CareerCup

Just find there is a number that exists within the min_max range divisible by options and their combinations
In this example that would be (3,4,7,10,11)

You don't need to calculate for each value in min-max range of cup. just calculate for minimum value, if the remainder is any one of the options+combinations array, we should be returning a true.

The solution can be optimised further as it's an O(n*k) still.

In order to figure out if the coffee can be filed, one of the following should lie in the min-max range of cup
- Multiples of pouring options or their combinations (3/4/7/10/11)
[10 and 11 are 3+7 and 4+7 (3+4 is already covered)]

all values >= 20 can be hit:

X % 4 == 0 ? X/4 * 4
X % 4 == 3 ? (X div 4 - 1) * 4 + 1 * 7
X % 4 == 2 ? (X div 4 - 3) * 4 + 2 * 7
X % 4 == 1 ? (X div 4 - 5) * 4 + 3 * 7

for values <20 simply write a table in O(1)

all values >= 20 can be hit:

X % 4 == 0 ? X/4 * 4
X % 4 == 3 ? (X div 4 - 1) * 4 + 1 * 7
X % 4 == 2 ? (X div 4 - 3) * 4 + 2 * 7
X % 4 == 1 ? (X div 4 - 5) * 4 + 3 * 7

for values <20 simply write a table in O(1)

For small min and max values brute force recursive just works:

def canFillTheCup(min: Int, max: Int, buttonAmounts: Seq[Int]): Boolean = {
    if (buttonAmounts.exists(ba => min <= ba && ba <= max)) {
      true
    } else {
      buttonAmounts.exists(ba => min - ba > 0 && canFillTheCup(min - ba, max - ba, buttonAmounts))
    }
  }