Problem D. Coin Change
Imagine you got N cents as combination of some coins, and the last added coin was C. Then, number of possible combinations dp[N] is equal to dp[N-C]. Consider that you can reach N cents by adding either c[0], c[1], ... ,c[m-1] cents, so number of possible combinations is:
dp[N] = dp[N-c[0]] + dp[N-c[1]] + ... + dp[N-c[m-1]]
Note that value of N-c[i] must be non-negative. Let's use tabulation: for values j from coin up to N: update dp[j] with adding dp[j-coin]; repeat for each coin.
12345678910def coinChange(n , coins): dp = [0 for _ in range(n+1)] dp[0] = 1 for i in range(len(coins)): for j in range(coins[i], n+1): dp[j] += dp[j-coins[i]] return dp[n] print(coinChange(14, [1,2,3,7])) print(coinChange(100, [2,3,5,7,11]))
Kiitos palautteestasi!
single
Kysy tekoälyä
Kysy tekoälyä
Kysy mitä tahansa tai kokeile jotakin ehdotetuista kysymyksistä aloittaaksesi keskustelumme
Tiivistä tämä luku
Explain code
Explain why doesn't solve task
Awesome!
Completion rate improved to 8.33Problem D. Coin Change
Pyyhkäise näyttääksesi valikon
Imagine you got N cents as combination of some coins, and the last added coin was C. Then, number of possible combinations dp[N] is equal to dp[N-C]. Consider that you can reach N cents by adding either c[0], c[1], ... ,c[m-1] cents, so number of possible combinations is:
dp[N] = dp[N-c[0]] + dp[N-c[1]] + ... + dp[N-c[m-1]]
Note that value of N-c[i] must be non-negative. Let's use tabulation: for values j from coin up to N: update dp[j] with adding dp[j-coin]; repeat for each coin.
12345678910def coinChange(n , coins): dp = [0 for _ in range(n+1)] dp[0] = 1 for i in range(len(coins)): for j in range(coins[i], n+1): dp[j] += dp[j-coins[i]] return dp[n] print(coinChange(14, [1,2,3,7])) print(coinChange(100, [2,3,5,7,11]))
Vaihda työpöytään todellista harjoitusta vartenJatka siitä, missä olet käyttämällä jotakin alla olevista vaihtoehdoistaKiitos palautteestasi!
single
