# Long Narrow City // My hard problem in Counter Code by HackerRank

It was great preparing a hard problem for Counter Code hosted by HackerRank. Competition was great, and during 24 hours 26 participants manage to solved it. This is the most detailed editorial that I have written so far.

The problem is available here: Long Narrow City

## Statement

Problem statement is quite clear, we are given a graph for which we have to count the number of its spanning trees, containing at most k distincted roads. $k$ can be any number in range $[0, 200]$ and there are at most $1000$ distincted roads given in the input.

However, the graph can be really huge! It can have up to $3 \cdot 10^{12}$ vertices - definitely way too big to store them in the memory.

The good news is that is has a very specific structure. As the name of the problem suggests, it is very long, but fortunately a narrow one

Specifically, it has $3$ layers: top, middle and bottom, each consisting of $N$ vertices. Layers along with graph edges are presented in the below drawing. Notice, that there are $N$ edges between consecutive layers and each layer contains $N - 1$ internal edges:

Now we know what we are dealing with, so let's start solving it. What I like often to do, is to start with some simpler problem, derive a solution to it, and then adapt it further to the original problem. Let's give it a try here!

## Simplified problem

First, let's assume that there are no distincted edges.

### Simplified graph

Moreover, let's assume, that the graph has only $2$ layers instead of $3$. It is even more narrow, isn't it? From now we are referring only to simplified graphs unless we call the graph original explicitly.

Our simplified graph looks like this:

Since we assumed that there are no distincted edges, our task is to find the number of all spanning trees of it.

First, let's define what a segment it our narrow graph is. We call a segment a set of $3$ vertices with numbers $i, i + N, i + 2N$ for original graph, and a set of $2$ vertices with numbers $i, i + N$ for a simplified one.

An example segments, for original graph and simplified graph, are presented it the below drawing. Green vertices belong to the current segments, while red ones belong to the previous ones. Edges connecting these segments are also presented:

Let tail be the last segment of the graph.

Now we are ready to count the number of spanning trees of simplified graph. A great way to do that, is to try to extend a solution for a smaller graph by adding a new segment to it as a tail.

Let $G_{N}$ be a simplified graph with $3 \cdot N$ vertices. We can extend $G_{N}$ to $G_{N + 1}$ by adding a new segment to it and connect it to the tail of $G_N$.

This is the incremental method to build our graphs.

In order to count the number of spanning trees of $G_{n + 1}$, let's assume that we know the number of spanning trees of $G_{N}$. How does it help? Well, let $T_{N}$ be any spanning tree of $G_N$. We can extend $T_N$ to a spanning tree of $G_{N+1}$ by adding some edges of the tail of $G_{N+1}$ to $T_N$.

The crucial observation is that we can count spanning trees incrementally as well.

But be aware here. Does any spanning tree of $G_{N+1}$ can be produced from a spanning tree of $G_{N}$? Let's consider the below drawing presenting a valid spanning tree of $G_3$.

If you look closely, you will notice that this spanning tree cannot be produced from any spanning tree of $G_4$, because vertices $2$ and $5$ are connected in it through the tail of $G_5$.

However, the good news is that it can be producted from a spanning forest of $G_4$ with two connected components in which the top layer is the first connected component and the bottom layer is the second connected component. What is more important, the following fact is true:

Every spanning tree of $G_{N+1}$ can be produced by extending a spanning tree of $G_N$ or a spanning forest of $G_N$, in which top and bottom layers are different connected components, by adding a new layer and connecting it with some edges to the previous structure.

This is true, because in any valid spanning tree of $G_{N+1}$, vertices $N$ and $2N$ have to be connected by exactly one path. They can be connected by a path which does not use vertices $N + 1$ and $2N + 1$ or by a path which uses them. In the first case, the T_{N+1} is formed from $T_N$, while is the second case, it is formed from a forest of $G_N$ in which the top and the bottom layers are distinct connected components.

This is a good news, because we can count the number of spanning trees of $G_{N+1}$ by just knowing the number of spanning trees of $G_N$ and the number of specific spanning forests of $G_N$.

Let $T(N)$ be the number of spanning trees of $G_N$ and $F_N$ be the number of spanning forests of $G_N$ in which top and bottom layers are distinct connected components. In order to get the exact equation for $G_N$, we have to examine below cases and count the number of possible extensions in each case.

