A rectangular board with three rows and $n$ columns is filled with $3n$ counters, of which $n$ are red, $n$ are white, and $n$ are blue. The object is to rearrange the counters to have counters of each of the three different colors in every column. The only operation allowed is to swap counters in the same row. Design an algorithm to accomplish such a task or prove that such an algorithm does not exist.

Let $\mathcal{O}(1)$ be the cost of swapping the counters in the same row.

Iterate over each column. Let us consider that up to column $i-1$, we are done. Now we are at column $i$, there will be three counters, if they are of different colors then we are done, else fix one counter ( say white ) and scan other two rows to get two different color counters ( red and blue ). Running the previous step for a constant number of times ( six ) will solve the problem. More precisely there will be six possibilities in each column (123,132,.....,321), so I will try each of those.

The runtime of the above algorithm will be $\mathcal{O}(n^3)$ as there are $n$ columns and time spent on each column is $\mathcal{O}(n^2)$.

Question: Is it possible to solve the above-described problem linear time $\mathcal{O}(n)$ time as input here has size $\mathcal{O}(n)$? I am not even able to come up with an algorithm faster than $\mathcal{O}(n^3)$ time.

I have taken this question from the book titled Algorithmic Puzzles ( problem 46 )

  • $\begingroup$ You say counter, is that just something like a pebble or also something that holds a number? Also, what is the initial configuration - random? $\endgroup$
    – orlp
    Commented Dec 30, 2018 at 10:58
  • $\begingroup$ @ orlp Each type of a counter is a fixed number. For an example red counter have value 2 and so on $\endgroup$
    – user35837
    Commented Dec 30, 2018 at 12:05

2 Answers 2


An idea can be to reduce this problem to sorting.

Consider a final configuration with each colour appearing exactly once per column. You can move the columns around to obtain a final configuration that looks like that :


Where $a, b$ and $c$ are the three distinct colours. We are now going to create that kind of configuration. Let $a_1$ (resp. $b_1$ and $c_1$) denote the number of $a$'s (resp. $b$'s and $c$'s on the first row), same with $2$ on the second row. We can find those by counting, a simple $\mathcal{O}(n)$ first pass.

Now let $x_*, y_*$ denote the size of the first and second chunk of the colour $*$ on the second row (i.e. we have $x_a, x_b$, etc...). You can see it that way for the first two rows :

|    a_1    |     b_1   |    c_1    |
| x_b | x_c | y_c | x_a | y_a | y_b |

We then have the following equations :

  • $a_1 = x_b + x_c$
  • $b_1 = y_c + x_a$
  • $c_1 = y_a + y_b$
  • $x_a + y_a = a_2$ and so on.
    $6$ equations with $6$ unknowns : you can find the size of each block - you only need to know the number of times each colours appears on rows 1 and 2.

Now create 3 arrays of weights, one for each row :

  • On the first row, give weight $0$ to elements of colour $a$, $1$ to elements $b$, $2$ to elements of colour $c$.
  • On the second row, give weight $0$ to $x_b$ elements of colour $b$, weight $1$ to $x_c$ elements of colour $c$, etc.
  • Same on the third row, but with the other colours.

This an be done in $\mathcal{O}(n)$ by simply iterating the arrays (First row and weight array for the first row) and keeping track of how many elements of each colour we have seen so far.

Then sort each row : we obtain the solution described above. The complexity is then $\mathcal{O}(n +f(n))$ where the sorting you use is $\mathcal{O}(f(n))$.

So, now, all you have to do is to find a sorting algorithm that is efficient and that fits the problem : it has to sort by swapping elements. At least quicksort works (and you can find the median easily) so you have $O(n\log(n))$ complexity.

I think you could extend it to linear time using a sorting algorithm that counts.

