[DP] Minimum number of coins need for a specific amount
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
Explanation:
We use dynamic programming for this problem: First, we try to resolve the smaller problem!
Suppose we have numbers type of coins = [1,2,5] We need to get minimum coins for amount 2 The smallest amount is 0 then we need 0 coin With amount 1 we need 1 coin With amount 2: For case: coin 1 < 2, mean coin 1 can be used, then 1 coin We need to calculate the coins for remain amount The reimain amount is 2-1 = 1 No. of coin for amount 1 is 1 => for this case we need 2 coins. For case: coin 2 <= 2, mean coin 2 can be used, then we have 1 coin We need to calculate the coins for remain amount The reimain amount is 2 - 2 = 0 No. of coin for amount 0 is 0 => for this case we need 1 coins. compare with the previouse case, 1<2 then the answer is 1 For case 5: 5 > 2, cannot use Then we just continue until the amount we're looking for.
Solution:
public int coinChange(int[] coins, int amount) { int[] dp = new int[amount+1]; Arrays.fill(dp, amount + 1); dp[0] = 0; Arrays.sort(coins); for (int a = 1; a <= amount; a++){ for(int i = 0; i<coins.length; i++){ if (coins[i] <= a){ int remain = a - coins[i]; int numOfCoin = 1 + dp[remain]; dp[a] = Math.min(dp[a], numOfCoin); } } } return dp[amount] > amount? -1: dp[amount]; }
0 comments:
Post a Comment