Formatted question description: https://leetcode.ca/all/518.html

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/

// 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/

// 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/

// 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/

// 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];
    }
};

Java

  • 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 = coin; i <= amount; i++)
                    dp[i] += dp[i - coin];
            }
            return dp[amount];
        }
    }
    
  • // OJ: https://leetcode.com/problems/coin-change-2/
    // 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);
        }
    };
    
  • class Solution(object):
        def change(self, amount, coins):
            """
            :type amount: int
            :type coins: List[int]
            :rtype: int
            """
            dp = [0] * (amount + 1)
            dp[0] = 1
            for coin in coins:
                for i in range(1, amount + 1):
                    if coin <= i:
                        dp[i] += dp[i - coin]
            return dp[amount]
    

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