You are given an integer array coins representing coins of different denominations and an integer amount representing a total amount of money.
Return the number of combinations that make up that amount. If that amount of money cannot be made up by any combination of the coins, return 0.
You may assume that you have an infinite number of each kind of coin.
The answer is guaranteed to fit into a signed 32-bit integer.
Example 1:
Input: amount = 5, coins = [1,2,5] Output: 4 Explanation: there are four ways to make up the amount: 5=5 5=2+2+1 5=2+1+1+1 5=1+1+1+1+1
Example 2:
Input: amount = 3, coins = [2] Output: 0 Explanation: the amount of 3 cannot be made up just with coins of 2.
Example 3:
Input: amount = 10, coins = [10] Output: 1
1 <= coins.length <= 3001 <= coins[i] <= 5000- All the values of
coinsare unique. 0 <= amount <= 5000
class Solution:
def change(self, amount: int, coins: List[int]) -> int:
dp = [0] * (amount + 1)
dp[0] = 1
for coin in coins:
for j in range(coin, amount + 1):
dp[j] += dp[j - coin]
return dp[amount]Time Complexity: O(n * m), where n is the amount and m is the number of coin denominations. This is because for each coin, we potentially update every entry in the dp array from the coin's value up to amount.
Space Complexity: O(n), where n is the amount. We need a 1D array of size amount + 1 to store the number of ways to make each amount from 0 to amount.