package kata

import java.math.BigInteger

/*
You know combinations: for example, if you take 5 cards from a 52 cards deck you have 2,598,960
different combinations.

In mathematics the number of x combinations you can take from a set of n elements is called the
binomial coefficient of n and x, or more often n choose x.
The formula to compute m = n choose x is: m = n! / (x! * (n - x)!) where ! is the factorial operator.

You are a renowned poster designer and painter.
You are asked to provide 6 posters all having the same design each in 2 colors.
Posters must all have a different color combination and you have the choice of 4 colors:
red, blue, yellow, green.
How many colors can you choose for each poster?

The answer is two since 4 choose 2 = 6.
The combinations will be: {red, blue}, {red, yellow}, {red, green}, {blue, yellow}, {blue, green}, {yellow, green}.

Now same question but you have 35 posters to provide and 7 colors available.
How many colors for each poster?
If you take combinations 7 choose 2 you will get 21 with the above formula.
But 21 schemes aren't enough for 35 posters.
If you take 7 choose 5 combinations you will get 21 too.
Fortunately if you take 7 choose 3 or 7 choose 4 combinations you get 35 and so each poster
will have a different combination of 3 colors or 5 colors.
You will take 3 colors because it's less expensive.

Hence the problem is:

knowing m (number of posters to design),
knowing n (total number of available colors),
let us search x (number of colors for each poster so that each poster has a unique combination of
colors and the number of combinations is exactly the same as the number of posters).

In other words we must find x such as n choose x = m (1) for a given m and a given n; m >= 0 and n > 0.
If many x are solutions give as result the smallest x.
It can happen that when m is given at random there are no x satisfying equation (1) then return -1.

Examples:

checkchoose(6, 4) --> 2
checkchoose(4, 4) --> 1
checkchoose(4, 2) --> -1
checkchoose(35, 7) --> 3
checkchoose(36, 7) --> -1

a = 47129212243960
checkchoose(a, 50) --> 20
checkchoose(a + 1, 50) --> -1

clojure: Use big integers in the calculation of n choose k!

*/
fun fact(num: Int): BigInteger {
    var factorial: BigInteger = BigInteger.ONE
    for (i in 1..num) {
        factorial = factorial.multiply(BigInteger.valueOf(i.toLong()))
    }
    return factorial
}

fun nChooseX(n: Int, x: Int): BigInteger {
    // m = n! / (x! * (n - x)!)
    return fact(n).div(fact(x).multiply(fact(n - x)))
}

fun checkchoose(m: Long, n: Int): Long {
    for (x in 1..n) {
        val mCurrent = nChooseX(n, x)
        if (mCurrent.longValueExact() == m) {
            return x.toLong()
        }
    }
    return -1
}
package kata

import java.math.BigInteger

/*
You know combinations: for example, if you take 5 cards from a 52 cards deck you have 2,598,960
different combinations.

In mathematics the number of x combinations you can take from a set of n elements is called the
binomial coefficient of n and x, or more often n choose x.
The formula to compute m = n choose x is: m = n! / (x! * (n - x)!) where ! is the factorial operator.

You are a renowned poster designer and painter.
You are asked to provide 6 posters all having the same design each in 2 colors.
Posters must all have a different color combination and you have the choice of 4 colors:
red, blue, yellow, green.
How many colors can you choose for each poster?

The answer is two since 4 choose 2 = 6.
The combinations will be: {red, blue}, {red, yellow}, {red, green}, {blue, yellow}, {blue, green}, {yellow, green}.

Now same question but you have 35 posters to provide and 7 colors available.
How many colors for each poster?
If you take combinations 7 choose 2 you will get 21 with the above formula.
But 21 schemes aren't enough for 35 posters.
If you take 7 choose 5 combinations you will get 21 too.
Fortunately if you take 7 choose 3 or 7 choose 4 combinations you get 35 and so each poster
will have a different combination of 3 colors or 5 colors.
You will take 3 colors because it's less expensive.

Hence the problem is:

knowing m (number of posters to design),
knowing n (total number of available colors),
let us search x (number of colors for each poster so that each poster has a unique combination of
colors and the number of combinations is exactly the same as the number of posters).

In other words we must find x such as n choose x = m (1) for a given m and a given n; m >= 0 and n > 0.
If many x are solutions give as result the smallest x.
It can happen that when m is given at random there are no x satisfying equation (1) then return -1.

Examples:

checkchoose(6, 4) --> 2
checkchoose(4, 4) --> 1
checkchoose(4, 2) --> -1
checkchoose(35, 7) --> 3
checkchoose(36, 7) --> -1

a = 47129212243960
checkchoose(a, 50) --> 20
checkchoose(a + 1, 50) --> -1

clojure: Use big integers in the calculation of n choose k!

*/
fun fact(num: Int): BigInteger {
    var factorial: BigInteger = BigInteger.ONE
    for (i in 1..num) {
        factorial = factorial.multiply(BigInteger.valueOf(i.toLong()))
    }
    return factorial
}

fun nChooseX(n: Int, x: Int): BigInteger {
    // m = n! / (x! * (n - x)!)
    return fact(n).div(fact(x).multiply(fact(n - x)))
}

fun checkchoose(m: Long, n: Int): Long {
    for (x in 1..n) {
        val mCurrent = nChooseX(n, x)
        if (mCurrent.longValueExact() == m) {
            return x.toLong()
        }
    }
    return -1
}