-
Notifications
You must be signed in to change notification settings - Fork 2
/
Coin Change 2.java
64 lines (52 loc) · 2.12 KB
/
Coin Change 2.java
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
/*
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
*/
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 way to make up amount j - coins[i] by using first i coins(including ith), which is dp[i][j-coins[i]]
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];
}