Nice and easy, the answer is $199$ and in general, for $n$ colors, the answer is $2n-1$ Obviously, every color is used at least once. Observe that it is impossible for more than one color to be used exactly once. This is because, one of them, say $c_1$ must be to the left of $c_2$, but then $c_2$ will never appear to the left of $c_1$ since both these colors appear only once. So, at most one color can be used once and all others must be used at least twice. So, the minimum is at least $1 + 2(n-1) = 2n-1$ The construction is just $1,2,3...,n-1,n,n-1,n-2....,3,2,1$

Cute For $k$ colours ,the answer is $2k-1$. Let the colours be denoted by numbers. By a contradiction argument ,we can say that at most $1$ colour can appear exactly once. The construction,is quite obvious. $1,2,3...k-1,k,1,2,3....k-1$