LeetCode 322. Coin Change (Iteration in the recurrence relation)

 322. Coin Change

You are given an integer array coins representing coins of different denominations and an integer amount representing a total amount of money.

Return the fewest number of coins that you need to make up that amount. If that amount of money cannot be made up by any combination of the coins, return -1.

You may assume that you have an infinite number of each kind of coin.

 

Example 1:

Input: coins = [1,2,5], amount = 11
Output: 3
Explanation: 11 = 5 + 5 + 1

Example 2:

Input: coins = [2], amount = 3
Output: -1

Example 3:

Input: coins = [1], amount = 0
Output: 0

 

Constraints:

• 1 <= coins.length <= 12
• 1 <= coins[i] <= 231 - 1
• 0 <= amount <= 104

	class Solution {
	public:
	/**
	Framework(top-down) Iteration in the recurrence relation
	1. function
	i: min num of i coin
	dp(amount)
	2. relation
	dp[i] = min(dp(i - coin)) coin: avaible
	3. base
	dp(0) return 0
	*/
	int n;
	vector<int> con;
	vector<int> memo;
	int dp(int a){
	// base
	if(a == 0)
	return 0;
	if(memo[a] == 0){
	// iteration for each coin
	int res = INT_MAX;
	for(auto c: con){
	if(a >= c and dp(a - c) != -1)
	res = min(res, dp(a - c) + 1);
	}
	if(res == INT_MAX)
	memo[a] = -1;
	else
	memo[a] = res;
	}
	return memo[a];
	}
	int coinChange(vector<int>& coins, int amount) {
	// make global
	con = coins;
	memo.resize(amount + 1);
	return dp(amount);
	}
	};

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