518. Coin Change 2

518. Coin Change 2 (Medium)

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 1. DFS + Memo

// OJ: https://leetcode.com/problems/coin-change-2/
// Author: github.com/lzl124631x
// Time: O(A * C^2)
// Space: O(AC)
class Solution {
private:
    unordered_map<int, int> memo;
    int change(int amount, vector<int>& coins, int start) {
        if (!amount) return 1;
        if (start >= coins.size()) return 0;
        int key = amount * 1000 + start;
        if (memo.find(key) != memo.end()) return memo[key];
        int ans = 0;
        for (int i = start; i < coins.size(); ++i) {
            if (amount >= coins[i]) ans += change(amount - coins[i], coins, i);
        }
        return memo[key] = ans;
    }
public:
    int change(int amount, vector<int>& coins) {
        return change(amount, coins, 0);
    }
};

Solution 2. DP (Unbounded knapsack problem), Naive version

This is a typical unbounded knapsack problem where you can pick item unlimited times (unbounded).

Let dp[i+1][T] be the ways to form T value using A[0] ... A[i].

dp[i+1][T]      = dp[i][T] + dp[i][T-A[i]] + dp[i][T-2*A[i]] ...

dp[i][0] = 1

We can directly apply this formula.

This is the naive solution, and in the next solution we find way to optimize it.

// OJ: https://leetcode.com/problems/coin-change-2/
// Author: github.com/lzl124631x
// Time: O(N^2 * T)
// Space: O(NT)
class Solution {
public:
    int change(int T, vector<int>& A) {
        int N = A.size();
        vector<vector<int>> dp(N + 1, vector<int>(T + 1));
        for (int i = 0; i <= N; ++i) dp[i][0] = 1;
        for (int i = 0; i < N; ++i) {
            for (int t = 1; t <= T; ++t) {
                for (int k = 0; t - k * A[i] >= 0; ++k) {
                    dp[i + 1][t] += dp[i][t - k * A[i]];
                }
            }
        }
        return dp[N][T];
    }
};

Solution 3. DP (Unbounded knapsack problem)

Let dp[i+1][T] be the ways to form T value using A[0] ... A[i].

dp[i+1][T]      = dp[i][T] + dp[i][T-A[i]] + dp[i][T-2*A[i]] ...

dp[i+1][T-A[i]] =            dp[i][T-A[i]] + dp[i][T-2*A[i]] ...

// so
dp[i+1][T] = dp[i+1][T-A[i]] + dp[i][T]

dp[i][0] = 1

// OJ: https://leetcode.com/problems/coin-change-2/
// Author: github.com/lzl124631x
// Time: O(NT)
// Space: O(NT)
class Solution {
public:
    int change(int T, vector<int>& A) {
        int N = A.size();
        vector<vector<int>> dp(N + 1, vector<int>(T + 1));
        for (int i = 0; i <= N; ++i) dp[i][0] = 1;
        for (int i = 0; i < N; ++i) {
            for (int t = 1; t <= T; ++t) {
                dp[i + 1][t] = (t - A[i] >= 0 ? dp[i + 1][t - A[i]] : 0) + dp[i][t];
            }
        }
        return dp[N][T];
    }
};

Solution 4. DP (Unbounded knapsack problem) with Space Optimization

Since dp[i+1][T] only depends on dp[i+1][T-A[i]] and dp[i][T], we can reduce the dp array from N * T to 1 * T.

// OJ: https://leetcode.com/problems/coin-change-2/
// Author: github.com/lzl124631x
// Time: O(NT)
// Space: O(T)
class Solution {
public:
    int change(int T, vector<int>& A) {
        int N = A.size();
        vector<int> dp(T + 1);
        dp[0] = 1;
        for (int i = 0; i < N; ++i) {
            for (int t = 1; t <= T; ++t) {
                if (t - A[i] >= 0) dp[t] += dp[t - A[i]];
            }
        }
        return dp[T];
    }
};