  • $\begingroup$ Thanks for the answer. I have understood your idea. So now the algorithm will sort each ( suppose each counter have some value ). Note that sorting can be done in $\mathcal{O}(n)$ time. Iterate over each column and now the things will be easy by using some pointers. Note that sorting in our case can be done in $\mathcal{O}(n)$ time in our case. $\endgroup$
    – user35837
    Commented Dec 30, 2018 at 11:48

Here is an algorithm with $O(n)$ time-complexity.

First, count how many counters of each color there are on each row. With $O(n)$ time, we will obtain the number of red counters $r_i$, the number of white counters $w_i$, the number of blue counters $b_i$ on row $i$, $i=1,2,3$.

Second, we can solve the following equations for nonnegative integer variables $a_k$, $1\le k\le6$ in $O(1)$ time.

$$r_1 = a_1 + a_2$$ $$w_1 = a_3 + a_4$$ $$b_1 = a_5 + a_6$$ $$r_2 = a_4 + a_5$$ $$w_2 = a_1 + a_6$$ $$b_2 = a_2 + a_3$$ $$r_3 = a_3 + a_6$$ $$w_3 = a_2 + a_5$$ $$b_3 = a_1 + a_4$$

Third, we arrange the three rows into the following configuration. $$ \begin{array}{|l|l|l|l|l|l|} \hline a_1\text{ red} & a_2\text{ red} & a_3\text{ white} & a_4\text{ white} & a_5\text{ blue} & a_6\text{ blue} \\\hline a_1\text{ white} & a_2\text{ blue} & a_3\text{ blue} & a_4\text{ red} & a_5\text{ red} & a_6\text{ white} \\\hline a_1\text{ blue} & a_2\text{ white} & a_3\text{ red} & a_4\text{ blue} & a_5\text{ white} & a_6\text{ red} \\\hline \end{array} $$

We are done.

Here are a couple of exercises that fill the gap in the above specification of the algorithm.

Exercise 1. Find an algorithm that solves in $O(1)$ time that system of 9 equations for nonnegative integer variables $a_k$, $1\le k\le6$, given the condition that $r_i+w_i+b_i=n$ for all $i$ and $\sum_{i=1}^3r_i=\sum_{i=1}^3w_i=\sum_{i=1}^3b_i=n$. Show that there is always a solution.

Exercise 2. Check that arrangement in third step can be done in $O(n)$ time by swapping the counters in each row.

Here is a problem that may not be easy to do.

Problem 1. Generalize the problem of tricolor arrangement to $m$ rows and $m$ colors, $m>3$. Solve it with an $O(n)$ algorithm, assuming $m$ is a constant. (Hint, use Hall's marriage theorem to reduce the sum of the number of distinct colors in each row.)

  • $\begingroup$ The rearranging is massively simplified and easy to see how it can be done in $O(n)$ time with Dutch national flag problem. The red/blue/white color scheme is also a reference to this. $\endgroup$
    – orlp
    Commented Dec 30, 2018 at 11:15
  • $\begingroup$ Thanks for the answer. I am trying to understand your answer. Are you solving the decision version which means yes/no output at the end or also ouput the one of the configuration. $\endgroup$
    – user35837
    Commented Dec 30, 2018 at 12:02
  • $\begingroup$ Output all the columns, of course. For example, the first row. Iterate through the counters, moving the red ones to the front. Iterate again, moving the whites one after the red ones. Not terribly efficient, but enough to prove $O(n)$. No need to use sophisticated algorithm such as the one in Dutch national flag problem $\endgroup$
    – John L.
    Commented Dec 30, 2018 at 12:05
  • $\begingroup$ The algorithm for the decision version of the problem is just one phrase, "Return yes". $\endgroup$
    – John L.
    Commented Dec 30, 2018 at 12:12
  • $\begingroup$ @orlp, yes, it is always doable. Notice that we have same number of counter of each color initially. You may want to do my exercise 1. $\endgroup$
    – John L.
    Commented Dec 30, 2018 at 12:38

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