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Minimum number of coins that make a given value
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 is the 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 the 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.
Input and Output
Input: The required value. Say 48 Output: Minimum required coins. Here the output is 7. 48 = 10 + 10 + 10 + 10 + 5 + 2 + 1
Algorithm
minCoins(coinList, n, value)
Input: list of different coins, number of coins, given value.
Output: Minimum number of coins to get given value.
Begin if value = 0, then return 0 define coins array of size value + 1, fill with ∞ coins[0] := 0 for i := 1 to value, do for j := 0 to n, do if coinList[j] <= i, then tempCoins := coins[i-coinList[j]] if tempCoins ≠ ∞ and (tempCoins + 1) < coins[i], then coins[i] := tempCoins + 1 done done return coins[value] End
Example
#include<iostream>
using namespace std;
int minCoins(int coinList[], int n, int value) {
int coins[value+1]; //store minimum coins for value i
if(value == 0)
return 0; //for value 0, it needs 0 coin
coins[0] = 0;
for (int i=1; i<=value; i++)
coins[i] = INT_MAX; //initially all values are infinity except 0 value
for (int i=1; i<=value; i++) { //for all values 1 to value, find minimum values
for (int j=0; j<n; j++)
if (coinList[j] <= i) {
int tempCoins = coins[i-coinList[j]];
if (tempCoins != INT_MAX && tempCoins + 1 < coins[i])
coins[i] = tempCoins + 1;
}
}
return coins[value]; //number of coins for value
}
int main() {
int coins[] = {1, 2, 5, 10};
int n = 4, value;
cout << "Enter Value: "; cin >> value;
cout << "Minimum "<<minCoins(coins, n, value)<<" coins required.";
return 0;
}Output
Enter Value: 48 Minimum 7 coins required.
Updated on: 17-Jun-2020
2K+ Views
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