Problem 18: Maximum path sum I

(see projecteuler.net/problem=18)

By starting at the top of the triangle below and moving to adjacent numbers on the row below, the maximum total from top to bottom is 23.

3
7 4
2 4 6
8 5 9 3

That is, 3 + 7 + 4 + 9 = 23.

Find the maximum total from top to bottom of the triangle below:

75
95 64
17 47 82
18 35 87 10
20 04 82 47 65
19 01 23 75 03 34
88 02 77 73 07 63 67
99 65 04 28 06 16 70 92
41 41 26 56 83 40 80 70 33
41 48 72 33 47 32 37 16 94 29
53 71 44 65 25 43 91 52 97 51 14
70 11 33 28 77 73 17 78 39 68 17 57
91 71 52 38 17 14 91 43 58 50 27 29 48
63 66 04 68 89 53 67 30 73 16 69 87 40 31
04 62 98 27 23 09 70 98 73 93 38 53 60 04 23

NOTE: As there are only 16384 routes, it is possible to solve this problem by trying every route.
However, Problem 67, is the same challenge with a triangle containing one-hundred rows; it cannot be solved by brute force, and requires a clever method! ;o)

My Algorithm

The main idea is to build a data structure similar to the input data:
but instead of just storing the raw input we store the biggest sum up to this point.

All data is processed row-by-row

Of course, the first row consists of a single number and it has no "parents", that means no rows above it.
Therefore the "sum" is the number itself.
This row now becomes my "parent row" called last.

For each element of the next rows I have to find its parents (some have one, some have two),
figure out which parent is bigger and then add the current input to it.
This sum is stored in current.

When a row is fully processed, current becomes last.
When all rows are processed, the largest element in last is the result of the algorithm.

Example:

1
2 3
4 5 6
initialize:
last[0] = 1;

read second line:
current[0] = 2 + last[0] = 3
current[1] = 3 + last[0] = 4
copy current to last (which becomes { 3, 4 })

read third line:
current[0] = 4 + last[0] = 7
current[1] = 5 + max(last[0], last[1]) = 9
current[2] = 6 + last[1] = 10
copy current to last (which becomes { 7, 9, 10 })

finally:
print max(last) = 10

Note

Exactly the same algorithm is used for problem 67.

Interactive test

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

This live test is based on the Hackerrank problem.

Number of test cases (1-5):

Input data (separated by spaces or newlines):

This is equivalent to
echo "1 4 3 7 4 2 4 6 8 5 9 3" | ./18

Output:

(please click 'Go !')

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

My code

… was written in C++ and can be compiled with G++, Clang++, Visual C++. You can download it, as well as the input data, 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>
 
int main()
{
unsigned int tests = 1;
 
//#define ORIGINAL
#ifndef ORIGINAL
std::cin >> tests;
#endif
 
while (tests--)
{
unsigned int numRows = 15;
#ifndef ORIGINAL
std::cin >> numRows;
#endif
 
// process input row-by-row
// each time a number is read we add it to the two numbers above it
// choose the bigger sum and store it
// if all rows are finished, find the largest number in the last row
 
// read first line, just one number
std::vector<unsigned int> last(1);
std::cin >> last[0];
 
// read the remaining lines
for (unsigned int row = 1; row < numRows; row++)
{
// prepare array for new row
unsigned int numElements = row + 1;
std::vector<unsigned int> current;
 
// read all numbers of current row
for (unsigned int column = 0; column < numElements; column++)
{
unsigned int x;
std::cin >> x;
 
// find sum of elements in row above (going a half step to the left)
unsigned int leftParent = 0;
// only if left parent is available
if (column > 0)
leftParent = last[column - 1];
 
// find sum of elements in row above (going a half step to the right)
unsigned int rightParent = 0;
// only if right parent is available
if (column < last.size())
rightParent = last[column];
 
// add larger parent to current input
unsigned int sum = x + std::max(leftParent, rightParent);
// and store this sum
current.push_back(sum);
}
 
// row is finished, it become the "parent" row
last = current;
}
 
// find largest sum in final row
std::cout << *std::max_element(last.begin(), last.end()) << std::endl;
}
 
return 0;
}

This solution contains 13 empty lines, 17 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 24, 2017 submitted solution
April 3, 2017 added comments

Hackerrank

see https://www.hackerrank.com/contests/projecteuler/challenges/euler018

My code solves 6 out of 6 test cases (score: 100%)

Difficulty

5% 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
  the flashing problem is the one I solved most recently

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The 289 solved problems (level 11) had an average difficulty of 32.1% at Project Euler and
I scored 13,486 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.
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