Maximize count of array elements required to obtain given sum

Given an integer V and an array arr[] consisting of N integers, the task is to find the maximum number of array elements that can be selected from array arr[] to obtain the sum V. Each array element can be chosen any number of times. If the sum cannot be obtained, print -1.

Examples:

Input: arr[] = {25, 10, 5}, V = 30
Output: 6
Explanation:
To obtain sum 30, select arr[2] (= 5), 6 times.
Therefore, the count is 6.

Input: arr[] = {9, 6, 4, 3}, V = 11
Output: 3
Explanation:
To obtain sum 11, possible combinatio  is : 4,4,3
Therefore, the count is 3

Naive Approach: The simplest approach is to recursively find the maximum number of array elements to generate the sum V using array elements from indices 0 to j before finding the maximum number of elements required to generate V using elements from indices 0 to i where j < i < N. After completing the above steps, print the count of array elements required to obtain the given sum V



Time Complexity: O(VN)
Auxiliary Space: O(N)

Efficient Approach: To optimize the above approach, the idea is to use Dynamic Programming. Follow the below steps to solve the problem:

  • Initialize an array table[] of size V + 1 where table[i] will store the optimal solution to obtain sum i.
  • Initialize table[] with -1 and table[0] with 0 as 0 array elements are required to obtain the value 0.
  • For each value from i = 0 to V, calculate the maximum number of elements from the array required by the following DP transition:

table[i] = Max( table[i – arr[j]], table[i]), Where table[i-arr[j]]!=-1 
where 1 ≤ i ≤ V and 0 ≤ j ≤ N

  • After completing the above steps, print the value of the table[V] which is the required answer.

Below is the implementation of the above approach:

C++14

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// C++14 program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function that count the maximum
// number of elements to obtain sum V
int maxCount(vector<int> arr, int m, int V)
{
 
    // Stores the maximum number of
    // elements required to obtain V
    vector<int> table(V + 1);
 
    // Base Case
    table[0] = 0;
 
    // Initialize all table values
    // as Infinite
    for (int i = 1; i <= V; i++)
        table[i] = -1;
 
    // Find the max arr required
    // for all values from 1 to V
    for (int i = 1; i <= V; i++) {
 
        // Go through all arr
        // smaller than i
        for (int j = 0; j < m; j++) {
 
            // If current coin value
            // is less than i
            if (arr[j] <= i) {
                int sub_res = table[i - arr[j]];
 
                // Update table[i]
                if (sub_res != -1 && sub_res + 1 > table[i])
                    table[i] = sub_res + 1;
            }
        }
    }
 
    // Return the final count
    return table[V];
}
 
// Driver Code
int main()
{
 
    // Given array
    vector<int> arr = { 25, 10, 5 };
    int m = arr.size();
 
    // Given sum V
    int V = 30;
 
    // Function call
    cout << (maxCount(arr, m, V));
 
    return 0;
}
 
// This code is contributed by mohit kumar 29

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Java

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// Java program for the above approach
import java.io.*;
 
class GFG {
 
    // Function that count the maximum
    // number of elements to obtain sum V
    static int maxCount(int arr[], int m, int V)
    {
        // Stores the maximum number of
        // elements required to obtain V
        int table[] = new int[V + 1];
 
        // Base Case
        table[0] = 0;
 
        // Initialize all table values
        // as Infinite
        for (int i = 1; i <= V; i++)
            table[i] = -1;
 
        // Find the max arr required
        // for all values from 1 to V
        for (int i = 1; i <= V; i++) {
 
            // Go through all arr
            // smaller than i
            for (int j = 0; j < m; j++) {
 
                // If current coin value
                // is less than i
                if (arr[j] <= i) {
 
                    int sub_res = table[i - arr[j]];
 
                    // Update table[i]
                    if (sub_res != -1
                        && sub_res + 1 > table[i])
                        table[i] = sub_res + 1;
                }
            }
        }
 
        // Return the final count
        return table[V];
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        // Given array
        int arr[] = { 25, 10, 5 };
        int m = arr.length;
 
