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]))
Danke für Ihr Feedback!
single
Fragen Sie AI
Fragen Sie AI
Fragen Sie alles oder probieren Sie eine der vorgeschlagenen Fragen, um unser Gespräch zu beginnen
Awesome!
Completion rate improved to 8.33Problem D. Coin Change
Swipe um das Menü anzuzeigen
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]))
Wechseln Sie zum Desktop, um in der realen Welt zu übenFahren Sie dort fort, wo Sie sind, indem Sie eine der folgenden Optionen verwendenDanke für Ihr Feedback!
Awesome!
Completion rate improved to 8.33single
