Largest subset of Array having sum at least 0

Given an array arr[] that contains N integers, the task is to find the largest subset having sum at least 0.

Examples:

Input: arr[] = {5, -7, 0, -5, -3, -1}
Output: 4
Explanation: The largest subset that can be selected is {5, 0, -3, -1}. It has size 4

Input: arr[] = {1, -4, -2, -3}
Output: 1

Naive Approach: The basic idea to solve the problem is by using Recursion based on the following idea:

At every index, there are two choices, either select that element or not. If sum is becoming negative then don't pick it otherwise pick it. And from every recursion return the size of the largest possible subset between the two choices.

Follow the steps mentioned below:

• Use a recursive function and for each index there are two choices either select that element or not.
• Avoid selecting that element, whose value make the sum negative.  
• Return the count of maximum picked elements out of both choices.
• The maximum among all of these is the required subset size.

Below is the implementation of the above approach:

// C++ code to implement the approach

#include <bits/stdc++.h>
using namespace std;

// Function to return maximum count
int pick_max_elements(int pos, int sum,
                      int n, int arr[])
{

    // Return if the end of the array
    // is reached
    if (pos == n)
        return 0;

    int taken = INT_MIN;

    // If we select element at index pos
    if (sum + arr[pos] >= 0)
        taken = 1
                + pick_max_elements(pos + 1,
                                    sum + arr[pos],
                                    n, arr);

    int not_taken
        = pick_max_elements(pos + 1,
                            sum, n, arr);

    // Return the maximize steps taken
    return max(taken, not_taken);
}

// Driver code
int main()
{
    int arr[] = { 1, -4, -2, -3 };
    int N = sizeof(arr) / sizeof(arr[0]);

    // Function to pick maximum number
    // of elements
    cout << pick_max_elements(0, 0, N, arr);
    return 0;
}

// Java code to implement the approach
import java.util.*;
class GFG 
{

  // Function to return maximum count
  public static int pick_max_elements(int pos, int sum,
                                      int n, int arr[])
  {

    // Return if the end of the array
    // is reached
    if (pos == n)
      return 0;

    int taken = Integer.MIN_VALUE;

    // If we select element at index pos
    if (sum + arr[pos] >= 0)
      taken = 1
      + pick_max_elements(
      pos + 1, sum + arr[pos], n, arr);

    int not_taken
      = pick_max_elements(pos + 1, sum, n, arr);

    // Return the maximize steps taken
    return Math.max(taken, not_taken);
  }

  // Driver code
  public static void main(String[] args)
  {
    int arr[] = { 1, -4, -2, -3 };
    int N = arr.length;

    // Function to pick maximum number
    // of elements
    System.out.print(pick_max_elements(0, 0, N, arr));
  }
}

// This code is contributed by Taranpreet

# Python code to implement the approach
INT_MIN = -(1e9 + 7)

# Function to return maximum count
def pick_max_elements(pos, sum, n, arr):

    # Return if the end of the array
    # is reached
    if (pos == n):
        return 0

    taken = INT_MIN

    # If we select element at index pos
    if (sum + arr[pos] >= 0):
        taken = 1 + pick_max_elements(pos + 1, sum + arr[pos], n, arr)

    not_taken = pick_max_elements(pos + 1, sum, n, arr)

    # Return the maximize steps taken
    return max(taken, not_taken)

# Driver code
arr = [1, -4, -2, -3]
N = len(arr)

# Function to pick maximum number
# of elements
print(pick_max_elements(0, 0, N, arr))

# This code is contributed by Samim Hossain Mondal.

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

  // Function to return maximum count
  public static int pick_max_elements(int pos, int sum,
                                      int n, int[] arr)
  {

    // Return if the end of the array
    // is reached
    if (pos == n)
      return 0;

    int taken = Int32.MinValue;

    // If we select element at index pos
    if (sum + arr[pos] >= 0)
      taken = 1
      + pick_max_elements(
      pos + 1, sum + arr[pos], n, arr);

    int not_taken
      = pick_max_elements(pos + 1, sum, n, arr);

    // Return the maximize steps taken
    return Math.Max(taken, not_taken);
  }

  // Driver code
  public static void Main(string[] args)
  {
    int[] arr = { 1, -4, -2, -3 };
    int N = arr.Length;

    // Function to pick maximum number
    // of elements
    Console.Write(pick_max_elements(0, 0, N, arr));
  }
}

// This code is contributed by ukasp.

    <script>
        // JavaScript code for the above approach

        // Function to return maximum count
        function pick_max_elements(pos, sum,
            n, arr) {

            // Return if the end of the array
            // is reached
            if (pos == n)
                return 0;

            let taken = Number.MIN_VALUE;

            // If we select element at index pos
            if (sum + arr[pos] >= 0)
                taken = 1
                    + pick_max_elements(pos + 1,
                        sum + arr[pos],
                        n, arr);

            let not_taken
                = pick_max_elements(pos + 1,
                    sum, n, arr);

            // Return the maximize steps taken
            return Math.max(taken, not_taken);
        }

        // Driver code

        let arr = [1, -4, -2, -3];
        let N = arr.length;

        // Function to pick maximum number
        // of elements
        document.write(pick_max_elements(0, 0, N, arr));

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

1

Time Complexity: O( 2^N) 
Auxiliary Space: O( N )

Efficient Approach: The efficient approach is using multiset based on the following idea:

Traverse from start of array and if at any index the sum till now becomes negative then erase the minimum element till current index from the subset. This will increase the subset sum. 
To efficiently find the minimum multiset is used.

Follow the illustration given below for a better understanding.

