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Problem 31: Coin sums

In England the currency is made up of pound, £, and pence, p, and there are eight coins in general circulation:

1p, 2p, 5p, 10p, 20p, 50p, £1 (100p) and £2 (200p).

It is possible to make £2 in the following way:

1x £1 + 1x 50p + 2x 20p + 1x 5p + 1x 2p + 3x 1p

How many different ways can £2 be made using any number of coins?

My Algorithm

My program creates a table history that contains the number of combinations for a given sum of money:

• its entry history[0]  refers to £0
• its entry history[1] refers to £0.01
• its entry history[2] refers to £0.02
• its entry history[3] refers to £0.03
• ...
• its entry history[200] refers to £2.00
There are 8 different coins and therefore each entry of history is a std::vector itself with 8 elements:
it tells how many combinations exist if only the current coin or smaller coins are used.

For example, there is always one way/combination to pay a certain amout if you only have single pennies.
That means, the first element is always 1.

Moreover, each of the next element is at least 1, too, because I said: "current coin or smaller coins".
If we would like to pay £0.01 then history[1] = { 1,1,1,1,1,1,1,1 }.

Now comes the only part that isn't obvious: there is one combination of paying zero pounds, too:
history[0] = { 1,1,1,1,1,1,1,1 }. From now on, everything comes natural, trust me, ...

If we would like to pay £0.02 then there are two ways: pay with two single pennies or a 2p coin.
What we do is:
1. try not to use the current coin (2p in our case), only smaller coins → there is one combination
2. try to use the current coin (2p in our case) → then there are 0.00 £ left which is possible in one way

So far we had history[2] = { 1,?,?,?,?,?,?,? }
Step 1 is the same as history[2][currentCoinId - 1] = history[2][0] = 1.
Step 2 is the same as history[2 - currentCoinValue][currentCoinId] = history[0][1] = 1.
Therefore we have 1+1=2 combinations (as expected:) history[2] = { 1,2,?,?,?,?,?,? }.

The next coin, it's the 5p coin, can't be used because it's bigger than the total of £0.02. In software terms currentCoinValue > total.
Only step 1 applies to all remaining elements: history[2] = { 1,2,2,2,2,2,2,2 }.

What does it mean ? There are 2 ways to pay 0.02 £ with 1p and 2p. And there are still only two ways if you use all coins up to £2.
When the program computes history[200] then the result of the problem is stored in the last element (history[200][7]).

Modifications by HackerRank

There are multiple test cases. My program computes all combinations up to the input values and stores them in history.
If a test case's input is smaller than something we had before then no computation is required at all, it will become a basic table lookup.

The results may exceed 32 bits and thus I compute mod 10^9+7 whenever possible (as requested by their modified problem statement).

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):
Note: please enter in pence, not pound

This is equivalent to
echo "1 10" | ./31

Output:

(please click 'Go !')

Note: the original problem's input 200 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. Or just jump to my GitHub repository.

       #include <iostream>
#include <vector>

const unsigned int NumCoins = 8;
// face value of all coins in cents
const unsigned int Coins[NumCoins] = { 1,2,5,10,20,50,100,200 };

// store number of combinations in [x] if coin[x] is allowed:
// [0] => combinations if only pennies are allowed
// [1] => 1 cent and 2 cents are allowed, nothing more
// [2] => 1 cent, 2 cents and 5 cents are allowed, nothing more
// ...
// [6] => all but 2 pounds (= 200 cents) are allowed
// [7] => using all coins if possible
typedef std::vector<unsigned long long> Combinations;

int main()
{
// remember combinations for all prices from 1 cent up to 200 cents (2 pounds)
std::vector<Combinations> history;

unsigned int tests;
std::cin >> tests;
while (tests--)
{
unsigned int total;
std::cin >> total;

// initially we start at zero
// but if there are previous test cases then we can re-use the old results
for (unsigned int cents = history.size(); cents <= total; cents++)
{
// count all combinations of those 8 coins
Combinations ways(NumCoins);

// one combination if using only 1p coins (single pennys)
ways[0] = 1;

// use larger coins, too
for (size_t i = 1; i < ways.size(); i++)
{
// first, pretend not to use that coin (only smaller coins)
ways[i] = ways[i - 1];

// now use that coin once (if possible)
auto currentCoin = Coins[i];
if (cents >= currentCoin)
{
auto remaining = cents - currentCoin;
ways[i] += history[remaining][i];
}

// not needed for the original problem, only for Hackerrank's modified problem
ways[i] %= 1000000007;
}

// store information for future use
history.push_back(ways);
}

// look up combinations
auto result = history[total];
// the last column (allow all coins) contains the desired value
auto combinations = result.back();
combinations %= 1000000007; // for Hackerrank only
std::cout << combinations << std::endl;
}

return 0;
}


This solution contains 12 empty lines, 20 comments and 2 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 9 out of 9 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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