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  1. Prepare
  2. Algorithms
  3. Greedy
  4. Greedy Florist
  5. Discussions

Greedy Florist

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  • schmiddla
     + 0 comments

    JS

    function getMinimumCost(k, c) {
  let min = 0;

  // sort flowers - highest to lowest
  c.sort((a, b) => (a - b < 0 ? 1 : -1));

  for (let buyerIdx = 0; buyerIdx < k; buyerIdx++) {  // e.g. 3 friends = 3 buyers
    // with every purchase the next purchased flower gets multiplied
    let multiplier = 1;

    // to have the lowest possible multiplier on the hightest prices,
    //  -> buy every "k" flower (flowerIdx += k) in the price-sorted array
    for (let flowerIdx = buyerIdx; flowerIdx < c.length; flowerIdx += k) {
      min += c[flowerIdx] * multiplier++;
    }
  }

  return min;
}

  • ercanbeyen
     + 0 comments

    Java Solution

    import java.util.*;

public class Solution {

    static int getMinimumCost(int numberOfFriends, Integer[] prices) {
        Arrays.sort(prices, Collections.reverseOrder());

        int i = 0;
        int numberOfBoughtFlowers = 0;
        int minimumCost = 0;
        
        while (i < prices.length) {
            for (int j = 0; j < numberOfFriends && i < prices.length; j++) {
                minimumCost += (1 + numberOfBoughtFlowers) * prices[i];
                i++;
            }
            
            numberOfBoughtFlowers++;
        }
        
        return minimumCost;
    }

    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        final String SPACE = " ";
        String[] amounts = scanner.nextLine().split(SPACE);
        int numberOfFlowers = Integer.parseInt(amounts[0]);
        int numberOfFriends = Integer.parseInt(amounts[1]);

        Integer[] prices = new Integer[numberOfFlowers];
        String[] priceItems = scanner.nextLine().split(SPACE);

        for (int i = 0; i < numberOfFlowers; i++) {
            int price = Integer.parseInt(priceItems[i]);
            prices[i] = price;
        }

        int minimumCost = getMinimumCost(numberOfFriends, prices);
        System.out.println(minimumCost);
        
        scanner.close();
    }
}

  • michal_dwojak92
     + 0 comments

    c# solution:

    static int getMinimumCost(int k, int[] c) {

     int howManyFlowers = c.Count();
     Array.Sort(c);
    int cost = 0;
    int multipleCost = 1;
    int couterForMultipleCost = 0;
    if (howManyFlowers <= k)
    {
        //cost = c.Sum();
        return c.Sum();
    }
    else
    {
        do
        {
        couterForMultipleCost++;
        if (couterForMultipleCost > k)
        {
            couterForMultipleCost = 1;
            multipleCost++;
        }
        cost += multipleCost * c[howManyFlowers - 1];
        howManyFlowers -= 1;
        } while (howManyFlowers != 0);
    }
    return cost;
}

  • shrutimadan649
     + 0 comments

    Solution in JAVA

    // Complete the getMinimumCost function below. static int getMinimumCost(int k, int[] c) {

        Arrays.sort(c);
    int res=0;
    int m=0;int counter=0;
    int i=c.length-1;
    for(;i>=0;i--)
    {

        if(counter%k==0 && counter>0){
            counter=0;
            m++;
        }
        System.out.println(i+" "+m+" "+counter);
        res=res+(m+1)*c[i];
        counter++;
        System.out.println(res);
    }
    return res;


}

  • kalyaniarla
     + 0 comments

    Simple Java solution

     static int getMinimumCost(int k, int[] c) {
        Arrays.sort(c);
        int price = 0;
        int count = 0;
        for(int i=c.length-1 ;i>=0; i--){
            price += (count/k+1)*c[i];
            count++;
        }
        return price;
    }