Progress with InterviewCake's coin problem
I'm writing a function that returns the total number of ways a cashier can make change given the `amount` of change that the customer needs and an array of `coins` from which to create the change. My solution conceptually works but it actually does not return the results I am expecting because I cannot create a Set of Map<A, B> in JavaScript. I'm also somewhat sure that InterviewCake is expecting a less computationally expensive answer.
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								scratch/deepmind/part_two/coin.ts
									
										
									
									
									
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								scratch/deepmind/part_two/coin.ts
									
										
									
									
									
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							|  | @ -0,0 +1,102 @@ | |||
| // The denomination of a coin.
 | ||||
| type Coin = number; | ||||
| 
 | ||||
| // The amount of change remaining.
 | ||||
| type Amount = number; | ||||
| 
 | ||||
| // Mapping of Coin -> Int
 | ||||
| type CoinBag = Map<Coin, number>; | ||||
| 
 | ||||
| function createCoinBag(coins: Coin[]): CoinBag { | ||||
|   const result = new Map(); | ||||
| 
 | ||||
|   for (const coin of coins) { | ||||
|     result.set(coin, 0); | ||||
|   } | ||||
| 
 | ||||
|   return result; | ||||
| } | ||||
| 
 | ||||
| // This algorithm should work conceptual, but it does not actually
 | ||||
| // work. JavaScript uses reference equality when constructing a Set<Map<A,B>>,
 | ||||
| // so my result.size returns a higher number than I expect because it contains
 | ||||
| // many duplicate entries.
 | ||||
| //
 | ||||
| // Conceptually, I'm not sure this solution is optimal either -- even after I
 | ||||
| // can dedupe the entries in `result`.
 | ||||
| function changePossibilities(amt: Amount, coins: Coin[]): number { | ||||
|   if (amt === 0) { | ||||
|     return 1; | ||||
|   } | ||||
|   const result: Set<CoinBag> = new Set(); | ||||
| 
 | ||||
|   const q: [Coin, Amount, CoinBag][] = []; | ||||
| 
 | ||||
|   for (const coin of coins) { | ||||
|     const bag = createCoinBag(coins); | ||||
|     bag.set(coin, 1); | ||||
|     q.push([coin, amt - coin, bag]); | ||||
|   } | ||||
| 
 | ||||
|   while (q.length > 0) { | ||||
|     const [coin, amt, bag] = q.shift(); | ||||
| 
 | ||||
|     console.log([coin, amt, bag]); | ||||
| 
 | ||||
|     if (amt === 0) { | ||||
|       result.add(bag); | ||||
|     } else if (amt < 0) { | ||||
|       continue; | ||||
|     } else { | ||||
|       for (const c of coins) { | ||||
|         const bagCopy = new Map(bag); | ||||
|         const value = bagCopy.get(c); | ||||
|         bagCopy.set(c, value + 1); | ||||
|         q.push([c, amt - c, bagCopy]); | ||||
|       } | ||||
|     } | ||||
|   } | ||||
|   console.log(result); | ||||
|   return result.size; | ||||
| } | ||||
| 
 | ||||
| // Tests
 | ||||
| let desc = "sample input"; | ||||
| let actual = changePossibilities(4, [1, 2, 3]); | ||||
| let expected = 4; | ||||
| assertEqual(actual, expected, desc); | ||||
| 
 | ||||
| desc = "one way to make zero cents"; | ||||
| actual = changePossibilities(0, [1, 2]); | ||||
| expected = 1; | ||||
| assertEqual(actual, expected, desc); | ||||
| 
 | ||||
| desc = "no ways if no coins"; | ||||
| actual = changePossibilities(1, []); | ||||
| expected = 0; | ||||
| assertEqual(actual, expected, desc); | ||||
| 
 | ||||
| desc = "big coin value"; | ||||
| actual = changePossibilities(5, [25, 50]); | ||||
| expected = 0; | ||||
| assertEqual(actual, expected, desc); | ||||
| 
 | ||||
| desc = "big target amount"; | ||||
| actual = changePossibilities(50, [5, 10]); | ||||
| expected = 6; | ||||
| assertEqual(actual, expected, desc); | ||||
| 
 | ||||
| // I think InterviewCake designed this assertion to be computationally
 | ||||
| // expensive.
 | ||||
| desc = "change for one dollar"; | ||||
| actual = changePossibilities(100, [1, 5, 10, 25, 50]); | ||||
| expected = 292; | ||||
| assertEqual(actual, expected, desc); | ||||
| 
 | ||||
| function assertEqual(a, b, desc) { | ||||
|   if (a === b) { | ||||
|     console.log(`${desc} ... PASS`); | ||||
|   } else { | ||||
|     console.log(`${desc} ... FAIL: ${a} != ${b}`); | ||||
|   } | ||||
| } | ||||
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