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# Problem 31: Coin sums

(see projecteuler.net/problem=31)

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?

# 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

3. add 1.+2.

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).

# My code

… was written in C++11 and can be compiled with G++, Clang++, Visual C++. You can download it, too.

#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.

# Interactive test

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

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

Output:

*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)*

# Benchmark

The correct solution to the original Project Euler problem was found in **less than 0.01** seconds on a Intel® Core™ i7-2600K CPU @ 3.40GHz.

(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

February 23, 2017 submitted solution

April 6, 2017 added comments

# Hackerrank

see https://www.hackerrank.com/contests/projecteuler/challenges/euler031

My code solved **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 never an option.

# Links

projecteuler.net/thread=31 - **the** best forum on the subject (*note:* you have to submit the correct solution first)

Code in various languages:

Python: www.mathblog.dk/project-euler-31-combinations-english-currency-denominations/ (written by Kristian Edlund)

Haskell: github.com/nayuki/Project-Euler-solutions/blob/master/haskell/p031.hs (written by Nayuki)

Java: github.com/nayuki/Project-Euler-solutions/blob/master/java/p031.java (written by Nayuki)

Mathematica: github.com/nayuki/Project-Euler-solutions/blob/master/mathematica/p031.mathematica (written by Nayuki)

C: github.com/eagletmt/project-euler-c/blob/master/30-39/problem31.c (written by eagletmt)

Go: github.com/frrad/project-euler/blob/master/golang/Problem031.go (written by Frederick Robinson)

Scala: github.com/samskivert/euler-scala/blob/master/Euler031.scala (written by Michael Bayne)

# 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 already 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.

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

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