There is a list of coin C(c1, c2, ……Cn) is given and a value V is also given. Now the problem is to use the minimum number of coins to make the chance V.
Note − Assume there are an infinite number of coins C
In this problem, we will consider a set of different coins C{1, 2, 5, 10} are given, There is an infinite number of coins of each type. To make change the requested value we will try to take the minimum number of coins of any type.
As an example, for value 22 − we will choose {10, 10, 2}, 3 coins as the minimum.
The time complexity of this algorithm id O(V), where V is the value.
Input and Output
Input: A value, say 47 Output: Enter value: 47 Coins are: 10, 10, 10, 10, 5, 2
Algorithm
findMinCoin(value)
Input − The value to make the change
Output − Set of coins.
Begin coins set with value {1, 2, 5, 10} for all coins i as higher value to lower value do while value >= coins[i] do value := value – coins[i] add coins[i], in thecoin list done done print all entries in the coin list. End
Example
#include<iostream> #include<list> #define COINS 4 using namespace std; float coins[COINS] = {1, 2, 5, 10}; void findMinCoin(int cost) { list<int> coinList; for(int i = COINS-1; i>=0; i--) { while(cost >= coins[i]) { cost -= coins[i]; coinList.push_back(coins[i]); //add coin in the list } } list<int>::iterator it; for(it = coinList.begin(); it != coinList.end(); it++) { cout << *it << ", "; } } main() { int val; cout << "Enter value: "; cin >> val; cout << "Coins are: "; findMinCoin(val); cout << endl; }
Output
Enter value: 47 Coins are: 10, 10, 10, 10, 5, 2