Medium
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
Constraints:
1 <= coins.length <= 300
1 <= coins[i] <= 5000
coins
are unique.0 <= amount <= 5000
class Solution {
fun change(amount: Int, coins: IntArray): Int {
val dp = IntArray(amount + 1)
dp[0] = 1
for (coin in coins) {
for (i in coin..amount) {
dp[i] += dp[i - coin]
}
}
return dp[amount]
}
}