You are given a list of N coins of different denominations. you can pay an amount equivalent to any 1 coin and can acquire that coin. In addition, once you have paid for a coin, we can choose at most K more coins and can acquire those for free. The task is to find the minimum amount required to acquire all the N coins for a given value of K.

Examples:

Input : coin[] = {100, 20, 50, 10, 2, 5}, k = 3 Output : 7 Input : coin[] = {1, 2, 5, 10, 20, 50}, k = 3 Output : 3

As per question, we can see that at a cost of 1 coin, we can acquire at most K+1 coins. Therefore, in order to acquire all the n coins, we will be choosing ceil(n/(k+1)) coins and the cost of choosing coins will be minimum if we choose smallest ceil(n/(k+1)) ( Greedy approach). Smallest ceil(n/(k+1)) coins can be found by simply sorting all the N values in increasing order.

If we should check for time complexity (n log n) is for sorting element and (k) is for adding the total amount. So, finally Time Complexity : O(n log n).

// C++ program to acquire all n coins #include<bits/stdc++.h> using namespace std; // function to calculate min cost int minCost(int coin[], int n, int k) { // sort the coins value sort(coin, coin+n); // calculate no. of coins needed int coins_needed = ceil(1.0*n / (k+1)); // calculate sum of all selected coins int ans = 0; for (int i=0; i<=coins_needed-1; i++) ans += coin[i]; return ans; } // driver program int main() { int coin[] = {8, 5, 3, 10, 2, 1, 15, 25}; int n = sizeof(coin) / sizeof(coin[0]); int k = 3; cout << minCost(coin, n, k); return 0; }

Output:

3

Note that there are more efficient approaches to find given number of smallest values. For example, method 6 of m largest(or smallest) elements in an array can find m’th smallest element in (n-m) Log m + m Log m).

**How to handle multiple queries for a single predefined array?**

In the case, if you are asked to find the above answer for many different values of K, you have to compute it fast and our time complexity got increased as per number of queries for k. For the purpose to serve, we can maintain a prefix sum array after sorting all the N values and can answer queries easily and quickly.

Suppose

// C++ program to acquire all n coins at minimum cost // with multiple values of k. #include<bits/stdc++.h> using namespace std; // Converts coin[] to prefix sum array void preprocess(int coin[], int n) { // sort the coins value sort(coin, coin+n); // Maintain prefix sum array for (int i=1; i<=n-1; i++) coin[i] += coin[i-1]; } // Function to calculate min cost when we can // get k extra coins after paying cost of one. int minCost(int coin[], int n, int k) { // calculate no. of coins needed int coins_needed = ceil(1.0*n / (k+1)); // return sum of from prefix array return coin[coins_needed - 1]; } // driver program int main() { int coin[] = {8, 5, 3, 10, 2, 1, 15, 25}; int n = sizeof(coin) / sizeof(coin[0]); preprocess(coin, n); int k = 3; cout << minCost(coin, n, k) << endl; k = 7; cout << minCost(coin, n, k) << endl; return 0; }

Output:

3 1

After preprocessing, every query for a k takes O(1) time.

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