Greedy Florist


Problem Statement :


A group of friends want to buy a bouquet of flowers. The florist wants to maximize his number of new customers and the money he makes. To do this, he decides he'll multiply the price of each flower by the number of that customer's previously purchased flowers plus 1. The first flower will be original price, ( 0 + 1 ) * original price , the next will be  ( 1 + 1 ) * original price and so on.

Given the size of the group of friends, the number of flowers they want to purchase and the original prices of the flowers, determine the minimum cost to purchase all of the flowers.


Function Description

Complete the getMinimumCost function in the editor below. It should return the minimum cost to purchase all of the flowers.

getMinimumCost has the following parameter(s):

c: an array of integers representing the original price of each flower
k: an integer, the number of friends

Input Format

The first line contains two space-separated integers n and k, the number of flowers and the number of friends.
The second line contains n space-separated positive integers c[ i ] , the original price of each flower.

Constraints

1  <=  n. k  <= 100
1  <=  c[ i ]  <=  10^6
answer  <   2^31
0   <=  i  <  n

Output Format

Print the minimum cost to buy all n flowers.


Solution :



title-img 


                            Solution in C :

In  C :






#include <stdio.h>
#include <stdlib.h>

long long a[200],i,j,k,l,m,n;

int com(const void *xx, const void *yy)
{
if(*(long long *)xx > *(long long*)yy) return -1;
return 1;
}

int main()
{

scanf("%lld %lld\n",&n,&k);
for(i=0;i<n;i++) scanf("%lld",a+i);

qsort(a,n,sizeof(a[0]),com);

m = 0;

for(i=0;i<n;i++) m+= a[i]*(1+i/k);

printf("%lld\n",m);


return 0;
}
                        


                        Solution in C++ :

In  C++ :




#include <cstdio>
#include <algorithm>
using namespace std;

int a[102];

int main() {
int i,j,m,n;
long long c,ans=0;
	scanf("%d%d",&n,&m);
	for (i=0;i<n;++i) {
		scanf("%d",&a[i]);
	}
	sort(a,a+n);
	for (i=n-1,c=0,j=0;i>=0;--i) {
		if (j==0) {
			++c;
		}
		ans+=c*a[i];
		if (++j==m) {
			j=0;
		}
	}
	printf("%Ld\n",ans);
	return 0;
}
                    


                        Solution in Java :

In  Java :






import java.util.Arrays;
import java.util.Scanner;

public class Solution {

    public static void main(String args[]) {

        Scanner in = new Scanner(System.in);

        int n, k;
        n = in.nextInt();
        k = in.nextInt();

        int c[] = new int[n];
        for (int i = 0; i < n; i++) {
            c[i] = in.nextInt();
        }

        Arrays.sort(c);
        
        int result = 0;
        
        if(k >= n){
            for(int i=0;i<n;i++){
                result +=  c[i];
            }
            System.out.println(result);
        }else{
            //Processing
            int x = 0;            
            while(n > 0){                
                for(int i=0;i<k;i++){
                    result += c[n-1]*(x+1);
                    n--;
                    if(n == 0){
                        break;
                    }
                }                
                x++;
            }
            System.out.println(result);
        }        
    }
}
                    


                        Solution in Python : 
                            
In Python3 :




NK = input()

N = int(NK.partition(' ')[0])
K = int(NK.partition(' ')[2])

prices = [int(x) for x in input().strip().split(' ')]

if K == N:
    print(sum(prices))
else:
    prices.sort()
    friends = [[] for friend in range(K)]
    
    for friend in friends:
        friend.append(prices.pop())
    
    index = 0
    while prices:
        friends[index].append(prices.pop()*(1+len(friends[index])))
        index = (index + 1)%K
        
    total = 0
    for friend in friends:
        total += sum(friend)
        
    print(total)


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