518. Coin Change 2

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

knapsack問題，定義一個2d dp table dp[i][j]代表前i個硬幣組合成金額j的可能數量。 dp[i][j] = dp[i - 1][j] + dp[i][j - coins[i]] dp[i - 1][j]代表不使用第i個coin，dp[i][j - coins[i]]代表使用第i個coin。

// Dynamic Programming
int change(int amount, vector<int>& coins) { // time: O(# of coins * amount); space: O(# of coins * amount)
    // dp[i][j]: # of combinations to make up j amount by the first i coins
    vector<vector<int> > dp(coins.size() + 1, vector<int> (amount + 1, 0));
    dp[0][0] = 1;
    for (int i = 1; i <= coins.size(); ++i) {
        dp[i][0] = 1; // initialization
        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.back().back();
}

優化空間的使用，稍微把2D table畫出來就可以理解其實只需要一個row就能計算。

// Space-Optimized Dynamic Programming
int change(int amount, vector<int>& coins) { // time: O(# of coins * amount); space: O(amount)
    vector<int> dp(amount + 1, 0);
    dp[0] = 1;
    for (int coin : coins) {
        for (int i = coin; i <= amount; ++i) {
            dp[i] += dp[i - coin];
        }
    }
    return dp.back();
}