Codewars #1: Fusion Chamber Shutdown

Resource: Codewars
Kyu: 7
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Practices:

• 4.14.25

Working out the Problem

Given an array, the contents entail [C, H, O], quantities of carbon, hydrogen, and oxygen elements. These elements combine to form the molecules: water (H20), carbon dioxide (CO2), and methane (CH4). They are produced in order of affinity, meaning if there are 20 C, 120 H, 20 O, the H and O will combine to form 20 water molecules, leaving 20 C and 80 H. The remainder will combine into 20 methane molecules.

1. Water: H2O
   H = O x 2

   • eg. H = 20, O = 10 :: H = O x 2
     Water = O = 10
     H = 0, O = 0

   • eg. H = 17, O = 10 :: H < O x 2
     Water = (H - 1) / 2 = 8
     H = 1, O = 2

   • eg. H = 22, O = 10 :: H > O x 2
     Water = O = 10
     H = 2, O = 0

2. Carbon dioxide: CO2

   O = C x 2

3. Methane: CH4
   H = C x 4

const burner = (c, h, o) => {
    const mole = []

    if (h >= o * 2)
        mole.push(o)
    else
        mole.push(Math.floor(h / 2))
    o -= mole[0]
    h -= mole[0] * 2

    if (o >= c * 2)
        mole.push(c)
    else
        mole.push(Math.floor(o / 2))
    c -= mole[1]
    o -= mole[1] * 2

    if (h >= c * 4)
        mole.push(c)
    else
        mole.push(Math.floor(h / 4))

    return mole
}

Submitted:

function burner(c, h, o) {
  const water = h >= o * 2 ? o : Math.floor(h / 2)
  o -= water
  h -= water * 2

  const co2 = o >= c * 2 ? c : Math.floor(o / 2)
  c -= co2
  o -= co2 * 2

  const methane = h >= c * 4 ? c : Math.floor(h / 4)

  return [water, co2, methane];
}

Iterative

Rather than the math, use incrementation and count how many iterations before exceeding either number. For example, in water, there are 2H for every O, so incrementing H +2 and O +1, regardless of whether even or odd, and whichever is higher, the max number of iterations is the number of molecules produced. However, to use the end sums to determine remaining H and O, the sum prior to exceeding the max has to be used, so there is a check, else while incrementing for each iteration, decrement from the amount of H and O available.

let water = 0
do {
    o--
    h -= 2
    water++
} while (h - 2 > 0 && o - 1 > 0
// [5, 44, 1] should be [5, 45, 0]
// [1, 1, 352] should be [0, 0, 354]

for (h, o; h > 0 && o > 0; h -= 2, o--)
    water++
// [ 5, 44, +0 ] should be [ 5, 45, +0 ]
// [ +0, +0, 353 ] should be [ +0, +0, 354 ]

while (h - 2 >= 0 && o > 0) {
    water++
    h -= 2
    o--
}

Mininum

The thinking I missed: Water is always 2H for every O, and whichever number is lower naturally fits into the other and can is the number of molecules made.

I didn’t know this before: ~~ provides integer closer to 0

Refer clever solution

const water = Math.min(~~(H / 2), O)
const carbon = Math.min(~~((O - water) / 2), C)
const methane = Math.min(~~((H - water * 2) / 4), C - carbon)

Solutions

1. Compare and Decrement

    const burner = (c, h, o) => {
        const water = h >= o * 2 ? o : Math.floor(h / 2)
        o -= water
        h -= water * 2
   
        const co2 = o >= c * 2 ? c : Math.floor(o / 2)
        c -= co2
        o -= co2 * 2
   
        const methane = h >= c * 4 ? c : Math.floor(h / 4)
        return [water, co2, methane]
    }
   

2. Iterative

    const burner = (c, h, o) => {
        let water = 0, co2 = 0, methane = 0
        while (h - 2 >= 0 && o > 0) {
            water++
            h -= 2
            o--
        }
        while (o - 2 >= 0 && c > 0) {
            co2++
            o -= 2
            c--
        }
        while (h - 4 >= 0 && c > 0) {
            methane++
            h -= 4
            c--
        }
        return [water, co2, methane]
    }
   

3. Minimum

    const burner = (c, h, o) => {
        const water = Math.min(~~(h / 2), o)
        h -= water * 2
        o -= water
        const co2 = Math.min(~~(o / 2), c)
        c -= co2
        const methane = Math.min(~~(h / 4), c)
        return [water, co2, methane]
    }