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 fewest number of coins that you need to make up that amount. If that amount of money cannot be made up by any combination of the coins, return -1
.
You may assume that you have an infinite number of each kind of coin.
Example 1:
Input: coins = [1,2,5], amount = 11
Output: 3
Explanation: 11 = 5 + 5 + 1
Example 2:
Input: coins = [2], amount = 3
Output: -1
Example 3:
Input: coins = [1], amount = 0
Output: 0
Constraints:
1 <= coins.length <= 12
1 <= coins[i] <= 231 - 1
0 <= amount <= 104
object Solution {
def coinChange(coins: Array[Int], amount: Int): Int = {
val dp = Array.fill(amount + 1)(0)
dp(0) = 0
for (i <- 1 to amount) {
var minCoins = Int.MaxValue
for (coin <- coins) {
if (i - coin >= 0 && dp(i - coin) != -1) {
minCoins = math.min(minCoins, dp(i - coin) + 1)
}
}
dp(i) = if (minCoins == Int.MaxValue) -1 else minCoins
}
dp(amount)
}
}