# Why?

This is the beginning of a new year (and decade) ๐ so it's time for a new commitment, last year I challenged myself on 365 days of coding problems with DailyCodingProblems and...I gave up after only 2 or 3 weeks...

This year I will try again! For 31 days and why not more, but let's do 31 days for now.

*Why try again?*

To motivate myself I choose to publish my solution of the daily problem here because you will probably have other solutions and that would be great to discuss algorithmic here.

*Why do that?*

Because it's fun! It's a good technique to improve your logic and secondly next year I will have to pass a technical interview for the last internship of my studies and I want to be prepared!

# How this series will work

First of all, this is my first series of articles and my first real article so don't hesitate to comments, I want to improve my writing skills (and my English also).

Every two days (or one time a week) I will post a problem I received and my solution, you can find here the repo don't hesitate to check the repo ๐.

# Day 1

- Difficulty: Medium
- Programming language used: Kotlin

## Problem

Write a function, `throw_dice(N, faces, total)`

that determines how many ways it is possible to

throw `N`

dices with some number of `faces`

each to get a specific `total`

.

For example:

`throw_dice(3, 6, 7)`

should return 15.

## My solution

I used a concept called dynamic programming (DP) to solve this one.

Finding how many ways it is possible to throw N dices to get a specific amount (X) is equivalent to find (and sum) how many ways it is possible to throw N-1 dices to get an X-1 amount, N-1 to get X-2, N-1 to get X-3, etc..until X and N equals 1.

```
/**
* Determine how many ways it is possible to throw N dices with the number of faces each to get the total
*
* @param N, number of dices
* @param faces, number of faces of each dice
* @param total, the sum of the dices we would like to get
* @return Int, number of ways
*/
fun throw_dice(N: Int, faces: Int, total: Int): Int {
val subProblems = Array(N+1) { IntArray(total+1) { 0 } }
var i = 1
while (i < total && i < faces) {
subProblems[1][i] = 1
i++
}
for (w in 2..N) {
for (j in 1..total) {
for (k in 1..faces) {
if(k >= j)
break
subProblems[w][j] += subProblems[w - 1][j - k]
}
}
}
return subProblems[N][total]
}
```

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