Problema
Use any number of each supplied denomination to make amount exactly. Count combinations, not orderings: two arrangements using the same counts of each coin are the same way. Denominations are distinct and positive.
Implement coin_sums(amount, coins).
Input
amount: The total amount to make.
coins: The available distinct denominations.
Output
Return the number of unordered combinations.
Restrições
0 ≤ amount ≤ 200. There are 0 through 10 distinct denominations, each from 1 through 200. All counts fit signed 64-bit. Amount 0 has one way: choose no coins.
Exemplos
Entrada:
coins = [1, 2]
amount = 4
Saída esperada:
expected = 3Entrada:
std::vector<long long> coins = {1, 2};
long long amount = 4;
Saída esperada:
long long expected = 3;Entrada:
long[] coins = new long[] {1, 2};
long amount = 4;
Saída esperada:
long expected = 3;Explicação
The combinations are four 1s, two 1s with one 2, and two 2s.
Entrada:
coins = [2]
amount = 3
Saída esperada:
expected = 0Entrada:
std::vector<long long> coins = {2};
long long amount = 3;
Saída esperada:
long long expected = 0;Entrada:
long[] coins = new long[] {2};
long amount = 3;
Saída esperada:
long expected = 0;Explicação
Any sum of 2-valued coins is even, so 3 is impossible.
Como abordar
Tente sozinho primeiro e depois abra uma dica de cada vez.
Dica 1
The order of coins does not matter, so combinations should be counted by denomination, not by permutation.
Dica 2
Build the number of ways for each amount incrementally, letting each coin contribute to all reachable totals.
Dica 3
Be careful with amount zero, which always has one valid way: choose no coins.
Complexidade esperada
O(amount × number_of_coins) time, O(amount) extra space
Referências
Este desafio foi adaptado para o CodeUtility. Siga o enunciado e as restrições acima.