1. Any $T_N$ can be extended to $T_{N+1}$ in $3$ different ways, because we can add any $2$ of $3$ edges which are adjacent to vertices from the last segment of $G_{N+1}$
2. Any $F_N$ can be extended to $T_{N+1}$ in just one way, because we have to add all $3$ edges adjacent to vertices from the last segment to $G_{N+1}$.
3.  Any $T_N$ can be extended to $F_{N+1}$ in $2$ ways, because we have to add just one horizontal edge adjacent to vertices from the last segment of $G_{N+1}$ to $T_N$ and there are two ways to do that.
4. Any $F_N$ can be extended to $F_{N+1}$ in just one way, because we have to add $2$ horizontal edges adjacent to vertices from the last segment of $G_{N+1}$ to $F_N$ .

Let $T(N)$ be the number of valid graphs of type $T_N$ and $F(N)$ be the number of valid graphs of type $F_N$.

Clearly, $T(1) = 1$ and $F(1) = 1$. From the above analysis, we know that:

$T(N + 1) = 3 \cdot G(N) + F(N)$

$F(N + 1) = 2 \cdot G(N) + F(N)$

These two equations helps us to compute the number of spanning trees of simplified graphs.

### Original graph

That is great, but what does it have to do with our original graph with $3$ layers of vertices? Well, the same analysis works, but we have more cases to consider. From now we are referring to original, $3$ layer graphs, unless we call the graph explicitly simplified.

First some definition, which will helps us to refer to some things.

Let $i, i + N, i + 2N$ be vertices in the last layer of $G_N$. We define the following graphs:

Let $F^1_N$ be a spanning forest of $G_N$, in which vertex $i$ belongs to one connected component, vertices $i + N$ and $i + 2N$ belong to the other one, and there are only two connected components in this forest.

Let $F^2$ be a spanning forest of $G_N$, in which vertex $i + N$ belongs to one connected component, vertices $i$ and $i + 2N$ belong to the other one, and there are only two connected components in this forest.

Let $F^3_N$ be a spanning forest of $G_N$, in which vertex $i + 2N$ belongs to one connected component, vertices $i$ and $i + N$ belong to the other one, and there are only two connected components in this forest.

Let $F^0_N$ be a spanning forest of $G_N$, in which vertices $i, i + N, i + 2N$ belong to 3 different connected components, and there are 3 connected components in this forest

$T_N$ is defined as before, i.e. it is a spanning tree of $G_N$, so all vertices $i, i + N, i + 2N$ belong to the only connected component.

All $5$ above graphs are presented below - vertices belonging to the same connected component have the same color and there are as many connected components as colors of vertices in the last segments (notice that I did not draw any edges between these vertices just for generality - we don't assume by what path they are connected, we just indicate that they are connected - might be by direct edges and might be by longer paths):

Let also $F_1(N), F_2(N), F_3(N), F_0(N), G(N)$ be the number of graphs of corresponding types defined just above.

Based on these values, we can compute $G(N + 1)$ in a similar way that we did it in for simplified graphs. There are just more cases to consider and each of the above $5$ functions has its own recursive relation. The exact relations are presented below. You can compute them either by hand or write a small program to do it for you. I decided to do it by hand first and then double check it by a program:

$T(N + 1) = 8 \cdot T(N) + 3 \cdot F_1(N) + 4 \cdot F_2(N) + 3 \cdot F_3(N) + 1 \cdot F_0(N)$
$F_1(N + 1) = 4 \cdot T(N) + 3 \cdot F_1(N) + 2 \cdot F_2(N) + 2 \cdot F_3(N) + 1 \cdot F_0(N)$
$F_2(N + 1) = 1 \cdot T(N) + 0 \cdot F_1(N) + 1 \cdot F_2(N) + 0 \cdot F_3(N) + 0 \cdot F_0(N)$
$F_3(N + 1) = 4 \cdot T(N) + 2 \cdot F_1(N) + 2 \cdot F_2(N) + 3 \cdot F_3(N) + 1 \cdot F_0(N)$
$F_0(N + 1) = 3 \cdot T(N) + 2 \cdot F_1(N) + 2 \cdot F_2(N) + 2 \cdot F_3(N) + 1 \cdot F_0(N)$

You have to derive values to $T(0), F_1(0), F_2(0), F_3(0), F_0(0)$ also.

