518. Coin Change 2

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
 
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




https://leetcode.com/problems/coin-change-2/discuss/99212/Knapsack-problem-Java-solution-with-thinking-process-O(nm)-Time-and-O(m)-Space



https://www.youtube.com/watch?v=ZKAILBWl08g


dp[i][j] : the number of combinations to make up amount j by using the first i types of coins
State transition:

1. not using the ith coin, only using the first i-1 coins 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 ways to make up amount j - coins[i-1] by using first i coins(including ith), which is dp[i][j-coins[i-1]]

Initialization: dp[i][0] = 1
我们采用的方法是一个硬币一个硬币的增加，每增加一个硬币，都从1遍历到amount，对于遍历到的当前钱数j，组成方法就是不加上当前硬币的频发dp[i-1][j]，还要加上，去掉当前硬币值的钱数的组成方法，当然钱数j要大于当前硬币值，那么我们的递推公式也在上面的分析中得到了：
dp[i][j] = dp[i - 1][j] + (j >= coins[i - 1] ? dp[i][j - coins[i - 1]] : 0)
注意我们要初始化每行的第一个位置为0，参见代码如下：      

Once you figure out all these, it's easy to write out the code:

    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];
    }

Now we can see that 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.

    public int change(int amount, int[] coins) {
        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];
    }

 

This is a classic knapsack problem. Honestly, I'm not good at knapsack problem, it's really tough for me.

dp[i][j] : the number of combinations to make up amount j by using the first i types of coins
State transition:

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

Once you figure out all these, it's easy to write out the code:

    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];
    }
Now we can see that 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.

    public int change(int amount, int[] coins) {
        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];
    }