**Problem:**

Please find the problem here.

**Solution:**

A classic problem solved with dynamic programming. Depending on the denomination greedy algorithm may or may not work, so use dynamic programming here.

Assume you already know the minimal number of coin needed to solve any $ k < n $. To solve at least how many coins is need for $ n $, you pick any one of the coin in the list and use it as the first coin, then the rest is just recursively 'solving' for $ n - c $, where $ c $ is the first coin value. Within all choices, just pick the best one.

**Code:**

#include "stdafx.h" // https://leetcode.com/problems/coin-change/ #include "LEET_COIN_CHANGE.h" #include <algorithm> #include <iostream> #include <sstream> #include <vector> #include <string> using namespace std; namespace _LEET_COIN_CHANGE { class Solution { public: int coinChange(vector<int>& coins, int amount) { vector<int> answer; answer.resize(amount + 1); answer[0] = 0; for (int i = 1; i <= amount; i++) { int goal = i; int count = -1; for (unsigned int j = 0; j < coins.size(); j++) { int remainder = goal - coins[j]; if (remainder >= 0) { int recursive_solution = answer[remainder]; if (recursive_solution != -1) { if (count == -1) { count = recursive_solution + 1; } else { count = min(count, recursive_solution + 1); } } } } answer[i] = count; } return answer[amount]; } }; }; using namespace _LEET_COIN_CHANGE; int LEET_COIN_CHANGE() { Solution solution; vector<int> coin; coin.push_back(1); coin.push_back(2); coin.push_back(5); cout << solution.coinChange(coin, 11) << endl; return 0; }

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