Having these relations, we are able to compute the number of spanning trees of our graphs of any reasonable size. Even if $N$ is as big as $10^{12}$, we can still use matrix exponentiation to do that quickly.

So far so good, but remember that we simplified the problem to not having any distincted edges. Let's now get back to our original problem.

## Original problem

In the original problem, we have at most $1000$ distincted edges and we want to know the answer to how many spanning trees of $G_N$ are there, such that they have at most $k$ distincted edges, where $k$ is at most $200$.

How do deal with that problem? Well, do you remember how we extend, let's say $F^1_N$ to $T_{N+1}$? From the recursive equations derived before, we know that there are $3$ ways to do that i.e. there are $3$ subsets of edges forming a segment such that you can connect them to the tail of any $F^1_N$ to form a valid $T_{N+1}$. Let's call the st of these subsets $S$.

Let's also define $F^1_{N,a}$ as $F^1_N$ with exactly $a$ distincted edges. Similarly we define $T_{N,a}$ and so on. Let's pick any subset of edges in $S$, call it $s$, and assume that $s$ has $d$ distincted edges.

The crucial observation here is that we can form a valid $T_{N, a + d}$ from any F^1_{N, a} by adding edges from $s$ to $F^1_{N, a}$.

Based on these observations we can implement a dynamic programming solution to the the problem.

Let $\text{dp}_{T,k,N}$ be the number of spanning trees of $G_N$ with exactly $k$ distincted edges.

Similarly we define $\text{dp}_{F_1,k,N}$, $\text{dp}_{F_2,k,N}$, $\text{dp}_{F_3,k,N}$ and $\text{dp}_{F_0,k,N}$

Then we can compute any of above tables iterating over each segment in the graph, then iterating over each possible subset of edges in this segment and extending the solution to the larger graphs by examining how many distincted edges this subset has and to what kind of graph this subset extends our current graph (if it extends to any). You can precompute the extension table, i.e. precompute if a graph of type, let's say F_1 can be extended to, let's say F_3, using a particular subset of edges.

However, since $N$ is up to $10^{12}$ we are not allowed to iterate over all segments in the graph. Think for a second, do we really need to do this? Of course we don't. Notice that there are at most $1000$ distincted edges in the input, so there are also at most $1000$ segments with any distincted edge, which is like nothing comparing it to the $10^{12}$ possible segments. What we can do instead, is to assign distincted edges to segments to which they belong, sort this segments in increasing manner and process the graph from left to right. We have two cases to consider then:

1. The current segment has distincted edges.
In this case, we can use our dynamic programming method to extend the solution to the graph with one more segment.
2. The current segment is free of distincted edges.
In this case, we can see how far to the right is the next segment with distincted edges. Then we can use the recursive equation and matrix exponentiation, which are described in the simplifier problem, to extend the current solution to the segment just before the next segment with distincted edges. Since this method is pretty fast, we are able to skip over huge number of segments without dangerous edges very fast.

Notice that we are not allowed to store the whole $\text{dp}$ table anymore, because there are just too many segments. The good news is that we do not have to do this, because we just need to know the entry corresponding to the last examined segment. This reduces the space complexity a lot and is a common trick in these type of problems.

Since in the statement we are asked to answer questions on how many spanning trees with at most $k$ dangerous edges are for some values of $k$, we can now easily answer them based on our $\text{dp}$ tables, just by computing:

$\text{answer}_{N,k} = \sum_{m=0}^k \text{dp}_{T,N,m}$

where $\text{answer}_{N,k}$ is the number of spanning trees of the input graph with exactly $k$ distincted edges.

### Time complexity

First, you have to precompute the table of extensions, which stores the information if a graph of one type can be extended to a graph of another type using some particular subset of edges. Since there are $5$ types of graphs, $32$ subsets of edges, and you can compute the resulting graph for a given graph and subset of edges using $\text{dfs}$ this takes so small amount of time that we do not have to worry about it.

During the computations, sometimes you use skips by matrix exponentiation and since the matrix size if $5 \times 5$ and the longest skip has length $10^{12}$, one skip takes $O(5^3 \cdot \log(10^{12}))$.

