Défi · algorithmes moyen Python ·
Ways to make change
Count the ways to make amount from coins of the given distinct positive denominations, each usable any number of times. Order doesn't matter (1+2 and 2+1 are the same way). Return the count modulo 1,000,000,007. There is exactly one way to make 0. Up to 60 denominations and amounts up to 30,000.
Signature
def coin_ways(coins: list[int], amount: int) -> int:Exemples
coin_ways([1, 2, 5], 5) # -> 4
coin_ways([2], 3) # -> 0
coin_ways([10], 10) # -> 1Plus 7 tests masqués et 1 test de performance sous contrainte de temps ; leurs entrées ne sont pas affichées. Les défis sont soumis aux modèles en anglais.
#1Mistral Devstral 2 2512résolu · 11 sur 11 testsréponse en 906 ms · exécution 345 ms · $0.0003 ✓ coin_ways([1, 2, 5], 5)attendu 4 2 ms ✓ coin_ways([2], 3)attendu 0 <1 ms ✓ coin_ways([10], 10)attendu 1 <1 ms ✓ <1 ms✓ <1 ms✓ <1 ms✓ <1 ms✓ <1 ms✓ <1 ms✓ 15 ms⏱ 322 msCode
def coin_ways(coins: list[int], amount: int) -> int: MOD = 10**9 + 7 dp = [0] * (amount + 1) dp[0] = 1 for coin in coins: for i in range(coin, amount + 1): dp[i] = (dp[i] + dp[i - coin]) % MOD return dp[amount]#1Qwen3 Coder Nextrésolu · 11 sur 11 testsréponse en 2.8 s · exécution 352 ms · $0.0002 ✓ coin_ways([1, 2, 5], 5)attendu 4 <1 ms ✓ coin_ways([2], 3)attendu 0 <1 ms ✓ coin_ways([10], 10)attendu 1 2 ms ✓ 1 ms✓ <1 ms✓ 2 ms✓ 3 ms✓ <1 ms✓ <1 ms✓ 14 ms⏱ 327 msCode
def coin_ways(coins: list[int], amount: int) -> int: MOD = 1000000007 # dp[i] = number of ways to make amount i dp = [0] * (amount + 1) dp[0] = 1 # One way to make 0: use no coins # For each coin, update dp array for all amounts >= coin for coin in coins: for i in range(coin, amount + 1): dp[i] = (dp[i] + dp[i - coin]) % MOD return dp[amount]#1SpaceXAI Grok Build 0.1résolu · 11 sur 11 testsréponse en 5.4 s · exécution 354 ms · $0.0022 ✓ coin_ways([1, 2, 5], 5)attendu 4 <1 ms ✓ coin_ways([2], 3)attendu 0 <1 ms ✓ coin_ways([10], 10)attendu 1 <1 ms ✓ <1 ms✓ <1 ms✓ 1 ms✓ <1 ms✓ <1 ms✓ <1 ms✓ 14 ms⏱ 334 msCode
def coin_ways(coins: list[int], amount: int) -> int: MOD = 1000000007 dp = [0] * (amount + 1) dp[0] = 1 for coin in coins: for x in range(coin, amount + 1): dp[x] = (dp[x] + dp[x - coin]) % MOD return dp[amount]#1Anthropic Claude Haiku 5.5résolu · 11 sur 11 testsréponse en 1.7 s · exécution 355 ms · $0.0002 ✓ coin_ways([1, 2, 5], 5)attendu 4 <1 ms ✓ coin_ways([2], 3)attendu 0 2 ms ✓ coin_ways([10], 10)attendu 1 <1 ms ✓ <1 ms✓ <1 ms✓ 1 ms✓ 1 ms✓ 1 ms✓ 1 ms✓ 15 ms⏱ 330 msCode
def coin_ways(coins: list[int], amount: int) -> int: MOD = 1_000_000_007 if amount < 0: return 0 dp = [0] * (amount + 1) dp[0] = 1 for c in coins: if c <= 0 or c > amount: continue for a in range(c, amount + 1): dp[a] = (dp[a] + dp[a - c]) % MOD return dp[amount] % MOD
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