Minimum number of bottles required to fill K glasses

Given N glasses having water, and a list A of each of their capacity. The task is to find the minimum number of bottles required to fill out exactly K glasses. The capacity of each bottle is 100 units.

Examples:

Input: N = 4, K = 3, arr[] = {200, 150, 140, 300}
Output: 5
We have to fill out exactly 3 glasses.
So we fill out 140, 150, 200 whose sum is 490 so we need 5 bottles for this.

Input: N = 5, K = 4, arr[] = {1, 2, 3, 2, 1}
Output: 1



Approach: To fill out exactly K glasses, take the K glasses with least capacity. So for this sort the list of given capacities. The final answer will be the ceiling value of (Sum of capacities of 1st k glasses) / (Capacity of 1 bottle).

Below is the implementation of the above approach:

C/C++

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// C++ implementation of the above approach
#include <bits/stdc++.h>
using namespace std;
  
// function to calculate minimum glasses
int Min_glass(int n, int k, int a[])
{
    // sort the array based on
    // their capacity
    sort(a, a + n);
  
    int sum = 0;
  
    // taking sum of capacity
    // of first k glasses
    for (int i = 0; i < k; i++)
        sum += a[i];
  
    // calculate the answer
    int ans = ceil((double)sum / (double)100);
  
    return ans;
}
  
// Driver code
int main()
{
  
    int n = 4;
    int k = 3;
    int a[] = { 200, 150, 140, 300 };
  
    cout << Min_glass(n, k, a);
  
    return 0;
}

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Java

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// Java implementation of the 
// above approach
import java.util.*;
  
class GFG
{
// function to calculate minimum glasses
public static double Min_glass(int n, int k, 
                            int[] a)
{
    // sort the array based on
    // their capacity
  
    int sum = 0;
  
    // taking sum of capacity
    // of first k glasses
    for (int i = 0; i < k; i++)
        sum += a[i];
  
    // calculate the answer
    double ans = Math.ceil((double)sum /
                            (double)100);
  
    return ans;
}
  
// Driver code 
public static void main(String[] args)
{
    int n = 4;
    int k = 3;
    int[] a = { 200, 150, 140, 300 };
    Arrays.sort(a);
    System.out.println(Min_glass(n, k, a));
}
}
  
// This code is contributed by mits

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Python3

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# Python3 implementation of the above approach 
from math import ceil
  
# Function to calculate minimum glasses 
def Min_glass(n, k, a): 
   
    # sort the array based on their capacity 
    a.sort() 
  
    # calculate the answer 
    return ceil(sum(a[:k]) / 100
   
# Driver code 
if __name__ == "__main__"
   
    n, k = 4, 3 
    a = [200, 150, 140, 300]  
  
    print(Min_glass(n, k, a)) 
  
# This code is contributed by Rituraj Jain

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C#

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// C# implementation of the 
// above approach
using System;
  
class GFG
{
// function to calculate minimum glasses
public static double Min_glass(int n, int k, 
                               int []a)
{
    // sort the array based on
    // their capacity
  
    int sum = 0;
  
    // taking sum of capacity
    // of first k glasses
    for (int i = 0; i < k; i++)
        sum += a[i];
  
    // calculate the answer
    double ans = Math.Ceiling((double)sum /
                              (double)100);
  
    return ans;
}
  
// Driver code 
public static void Main()
{
    int n = 4;
    int k = 3;
    int[] a = { 200, 150, 140, 300 };
    Array.Sort(a);
    Console.WriteLine(Min_glass(n, k, a));
}
}
  
// This code is contributed
// by Soumik Mondal

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PHP

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<?php
// PHP implementation of the above approach 
  
// function to calculate minimum glasses 
function Min_glass($n, $k, $a
    // sort the array based on 
    // their capacity 
    sort($a); 
  
    $sum = 0; 
  
    // taking sum of capacity 
    // of first k glasses 
    for ($i = 0; $i < $k; $i++) 
        $sum += $a[$i]; 
  
    // calculate the answer 
    $ans = ceil($sum /100); 
  
    return $ans
  
// Driver code 
$n = 4; 
$k = 3; 
$a = array( 200, 150, 140, 300 ); 
  
echo Min_glass($n, $k, $a); 
  
// This code is contributed
// by akt_mit 
?>

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Output:

5


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