At most $1000$ times we are forced to compute the extension of a current graph by dynamic programming, and doing this for one segment takes $O(5 \cdot 5 \cdot K \cdot 32)$, where $K$ is maximum number of distincted roads in a query. This is true, because in $\text{dp}$ table an entry for a graph of one type is based on values of every of $5$ other types of smaller graph and we compute the extension for any possible number of distincted edges. Computing one such entry takes 32 subsets of edges to consider. However this could be reduced by just examining the valid subsets of edges for a particular type of graph i.e. these ones which extends it to some other graph.

# Hackerrank Weekly Challenge #6 // Problem 1

## Problem statement

The problem statement is really simple. You are given a set $S$ of $n$ binary strings each of length $m$. The task is to return the maximum number of 1's in binary representation of $a \vee b$, where $a, b \in S$ and $\vee$ is the logical OR operator. In addition, you have to return the number of pairs for which that number is maximal.

For example, if $S = \{1100, 0011, 1010\}$ the result is 4 and 1, because $1100 \vee 0011$ has 4 ones and no other pair of strings produces 4 ones.

## Constraints update

Initially the constraints were $2 \leq n \leq 3000$, $1 \leq m \leq 1000$, but in the middle of the contests hackerrank decides to change it to $n, m \leq 500$, for which a simple brute force solution works fine i.e. for any pair of strings $a, b$ count the number of 1's in $a \vee b$ and return the maximum. The time complexity of this approach is $O(n^2 \cdot m)$.

Many coders complained about changing constraints, because it happened in the middle of the contest, which is not fair at all, especially because the time of submission matters.

## Solution for initial constraints

I'll give here my solution for the initial constraints that works in $O(n \cdot m + n \cdot n / 2 \cdot 32)$, which is $\backsimeq 10^8$ in the worst case.

The idea is to divide each string into chunks of size 32, because in that case we can store these chunks as 32-bit integers and there is a very clever and fast way of computing number of ones of 32-bit integer. We know that $m \leq 1000$, so we need 32 chunks, because $32^2 > 1000$.

Let $s[i]$ be the $i$-th input string and $s[i][j]$ be the $j$-th character of that string.

Let $bm[i][k]$ be the value of the $k$-th chunk of the $i$-th string.

We can compute $bm$ table using that code:

for(int i = 0; i < n; ++i)
{
for(int k = 0; k < 32; ++k) bm[i][k] = 0;
}
for(int i = 0; i < n; ++i)
{
int k = 0;
for(int j = 0; j < m; ++j)
{
if(j > 0 && j % 32 == 0) ++k;
bm[i][k] *= 2;
bm[i][k] += (int)(a[i][j] - '0');
}
}


Now we can iterate over each pair of input strings. Let's assume that we want to compute the number of 1's in binary representation of $s[i] \vee s[j]$. In order to do that, we can compute the number of 1's in $bm[i][k] \vee bm[j][k]$ for any $k$ separately and add these partial results at the end. But if we want to achieve a speedup over the brute force solution, we have to do it very fast for a single chunk. You can find many ways to do it quickly. I used the below, which I found on stackoverflow. We can assume that it works in $O(1)$:

int BitCount(unsigned int u)
{
unsigned int uCount;

uCount = u - ((u >> 1) & 033333333333) - ((u >> 2) & 011111111111);
return ((uCount + (uCount >> 3)) & 030707070707) % 63;
}


Here is my main loop based on $bm$ table and $\texttt{BitCount}$ function:

int res = 0;
int res_cnt = 0;
for(int i = 0; i < n; ++i)
{
for(int j = i + 1; j < n; ++j)
{
int local = 0;
for(int k = 0; k < 32; ++k)
{
int tmp = bm[i][k] | bm[j][k];
int ones = BitCount(tmp);
local += ones;
}
if(local > res)
{
res = local;
res_cnt = 1;
}
else if(local == res)
{
res_cnt++;
}
}
}


## Summary

I generated a few big test files and tested them on hackerrank online judge as custom tests before constraints were changed and there was no time limit exceeded error, so my program works in the desire time limit. I wonder, how many submissions would pass final test cases if constraints weren't changed.