Problem
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
Solution
DP
class Solution {
public int change(int amount, int[] coins) {
int[] dp = new int[amount+1];
dp[0] = 1;
for (int coin: coins) {
for (int i = 1; i <= amount; i++) {
if (i - coin >= 0) dp[i] += dp[i-coin];
}
}
return dp[amount];
}
}
DFS – TLE
class Solution {
int count = 0;
public int change(int amount, int[] coins) {
if (amount == 0) return 1;
dfs(coins, 0, 0, amount);
return count;
}
private void dfs(int[] coins, int start, int sum, int amount) {
if (start == coins.length) return;
if (sum == amount) {
count++;
return;
}
for (int i = start; i < coins.length; i++) {
if (coins[i] + sum > amount) continue;
else dfs(coins, i, sum+coins[i], amount);
}
}
}