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: 1Solution 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); } } }
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