        // Given sum V
        int V = 30;
 
        // Function Call
        System.out.println(maxCount(arr, m, V));
    }
}

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Python3

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# Python program for the
# above approach
 
# Function that count
# the maximum number of
# elements to obtain sum V
def maxCount(arr, m, V):
   
  '''
  You can assume array elements as domination
  which are provided to you in infinte quantity
  just like in coin change problem.
  I made a small change in logic on coin change
  problem (minimum number of coins required).
  There we use to take min(table[i-arr[j]]+1,table[i]),
  here min is changed with max function.
  Dry run:
  assume : taget = 10, arr = [2,3,5]
     
  table   0  1  2  3  4  5  6  7  8  9  10
          0 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1
     
  taking first domination = 2
     
  table    0  1  2  3  4  5  6  7  8  9  10
           0 -1  1 -1  2 -1  3 -1  4 -1  5
 
  taking second domination = 3
     
  table    0  1  2  3  4  5  6  7  8  9  10 
           0 -1  1  1  2 -1  3 -1  4  3  5  
  here for i = 6 we have max(table[i-dom]+1,table[i])        
  hence
  => max(tabel[6-3]+1,tabel[6])
  => max(2,3) => 3
 
  taking third domination = 5
     
  table    0  1  2  3  4  5  6  7  8  9  10 
           0 -1  1  1  2  1  3 -1  4  3  5 
         
  Hence total 5 coins are required (2,2,2,2,2)
  '''
  # Stores the maximum
  # number of elements
  # required to obtain V
  table = [0 for i in range(V+1)]
 
  # Base Case
  table[0] = 0
 
  # Initialize all table
  # values as Infinite
  for i in range(1, V + 1, 1):
    table[i] = -1
 
  # Find the max arr required
  # for all values from 1 to V
  for i in range(1, V + 1, 1):
 
    # Go through all arr
    # smaller than i
    for j in range(0, m, 1):
 
      # If current coin value
      # is less than i
      if (arr[j] <= i):
        sub_res = table[i - arr[j]]
 
        # Update table[i]
        if (sub_res != -1 and sub_res + 1 > table[i]):
          table[i] = sub_res + 1
 
  # Return the final count
  return table[V];
 
# Driver Code
if __name__ == '__main__':
   
    # Given array
    arr = [25, 10, 5]
    m = len(arr)
 
    # Given sum V
    V = 30
 
    # Function Call
    print(f'Maximum number of array elements required : {maxCount(arr, m, V)}')
 
# This code is contributed by Aaryaman Sharma

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C#

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// C# program for the
// above approach
using System;
class GFG{
     
// Function that count the
// maximum number of elements
// to obtain sum V
static int maxCount(int[] arr,
                    int m, int V)
{
  // Stores the maximum number of
  // elements required to obtain V
  int[] table = new int[V + 1];
 
  // Base Case
  table[0] = 0;
 
  // Initialize all table
  // values as Infinite
  for (int i = 1; i <= V; i++)
    table[i] = -1;
 
  // Find the max arr required
  // for all values from 1 to V
  for (int i = 1; i <= V; i++)
  {
    // Go through all arr
    // smaller than i
    for (int j = 0; j < m; j++)
    {
      // If current coin value
      // is less than i
      if (arr[j] <= i)
      {
        int sub_res = table[i - arr[j]];
 
        // Update table[i]
        if (sub_res != -1 &&
            sub_res + 1 > table[i])
          table[i] = sub_res + 1;
      }
    }
  }
 
  // Return the final count
  return table[V];
}
     
// Driver code
static void Main()
{
  // Given array
  int[] arr = {25, 10, 5};
  int m = arr.Length;
 
  // Given sum V
  int V = 30;
 
  // Function Call
  Console.WriteLine(maxCount(arr,
                             m, V));
}
}
 
// This code is contributed by divyeshrabadiya07

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Output

6

Time Complexity: O(N * V)
Auxiliary Space: O(N)

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