Amazon

Bloomberg

2020-06-07

## 518. Coin Change 2

### Question:

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.

#### 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 = 
Output: 0
Explanation: the amount of 3 cannot be made up just with coins of 2.
``````

#### Example 3:

``````Input: amount = 10, coins = 
Output: 1
``````

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

### Solution:

Using DP to solve the problem, `DP[i][j]` means select ith coin with total sum to j.

Time: O(n * m)
Space: O(n * m)
`m` means amount

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

for(int i = 1; i <= coins.length; i++) {
dp[i] = 1;
int curr = coins[i - 1];
for (int j = 1; j <= amount; j++) {
if (j >= curr) {
dp[i][j] = dp[i - 1][j] + dp[i][j - curr];
} else {
dp[i][j] = dp[i - 1][j];
}
}
}
return dp[coins.length][amount];
}
}
``````

It can be reduced to 1-D array.

Time: O(n * m)
Space: O(m)

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

for(int i = 1; i <= coins.length; i++) {
int curr = coins[i - 1];
for (int j = 1; j <= amount; j++) {
if (j >= curr) {
dp[j] += dp[j - curr];
}
}
}
return dp[amount];
}
}
``````