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 )