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Maximum number of buckets that can be filled

  • Difficulty Level : Easy
  • Last Updated : 21 Jan, 2022

Given an array arr[] consisting of capacities of N buckets where arr[i] denotes the capacity of the ith bucket. If the total amount of water available is the sum of array indices (1-based indexing), the task is to find the maximum number of buckets that can be filled with the available water. The bucket will be considered filled if it has at least 1 liter in it.
 

Examples:

Input: arr[] = {1, 5, 3, 4, 7, 9}
Output: 4
Explanation:
Total available water = Sum of arrayindices of arr[] = 1 + 2 + 3 + 4 + 5 = 15.
Sorting the array in ascending order modifies the array to {1, 3, 4, 5, 7, 9}.
Fill the bucket having capacity 1, then . Now, available water = 14. 
Fill the bucket having capacity 3 . Now, available water = 11.
Fill bucket having capacity 4. Now, available water = 7.
Fill bucket having capacity 5. Now, available water = 2. in this case it is filled with more than 1 liter so this buket  is also considered.
Therefore, the total buckets that can be fully filled with water is 4.

Input: arr[] = {2, 5, 8, 3, 2, 10, 8}
Output: 5

 

Approach: The given problem can be solved Greedily. Follow the steps below to solve the given problem:

Below is the implementation of the above approach:

C++




// C++ program for the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the maximum number
// of buckets that can be filled with
// the amount of water available
int getBuckets(int arr[], int N)
{
    // Find the total available water
    int availableWater = N * (N - 1) / 2;
 
    // Sort the array in ascending order
    sort(arr, arr + N);
 
    int i = 0, sum = 0;
 
    // Check if bucket can be
    // filled with available water
    while (sum <= availableWater) {
        sum += arr[i];
        i++;
    }
 
    // Print count of buckets
    cout << i - 1;
}
 
// Driver Code
int main()
{
    int arr[] = { 1, 5, 3, 4, 7, 9 };
    int N = sizeof(arr) / sizeof(arr[0]);
 
    getBuckets(arr, N);
 
    return 0;
}

Java




// Java program for the above approach
import java.util.*;
public class GFG
{
   
  // Function to find the maximum number
  // of buckets that can be filled with
  // the amount of water available
  static void getBuckets(int[] arr, int N)
  {
    // Find the total available water
    int availableWater = N * (N - 1) / 2;
 
    // Sort the array in ascending order
    Arrays.sort(arr);
 
    int i = 0, sum = 0;
 
    // Check if bucket can be
    // filled with available water
    while (sum <= availableWater) {
      sum += arr[i];
      i++;
    }
 
    // Print count of buckets
    System.out.println(i - 1);
  }
 
  // Driver code
  public static void main(String[] args)
  {
    int[] arr = { 1, 5, 3, 4, 7, 9 };
    int N = arr.length;
 
    getBuckets(arr, N);
  }
}
 
// This code is contributed by divyesh072019.

Python3




# Python3 program for the above approach
 
# Function to find the maximum number
# of buckets that can be filled with
# the amount of water available
def getBuckets(arr, N) :
 
    # Find the total available water
    availableWater = N * (N - 1) // 2
 
    # Sort the array in ascending order
    arr.sort()
 
    i, Sum = 0, 0
 
    # Check if bucket can be
    # filled with available water
    while (Sum <= availableWater) :
        Sum += arr[i]
        i += 1
 
    # Print count of buckets
    print(i - 1, end = "")
 
arr = [ 1, 5, 3, 4, 7, 9 ]
N = len(arr)
 
getBuckets(arr, N);
 
# This code is contributed by divyeshrabadiya07.

C#




// C# program to implement
// the above approach
using System;
using System.Collections.Generic;
using System.Linq;
 
class GFG
{
   
  // Function to find the maximum number
  // of buckets that can be filled with
  // the amount of water available
  static void getBuckets(int[] arr, int N)
  {
     
    // Find the total available water
    int availableWater = N * (N - 1) / 2;
 
    // Sort the array in ascending order
    Array.Sort(arr);
    int i = 0, sum = 0;
 
    // Check if bucket can be
    // filled with available water
    while (sum <= availableWater)
    {
      sum += arr[i];
      i++;
    }
 
    // Print count of buckets
    Console.Write(i - 1);
  }
 
// Driver Code
public static void Main(String[] args)
{
    int[] arr = { 1, 5, 3, 4, 7, 9 };
    int N = arr.Length;
 
    getBuckets(arr, N);
}
}
 
// This code is contributed by splevel62.

Javascript




<script>
 
    // Javascript program to implement
    // the above approach
     
    // Function to find the maximum number
    // of buckets that can be filled with
    // the amount of water available
    function getBuckets(arr, N)
    {
 
      // Find the total available water
      let availableWater = N * (N - 1) / 2;
 
      // Sort the array in ascending order
      arr.sort(function(a, b){return a - b});
      let i = 0, sum = 0;
 
      // Check if bucket can be
      // filled with available water
      while (sum <= availableWater)
      {
        sum += arr[i];
        i++;
      }
 
      // Print count of buckets
      document.write(i - 1);
    }
     
    let arr = [ 1, 5, 3, 4, 7, 9 ];
    let N = arr.length;
  
    getBuckets(arr, N);
   
</script>
Output: 
4

 

Time Complexity: O(N*log N)
Auxiliary Space: O(1)


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