Maximize sum of chosen Array elements with value at most M

Given an array arr[] of N positive numbers and an integer M. The task is to maximize the value of M by adding array elements when arr[i] ≤ M. 

Note: Any array element can be added at most once.

Examples: 

Input: arr[] = {3, 9, 19, 5, 21}, M = 10
Output: 67
Explanation: One way to getthe value is
M > 3; 3 is added to M and it becomes 10+3 = 13
M > 9; 9 is added to M and it becomes 13+9 = 22
M > 19; 19 is added to M and it becomes 22+19 = 41
M > 5; 5 is added to M and it becomes 41+5 = 46
M > 21; 21 is added to M and it becomes 46+21 = 67
Thus, M = 67 at the end.

Input: arr[] = {2, 13, 4, 19}, M = 6
Output: 12
Explanation: One way to get the value is
M > 4; 4 is added to M and it becomes 6+4 = 10
M > 2; 2 is added to M and it becomes 10+2 = 12
No other value in the array is smaller or equal to M.
Thus, M is 12 at the end.

Approach: The solution is based on the concept of sorting. Follow the steps mentioned below:

• First, sort the array in increasing order.
• For every index i, from 0 to N-1, do the following:
  • If M ≥ arr[i], add arr[i] with M.
  • If M< arr[i], stop iteration.
• Return the final value of M as the answer.

Below is the implementation of the above approach.

// C++ code to implement the approach
#include <bits/stdc++.h>
using namespace std;

// Function to calculate
// the maximum value of M
// that can be obtained
int IsArrayHungry(int M, vector<int>& arr)
{
    // Sort the array in increasing order.
    sort(arr.begin(), arr.end());
    long long sum = M;
    int N = arr.size();

    for (int i = 0; i < N; i++) {
        if (sum >= arr[i])
            sum += arr[i];
        else
            break;
    }
    return sum;
}

// Driver code
int main()
{
    vector<int> arr{ 3, 9, 19, 5, 21 };
    int M = 10;
    int res = IsArrayHungry(M, arr);
    cout << res;
    return 0;
}

// Java program for the above approach
import java.io.*;
import java.lang.*;
import java.util.*;

class GFG
{

  // Function to calculate
  // the maximum value of M
  // that can be obtained
  static int IsArrayHungry(int M, int arr[ ])
  {

    // Sort the array in increasing order.
    Arrays.sort(arr);
    int sum = M;
    int N = arr.length;

    for (int i = 0; i < N; i++) {
      if (sum >= arr[i])
        sum += arr[i];
      else
        break;
    }
    return sum;
  }

  // Driver code
  public static void main (String[] args)
  {
    int arr[ ] = { 3, 9, 19, 5, 21 };
    int M = 10;
    int res = IsArrayHungry(M, arr);
    System.out.print(res);
  }
}

// This code is contributed by hrithikgarg03188.

# Python 3 code to implement the approach

# Function to calculate
# the maximum value of M
# that can be obtained
def IsArrayHungry(M,  arr):

    # Sort the array in increasing order.
    arr.sort()
    sum = M
    N = len(arr)

    for i in range(N):
        if (sum >= arr[i]):
            sum += arr[i]

        else:
            break

    return sum

# Driver code
if __name__ == "__main__":

    arr = [3, 9, 19, 5, 21]
    M = 10
    res = IsArrayHungry(M, arr)
    print(res)

    # This code is contributed by ukasp.

// C# code to implement above approach
using System;
class GFG
{

  // Function to calculate
  // the maximum value of M
  // that can be obtained
  static int IsArrayHungry(int M, int []arr)
  {

    // Sort the array in increasing order.
    Array.Sort(arr);
    int sum = M;
    int N = arr.Length;

    for (int i = 0; i < N; i++) {
      if (sum >= arr[i])
        sum += arr[i];
      else
        break;
    }
    return sum;
  }

  // Driver Code:
  public static void Main()
  {
    int []arr = { 3, 9, 19, 5, 21 };
    int M = 10;
    int res = IsArrayHungry(M, arr);
    Console.WriteLine(res);
  }
}

// This code is contributed by Samim Hossain Mondal.

<script>
        // JavaScript code for the above approach 

        // Function to calculate
        // the maximum value of M
        // that can be obtained
        function IsArrayHungry(M, arr)
        {
        
            // Sort the array in increasing order.
            arr.sort(function (a, b) { return a - b })
            let sum = M;
            let N = arr.length;

            for (let i = 0; i < N; i++) {
                if (sum >= arr[i])
                    sum += arr[i];
                else
                    break;
            }
            return sum;
        }

        // Driver code
        let arr = [3, 9, 19, 5, 21];
        let M = 10;
        let res = IsArrayHungry(M, arr);
        document.write(res);

  // This code is contributed by Potta Lokesh
    </script>

67

Time Complexity: O(N * logN)
Auxiliary Space: O(1), since no extra space has been added.