Answer:
There are two algorithms in which apply different optimal solutions.
They are: A Dynamic and Naive recursive programs
Explanation:
// A Naive recursive C++ program to find minimum of coins
// to make a given change V
#include<bits/stdc++.h>
using namespace std;
// m is size of coins array (number of different coins)
int minCoins(int coins[], int m, int V)
{
// base case
if (V == 0) return 0;
// Initialize result
int res = INT_MAX;
// Try every coin that has smaller value than V
for (int i=0; i<m; i++)
{
if (coins[i] <= V)
{
int sub_res = minCoins(coins, m, V-coins[i]);
// Check for INT_MAX to avoid overflow and see if
// result can minimized
if (sub_res != INT_MAX && sub_res + 1 < res)
res = sub_res + 1;
}
}
return res;
}
// Driver program to test above function
int main()
{
int coins[] = {9, 6, 5, 1};
int m = sizeof(coins)/sizeof(coins[0]);
int V = 11;
cout << "Minimum coins required is "
<< minCoins(coins, m, V);
return 0;
}
.........................................
// A Dynamic Programming based C++ program to find minimum of coins
// to make a given change V
#include<bits/stdc++.h>
using namespace std;
// m is size of coins array (number of different coins)
int minCoins(int coins[], int m, int V)
{
// table[i] will be storing the minimum number of coins
// required for i value. So table[V] will have result
int table[V+1];
// Base case (If given value V is 0)
table[0] = 0;
// Initialize all table values as Infinite
for (int i=1; i<=V; i++)
table[i] = INT_MAX;
// Compute minimum coins required for all
// values from 1 to V
for (int i=1; i<=V; i++)
{
// Go through all coins smaller than i
for (int j=0; j<m; j++)
if (coins[j] <= i)
{
int sub_res = table[i-coins[j]];
if (sub_res != INT_MAX && sub_res + 1 < table[i])
table[i] = sub_res + 1;
}
}
return table[V];
}
// Driver program to test above function
int main()
{
int coins[] = {9, 6, 5, 1};
int m = sizeof(coins)/sizeof(coins[0]);
int V = 11;
cout << "Minimum coins required is "
<< minCoins(coins, m, V);
return 0;
}
