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 = [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
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[0][0] = 1;
for(int i = 1; i <= coins.length; i++) {
dp[i][0] = 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[0] = 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];
}
}