518. Coin Change 2

You are given coins of different denominations and a total amount of money. Write a function to compute the number of combinations that make up that amount. You may assume that you have infinite number of each kind of coin.

Note:You can assume that

• 0 <= amount <= 5000
• 1 <= coin <= 5000
• the number of coins is less than 500
• the answer is guaranteed to fit into signed 32-bit integer

Thoughts:

1. dp[i][j]: the number of combinations to make up amount jby using the first itypes of coinsState transition:

   1. not using the ith coin, only using the first i-1coins to make up amount j, then we have dp[i-1][j] ways.
   2. using the ith coin, since we can use unlimited same coin, we need to know how many way to make up amount j - coins[i]by using first icoins( including ith), which is dp[i][j-coins[i]]
   3. Initialization:dp[i][0] = 1

2. dp[i][j]only rely on dp[i-1][j]and dp[i][j-coins[i]], then we can optimize the space by only using one-dimension array.

Code: 2D DP: O(K * N), S: O(K * N)

class Solution {
    public int change(int amount, int[] coins) {
        int[][] dp = new int[coins.length+1][amount+1];
        dp[0][0] = 1;

        for (int i = 1; i <= coins.length; i++) {
            dp[i][0] = 1;
            for (int j = 1; j <= amount; j++) {
                dp[i][j] = dp[i-1][j] + (j >= coins[i-1] ? dp[i][j-coins[i-1]] : 0);
            }
        }
        return dp[coins.length][amount];
    }
}

Code: 1D DP: O(K * N), S: O(N)

class Solution {
    public int change(int amount, int[] coins) {
        if(amount < 0) return 0;
        int dp[] = new int [amount + 1];
        dp[0] = 1;
        for(int coin: coins){
            for(int i = coin; i <= amount ; i++){
                dp[i]+= dp[i - coin];
            }
        }

        return dp[amount];
    }
}