带有非负和的最长子序列

给定长度为N的数组arr [] ，任务是找到具有非负和的最大子序列的长度。

例子：

Input: arr[] = {1, 2, -3}
Output: 3
The complete array has a non-negative sum.

Input: arr[] = {1, 2, -4}
Output: 2
{1, 2} is the required subsequence.

方法：想法是所有非负数都必须包含在子序列中，因为这样的数只会增加总和的值。
现在，不难看出负数，必须首先选择较大的数。因此，只要不将总和的值减小到0以下，就可以按负值的递增顺序将它们相加。可以在对数组进行排序后执行此操作。

下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
  
// Function to return the length of
// the largest subsequence
// with non-negative sum
int maxLen(int* arr, int n)
{
    // To store the current sum
    int c_sum = 0;
  
    // Sort the input array in
    // non-increasing order
    sort(arr, arr + n, greater());
  
    // Traverse through the array
    for (int i = 0; i < n; i++) {
  
        // Add the current element to the sum
        c_sum += arr[i];
  
        // Condition when c_sum falls
        // below zero
        if (c_sum < 0)
            return i;
    }
  
    // Complete array has a non-negative sum
    return n;
}
  
// Driver code
int main()
{
    int arr[] = { 3, 5, -6 };
    int n = sizeof(arr) / sizeof(int);
  
    cout << maxLen(arr, n);
  
    return 0;
}

Java

// Java implementation of the approach
import java.util.*;
  
class GFG 
{
  
// Function to return the length of
// the largest subsequence
// with non-negative sum
static int maxLen(int[] arr, int n)
{
    // To store the current sum
    int c_sum = 0;
  
    // Sort the input array in
    // non-increasing order
    Arrays.sort(arr);
  
    // Traverse through the array
    for (int i = n-1; i >=0; i--)
    {
  
        // Add the current element to the sum
        c_sum += arr[i];
  
        // Condition when c_sum falls
        // below zero
        if (c_sum < 0)
            return i;
    }
  
    // Complete array has a non-negative sum
    return n;
}
  
// Driver code
public static void main(String []args)
{
    int arr[] = { 3, 5, -6 };
    int n = arr.length;
  
    System.out.println(maxLen(arr, n));
}
}
  
// This code is contributed by Rajput-Ji

Python3

# Python3 implementation of the approach 
  
# Function to return the length of 
# the largest subsequence 
# with non-negative sum 
def maxLen(arr, n) : 
  
    # To store the current sum 
    c_sum = 0; 
  
    # Sort the input array in 
    # non-increasing order 
    arr.sort(reverse = True); 
  
    # Traverse through the array 
    for i in range(n) :
  
        # Add the current element to the sum 
        c_sum += arr[i]; 
  
        # Condition when c_sum falls 
        # below zero 
        if (c_sum < 0) :
            return i; 
  
    # Complete array has a non-negative sum 
    return n; 
  
# Driver code 
if __name__ == "__main__" : 
  
    arr = [ 3, 5, -6 ]; 
    n = len(arr); 
  
    print(maxLen(arr, n)); 
  
# This code is contributed by AnkitRai01

C#

// C# implementation of the approach
using System; 
  
class GFG 
{
  
// Function to return the length of
// the largest subsequence
// with non-negative sum
static int maxLen(int[] arr, int n)
{
    // To store the current sum
    int c_sum = 0;
  
    // Sort the input array in
    // non-increasing order
    Array.Sort(arr);
  
    // Traverse through the array
    for (int i = n - 1; i >= 0; i--)
    {
  
        // Add the current element to the sum
        c_sum += arr[i];
  
        // Condition when c_sum falls
        // below zero
        if (c_sum < 0)
            return i;
    }
  
    // Complete array has a non-negative sum
    return n;
}
  
// Driver code
public static void Main(String []args)
{
    int []arr = { 3, 5, -6 };
    int n = arr.Length;
  
    Console.WriteLine(maxLen(arr, n));
}
}
  
// This code is contributed by PrinciRaj1992

输出：

3