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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(); }