# Problem 150: Searching a triangular array for a sub-triangle having minimum-sum

In a triangular array of positive and negative integers, we wish to find a sub-triangle such that the sum of the numbers it contains is the smallest possible.

In the example below, it can be easily verified that the marked triangle satisfies this condition having a sum of −42. We wish to make such a triangular array with one thousand rows, so we generate 500500 pseudo-random numbers s_k in the range \pm 2^19,
using a type of random number generator (known as a Linear Congruential Generator) as follows:

t := 0
for k = 1 up to k = 500500:
 t := (615949*t + 797807) modulo 220
 s_k := t-219

Thus: s_1 = 273519, s_2 = -153582, s_3 = 450905 etc

Our triangular array is then formed using the pseudo-random numbers thus:
s_1
s_2 s_3
s_4 s_5 s_6
s_7 s_8 s_9 s_10
...

Sub-triangles can start at any element of the array and extend down as far as we like
(taking-in the two elements directly below it from the next row, the three elements directly below from the row after that, and so on).

The "sum of a sub-triangle" is defined as the sum of all the elements it contains.
Find the smallest possible sub-triangle sum.

# My Algorithm

My solution can be called a "tweaked brute-force approach".
I store the triangle in a 2D array, where the number of "columns" of each row varies: only one at the top and columns = rows at the bottom.
For each cell (x,y) I walk through all possible triangles (x, y),(x + increment, y),(x, y + increment).
For increment = 0, the sum is just the value found in triangle[x][y].
For increment = 1, I add all the two values of the next row to the result of increment = 0, thus triangle[x][y] + triangle[x][y+1] + triangle[x+1][y+1]
In general: for each increment > 0, I add all the two values of the next row to the result of increment - 1.

Whenever my sum is lower than the previously lowest sum, I store it (Hackerrank: store every sum).

Repeatedly adding numbers in the same row turned out to be a performance bottleneck.
In step 2 of my code, a second container called sums is created which contains sums[x][y] = sum(triangle[0..x][y]) (similar to a running total).
Then sum(triangle[x1..x2][y]) = sums[x2][y] - sums[x1 - 1][y] (except when x1 is zero).
In the end, this optimization reduced runtime from 20 seconds to just 0.4 seconds.

## Modifications by HackerRank

Hackerrank provides all input data (instead of the Linear Congruential Generator) and asks you to print
the n smallest sub-triangle sums.

Because of the large number of sums (= high memory usage), I repeatedly sort all values of results and erase all but the smallest n.

I am fully aware that the high number if #ifdef severely decreases readability of my code.

# Interactive test

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

Input data (separated by spaces or newlines):
Note: Enter the size of the triangle.

This is equivalent to
echo 10 | ./150

Output:

(please click 'Go !')

Note: the original problem's input 1000 cannot be entered
because just copying results is a soft skill reserved for idiots.

(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.

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 <iostream>
#include <vector>
#include <algorithm>

// pseudo-random number generator
int lcg()
{
static long long seed = 0;
seed = (615949 * seed + 797807) % (1 << 20);
return int(seed - (1 << 19));
}

int main()
{
// indexed: triangle[y][x] where y denotes the row
std::vector<std::vector<int>> triangle;

// step 1: initialize the triangle (either with random data or from STDIN)
#define ORIGINAL
#ifdef ORIGINAL
// 1000 rows
unsigned int maxSize = 1000;
std::cin >> maxSize;

// fill with (pseudo) random data
triangle.resize(maxSize);
for (size_t y = 0; y < triangle.size(); y++)
for (size_t x = 0; x <= y; x++)
triangle[y].push_back(lcg());

#else
size_t size;
size_t maxResults;
std::cin >> size >> maxResults;

triangle.resize(size);
for (size_t y = 0; y < size; y++)
for (size_t x = 0; x <= y; x++)
{
int current;
std::cin >> current;
triangle[y].push_back(current);
}
#endif

// step 2: build another triangle called "sums" where each cell is sum(triangle[y][0..x])
std::vector<std::vector<int>> sums(triangle.size());
for (size_t y = 0; y < triangle.size(); y++)
{
// same size
sums[y].resize(triangle[y].size());

// store running sum of all cells to the left of the current cell, including the current, too
int rowSum = 0;
for (size_t x = 0; x <= y; x++)
{
rowSum += triangle[y][x];
sums[y][x] = rowSum;
}
}

// store the smallest sums
#ifdef ORIGINAL
auto result = triangle; // simplify if only one sum
#else
std::vector<int> results;   // needs sorting, pruning, etc.
#endif

// analyze all positions
for (size_t y = 0; y < triangle.size(); y++)
{
for (size_t x = 0; x <= y; x++)
{
// smallest triangle consists of a single-cell
auto sum = triangle[y][x];

// better than before ?
#ifdef ORIGINAL
if (result > sum)
result = sum;
#else
results.push_back(sum);
#endif

auto maxSize = triangle.size() - y;
for (size_t current = 1; current < maxSize; current++)
{
// compute sum(triangle[y + rows][x ... x + rows])
// add singles values, "old way"
//for (size_t scan = x; scan <= x + current; scan++)
//  sum += triangle[y + current][scan];

// access pre-computed running sums from step 2, "new way"
auto rowSum = sums[y + current][x + current];
if (x > 0)
rowSum -= sums[y + current][x - 1];
sum += rowSum;

// better than before ?
#ifdef ORIGINAL
if (result > sum)
result = sum;
#else
results.push_back(sum);
#endif
}
}

#ifndef ORIGINAL
// prune sums
std::sort(results.begin(), results.end());
if (results.size() > maxResults)
results.erase(results.begin() + maxResults, results.end());
#endif
}

// show result
#ifdef ORIGINAL
std::cout << result << std::endl;
#else
for (auto x : results)
std::cout << x << std::endl;
#endif

return 0;
}


This solution contains 17 empty lines, 21 comments and 21 preprocessor commands.

# Benchmark

The correct solution to the original Project Euler problem was found in 0.4 seconds on an Intel® Core™ i7-2600K CPU @ 3.40GHz.
Peak memory usage was about 7 MByte.

(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

July 5, 2017 submitted solution

# Hackerrank

My code solves 10 out of 10 test cases (score: 100%)

# Difficulty

Project Euler ranks this problem at 55% (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.

more about me can be found on my homepage, especially in my coding blog.
some names mentioned on this site may be trademarks of their respective owners.
thanks to the KaTeX team for their great typesetting library !