Illustration:

Consider the array arr[] = {1, -4, -2, -3}  

multiset <int>  s,

-> Insert arr[0] in s. s = {1}. sum = sum + arr[0] = 1

-> Insert arr[1] in s. s = { -4, 1 }. sum = sum + arr[1] = -3 
-> Remove the smallest element (i.e. -4). sum = -3 - (-4) = 1.

-> Insert arr[2] in s. s = { -2, 1 }. sum = sum + arr[2] = -1 
-> Remove the smallest element (i.e. -2). sum = -1 - (-2) = 1.

-> Insert arr[3] in s. s = { -3, 1 }. sum = sum + arr[1] = -2 
-> Remove the smallest element (i.e. is -3). sum = -2 - (-3) = 1.

Total 1 element in the subset

 Follow the below steps to solve this problem: 

• Iterate from i = 0 to N 
  • Increment the count
  • Add the current element to the subset sum.
  • Insert arr[i] into the set.
  • If sum becomes negative then subtract the smallest value from the set and also remove the smallest element from the set
  • Decrement the count
• Return the final count.

Below is the implementation of the above approach:

// C++ code to implement the approach

#include <bits/stdc++.h>
using namespace std;

// Function to return max count
int pick_max_elements(int arr[], int n)
{
    int cnt = 0, sum = 0;

    // To store elements in sorted order
    multiset<int> s;

    for (int i = 0; i < n; i++) {
        sum += arr[i];

        // An element added,
        // so increase the cnt
        cnt++;
        s.insert(arr[i]);
        if (sum < 0) {
            sum = sum - *s.begin();

            // To remove the
            // smallest element
            s.erase(s.begin());

            // Removed an element,
            // so decrease the cnt
            cnt--;
        }
    }
    return cnt;
}

// Driver code
int main()
{
    int arr[] = { 1, -4, -2, -3 };

    // Size of array
    int N = sizeof(arr) / sizeof(arr[0]);

    // Function to pick
    // maximum number of elements
    cout << pick_max_elements(arr, N);
    return 0;
}

// JAVA code to implement the above approach
import java.util.*;
class GFG {

// Function to return max count
static int pick_max_elements(int arr[], int n)
{
    int cnt = 0, sum = 0;

    // To store elements in sorted order
    Vector<Integer> s = new Vector<>();

    for (int i = 0; i < n; i++) {
        sum += arr[i];

        // An element added,
        // so increase the cnt
        cnt++;
        s.add(arr[i]);
        if (sum < 0) {
            sum = sum - s.get(0);

            // To remove the
            // smallest element
            s.remove(0);

            // Removed an element,
            // so decrease the cnt
            cnt--;
        }
    }
    return cnt;
}

// Driver code
public static void main(String[] args)
{
    int arr[] = { 1, -4, -2, -3 };
    int N = arr.length;

    // Function to pick maximum number
    // of elements
    System.out.print(pick_max_elements(arr, N));
}
}

// This code is contributed by sanjoy_62.

# Python3 code to implement the approach

# using bisect.insort() to
# store elements in sorted form
import bisect

# Function to return max count
def pick_max_elements(arr, n) :
    cnt = 0
    sum = 0

    # To store elements in sorted order
    s = []

    for i in range(0,n) :
        sum += arr[i]

        # An element added,
        # so increase the cnt
        cnt += 1
        bisect.insort(s, arr[i])
        
        if sum < 0 :
            sum = sum - s[0]

            # To remove the
            # smallest element
            
            s.pop(0)

            # Removed an element,
            # so decrease the cnt
            cnt -= 1
    return cnt

# Driver code
if __name__ == "__main__":

    arr = [ 1, -4, -2, -3 ]
    N = len(arr)
    
    # Function to pick maximum number
    # of elements
    print(pick_max_elements(arr, N))

# This code is contributed by Pushpesh Raj

// Include namespace system
using System;
using System.Collections.Generic;

public class GFG
{
    // Function to return max count
    public static int pick_max_elements(int[] arr, int n)
    {
        var cnt = 0;
        var sum = 0;
        // To store elements in sorted order
        var s = new List<int>();
        for (int i = 0; i < n; i++)
        {
            sum += arr[i];
            // An element added,
            // so increase the cnt
            cnt++;
            s.Add(arr[i]);
            if (sum < 0)
            {
                sum = sum - s[0];
                // To remove the
                // smallest element
                s.RemoveAt(0);
                // Removed an element,
                // so decrease the cnt
                cnt--;
            }
        }
        return cnt;
    }
    // Driver code
    public static void Main(String[] args)
    {
        int[] arr = {1, -4, -2, -3};
        var N = arr.Length;
      
        // Function to pick maximum number
        // of elements
        Console.Write(GFG.pick_max_elements(arr, N));
    }
}

// This code is contributed by dhanshriborse561

// Javascript code to implement the approach

    // Function to return max count
    const pickMaxElements = (arr) => {
      let cnt = 0, sum = 0;
       // To store elements in sorted order
      const s = new Set();
    
      for (let i = 0; i < arr.length; i++) {
    sum += arr[i];
    // An element added,
    // so increase the cnt
    cnt++;
    s.add(arr[i]);
    if (sum < 0) {
      sum = sum - Math.min(...s);
      // To remove the
      // smallest element
      s.delete(Math.min(...s));
      // Removed an element,
      // so decrease the cnt
      cnt--;
    }
      }
      return cnt;
    }
    // Driver code
    const arr = [1, -4, -2, -3];
    
    // Function to pick
    // maximum number of elements
    console.log(pickMaxElements(arr));
    
    // This code is contributed by Utkarsh

1

Time complexity: O( N * log N )
Auxiliary Space: O(N )