<< problem 149 - Searching for a maximum-sum subsequence Paper sheets of standard sizes: an expected ... - problem 151 >>

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.

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[0][0]; // 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.

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:

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)

Benchmark

The correct solution to the original Project Euler problem was found in 0.4 seconds on a 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=c++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

green problems solve the original Project Euler problem and have a perfect score of 100% at Hackerrank, too.
yellow problems score less than 100% at Hackerrank (but still solve the original problem).
gray problems are already solved but I haven't published my solution yet.
blue problems are solved and there wasn't a Hackerrank version of it at the time I solved it or I didn't care about it because it differed too much.
red problems are solved but exceed the time limit of one minute or the memory limit of 256 MByte.

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

 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100
 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160 161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180 181 182 183 184 185 186 187 188 189 190 191 192 193 194 195 196 197 198 199 200
 201 202 203 204 205 206 207 208 209 210 211 212 213 214 215 216 217 218 219 220 221 222 223 224 225 226 227 228 229 230 231 232 233 234 235 236 237 238 239 240 241 242 243 244 245 246 247 248 249 250 251 252 253 254 255 256 257 258 259 260 261 262 263 264 265 266 267 268 269 270 271 272 273 274 275 276 277 278 279 280 281 282 283 284 285 286 287 288 289 290 291 292 293 294 295 296 297 298 299 300
 301 302 303 304 305 306 307 308 309 310 311 312 313 314 315 316 317 318 319 320 321 322 323 324 325 326 327 328 329 330 331 332 333 334 335 336 337 338 339 340 341 342 343 344 345 346 347 348 349 350 351 352 353 354 355 356 357 358 359 360 361 362 363 364 365 366 367 368 369 370 371 372 373 374 375 376 377 378 379 380 381 382 383 384 385 386 387 388 389 390 391 392 393 394 395 396 397 398 399 400
 401 402 403 404 405 406 407 408 409 410 411 412 413 414 415 416 417 418 419 420 421 422 423 424 425 426 427 428 429 430 431 432 433 434 435 436 437 438 439 440 441 442 443 444 445 446 447 448 449 450 451 452 453 454 455 456 457 458 459 460 461 462 463 464 465 466 467 468 469 470 471 472 473 474 475 476 477 478 479 480 481 482 483 484 485 486 487 488 489 490 491 492 493 494 495 496 497 498 499 500
 501 502 503 504 505 506 507 508 509 510 511 512 513 514 515 516 517 518 519 520 521 522 523 524 525 526 527 528 529 530 531 532 533 534 535 536 537 538 539 540 541 542 543 544 545 546 547 548 549 550
The 235 solved problems (level 9) had an average difficulty of 29.1% at Project Euler and
I scored 13,183 points (out of 15300 possible points, top rank was 17 out of ≈60000 in August 2017) at Hackerrank's Project Euler+.
Look at my progress and performance pages to get more details.

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

 << problem 149 - Searching for a maximum-sum subsequence Paper sheets of standard sizes: an expected ... - problem 151 >>
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 !