<< problem 14 - Longest Collatz sequence Power digit sum - problem 16 >>

# Problem 15: Lattice paths

Starting in the top left corner of a 2x2 grid, and only being able to move to the right and down,
there are exactly 6 routes to the bottom right corner.

How many such routes are there through a 20x20 grid?

# My Algorithm

Each grid consists of (width+1)*(height+1) points. The grid in the problem statement has (2+1)*(2+1)=9 points.
I chose my coordinates to be [0][0] in the top left corner and [width][height] in the bottom right corner.

For each point in a grid, the number of routes from the current point to the lower right corner
is the sum of all routes when going right plus all routes when going down:
grid[x][y] = grid[x+1][y] + grid[x][y+1];

There is a single route from the lower right corner to itself ("stay where you are"):
grid[width][height] = 1;

Now I perform a breadth-first search (en.wikipedia.org/wiki/Breadth-first_search) from the lower-right corner to the upper-left corner.
A queue named next contains the coordinates of the next points to analyze - initially it holds the point above the lower-right corner
and the point to the left of the lower-right corner.

A loop processes each point in next:
1. If this point was already processed, skip it
2. If it is possible to move right then get the number of routes stored in grid[x+1][y]
3. If it is possible to move down then get the number of routes stored in grid[x][y+1]
4. Write the sum of step 2 and 3 to grid[x][y], potentially avoid overflow (see Hackerrank modification)
5. If there is a left neighbor, enqueue it in next
6. If there is a neighbor above, enqueue it in next

Finally we have the result in grid[0][0].

## Alternative Approaches

I later found a posting on the internet that the result is

dfrac{(width + height)!}{width! height!}

but I'm not a math guy, I'm a programmer ...
however, the best explanation for that formula is as follows and unfortunately, it's about bits :-) :

If we take any route through the grid then a decision has to be made at each point: walk to the right or walk down.
That's similar to a binary number where each bit can be either 0 or 1. Let's say 0 means "right" and 1 means "down".

Then we have to make width + height decisions: exactly width times we have to go right and exactly height times we have to go down.
In our imaginary binary number, there are width + height bits with height times 0s and width times 1s.

All permutations of these bits are valid routes in the grid.

And as turns out, there are dfrac{(width + height)!}{width! height!} permutations, see en.wikipedia.org/wiki/Permutation

## Modifications by HackerRank

The result should be printed mod (10^9 + 7)

# Interactive test

You can submit your own input to my program and it will be instantly processed at my server:

Number of test cases (1-5):

Input data (separated by spaces or newlines):

This is equivalent to
echo "1 2 2" | ./15

Output:

(this interactive test is still under development, computations will be aborted after one second)

# My code

… was written in C++11 and can be compiled with G++, Clang++, Visual C++. You can download it, too. Or just jump to my GitHub repository.

The code contains #ifdefs to switch between the original problem and the Hackerrank version.
Enable #ifdef ORIGINAL to produce the result for the original problem (default setting for most problems).

       #include <vector>
#include <deque>
#include <utility>
#include <iostream>

int main()
{
unsigned int tests;
std::cin >> tests;
while (tests--)
{
unsigned int width, height;
std::cin >> width >> height;

// create a 2D array which contains the number of paths
// from the current lattice point to the lower-right corner
// there are (width + 1) * (height + 1) such points
// for the original problem, i.e. 21x21 numbers must be found
const unsigned long long Unknown = 0;
std::vector<std::vector<unsigned long long>> grid(width + 1);
for (auto& column : grid)
column.resize(height + 1, Unknown);

// one route if we are already at the goal
grid[width][height] = 1;

// enqueue the next unprocessed lattice points: left and upper neighbor of the lower-right corner
std::deque<std::pair<unsigned int, unsigned int>> next;
next.push_back(std::make_pair(width - 1, height));
next.push_back(std::make_pair(width, height - 1));

// as long as there are unprocessed points
while (!next.empty())
{
// get next point
auto current = next.front();
next.pop_front();

// I prefer names which are easier to read ...
auto x = current.first;
auto y = current.second;

if (grid[x][y] != Unknown)
continue;

// sum of all path when going right plus when going down
unsigned long long routes = 0;
if (x < width)  // can go right ?
routes += grid[x + 1][y];
if (y < height) // can go down ?
routes += grid[x][y + 1];

#define ORIGINAL
#ifndef ORIGINAL
routes %= 1000000007; // Hackerrank wants the result MOD 10^9 + 7
#endif
// solved number for current lattice point
grid[x][y] = routes;

// add left and upper neighbors for further processing
if (x > 0)
next.push_back(std::make_pair(x - 1, y));
if (y > 0)
next.push_back(std::make_pair(x, y - 1));
}

// we are done !
std::cout << grid[0][0] << std::endl;
}
return 0;
}


This solution contains 11 empty lines, 14 comments and 7 preprocessor commands.

# Benchmark

The correct solution to the original Project Euler problem was found in less than 0.01 seconds on an Intel® Core™ i7-2600K CPU @ 3.40GHz.
(compiled for x86_64 / Linux, GCC flags: -O3 -march=native -fno-exceptions -fno-rtti -std=gnu++11 -DORIGINAL)

See here for a comparison of all solutions.

Note: interactive tests run on a weaker (=slower) computer. Some interactive tests are compiled without -DORIGINAL.

# Changelog

February 23, 2017 submitted solution

# Hackerrank

My code solves 11 out of 11 test cases (score: 100%)

# Difficulty

Project Euler ranks this problem at 5% (out of 100%).

Hackerrank describes this problem as easy.

Note:
Hackerrank has strict execution time limits (typically 2 seconds for C++ code) and often a much wider input range than the original problem.
In my opinion, Hackerrank's modified problems are usually a lot harder to solve. As a rule thumb: brute-force is rarely an option.

# Heatmap

Please click on a problem's number to open my solution to that problem:

 green solutions solve the original Project Euler problem and have a perfect score of 100% at Hackerrank, too yellow solutions score less than 100% at Hackerrank (but still solve the original problem easily) gray problems are already solved but I haven't published my solution yet blue solutions are relevant for Project Euler only: there wasn't a Hackerrank version of it (at the time I solved it) or it differed too much orange problems are solved but exceed the time limit of one minute or the memory limit of 256 MByte red problems are not solved yet but I wrote a simulation to approximate the result or verified at least the given example - usually I sketched a few ideas, too black problems are solved but access to the solution is blocked for a few days until the next problem is published [new] the flashing problem is the one I solved most recently

I stopped working on Project Euler problems around the time they released 617.
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The 310 solved problems (that's level 12) had an average difficulty of 32.6% at Project Euler and
I scored 13526 points (out of 15700 possible points, top rank was 17 out of ≈60000 in August 2017) at Hackerrank's Project Euler+.

My username at Project Euler is stephanbrumme while it's stbrumme at Hackerrank.

Look at my progress and performance pages to get more details.

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