrectnesss as well$)$