Partition integers into $M$ sets

Question

Given a circular list of two numbers $-1,+1$.I have to partition them into $M$ sets each containing $K$ numbers. Each set has to consist of consecutive $K$ numbers from that list. The set can start from any point in the list.

1.A set is called positive if number of $+1$ are greater than $-1$.

2.A set is called negative if number of $-1$ are greater than $+1$.

Now I want to know if there's a way to partition them such that the number of positive sets are more than number of negative sets.

Take $K$ to be odd for simplicity so that a set can be either called positive or negative.

For eg- List $A$ = $(-1, -1, 1, 1, -1, 1, 1, -1, -1)$ and I have to partition it into $3$ sets with $3$ numbers in it. Then there are many ways : 1. {$A_3$,$A_4$,$A_5$}, {$A_6$,$A_7$,$A_8$} and {$A_9$,$A_1$,$A_2$}

2. {$A_2$,$A_3$,$A_4$}, {$A_5$,$A_6$,$A_7$} and {$A_8$,$A_9$,$A_1$}

Here $2$ positive sets and $1$ negative set. So I can partition it.

I can only think of making every possible combination. Is there a way to do it efficiently and optimally.

Answer 1

Let $N=KM$. Let $S_i = A_i+\ldots +A_{i+K-1}$, $A_{N+i}=A_{i}$. Obviously, $\{A_i,\ldots ,A_{i+K-1}\}$ is positive if $S_i$ is positive, and $\{A_i,\ldots ,A_{i+K-1}\}$ is negative if $S_i$ is negative.

The $S_i$ can be calculated recursively with little effort: $$ S_{i+1} = S_i+A_{i+K}-A_{i} $$ For each $j\in \{0,\ldots ,K-1\}$, introduce a counter $p_j$ ("positive") and an counter $n_j$ ("negative").

Start with initializing your counters and calculating $S_1$. Then calculate each $S_i, \; i=1,\ldots, N$ with the recursive formula shown above. If $S_i >0$, increase the $p_j$ for which $j\equiv i\pmod{K}$. If $S_i<0$, increase the $n_j$ for which $j\equiv i\pmod{K}$. At the end, find an index $j$ for which $p_j$ is greater than $n_j$. This $j$ gives you the starting position in the list. Use the biggest $p_j$ if you want to have the optimum.

If we apply this method to the given example, we get (one after another) $$ (S)_{i\in\{1,\ldots,9\}} = (-1,\,1,\,1,\,1,\,1,\,1,\,-1,\,-3,\,-3) $$ For the counters $p_0$ and $n_0$, we have taken the elements $S_3$, $S_6$ and $S_9$ into account (while we calculated the $S_i$). We find that two of them are positive and one is negative, therefore $p_0=2$ and $n_0=1$ in the end.

For the counters $p_1$ and $n_1$, we have taken the elements $S_1$, $S_4$ and $S_7$ into account. We find that one of them is positive and two are negative, therefore $p_1 = 1$ and $n_1=2$.

For the counters $p_2$ and $n_2$, we have taken the elements $S_2$, $S_5$ and $S_8$ into account. We find that two of them are positive and one is negative, therefore $p_2=2$ and $n_2=1$.

For $j=0$ and $j=2$, $p_j>n_j$ holds. Therefore we can split the list either in front of the elements with indices $\equiv 0\pmod{3}$ or in front of the elements with indices $\equiv 2\pmod{3}$.