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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];
}
};

• 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];
}
}

############

class Solution {
public int change(int amount, int[] coins) {
int[] dp = new int[amount + 1];
dp[0] = 1;
for (int coin : coins) {
for (int j = coin; j <= amount; j++) {
dp[j] += dp[j - 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:
def change(self, amount: int, coins: List[int]) -> int:
dp = [0] * (amount + 1)
dp[0] = 1
for coin in coins:
for j in range(coin, amount + 1):
dp[j] += dp[j - coin]
return dp[-1]

############

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]

• func change(amount int, coins []int) int {
dp := make([]int, amount+1)
dp[0] = 1
for _, coin := range coins {
for j := coin; j <= amount; j++ {
dp[j] += dp[j-coin]
}
}
return dp[amount]
}


• function change(amount: number, coins: number[]): number {
let dp = new Array(amount + 1).fill(0);
dp[0] = 1;
for (let coin of coins) {
for (let i = coin; i <= amount; ++i) {
dp[i] += dp[i - coin];
}
}
return dp.pop();
}