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📜  具有相等数量的正负元素的最大和子集

📅  最后修改于: 2021-05-13 23:34:39             🧑  作者: Mango

给定数组arr [] ，任务是找到包含相等数量的正负元素的最大和子集。
例子：

方法：想法是将负元素和正元素存储到两个不同的数组中，然后按升序对它们分别进行排序。然后使用两个指针，每个指针从每个数组的最高元素开始，包括总和大于0的那些对。否则，如果该对的总和小于0，则停止查找更多元素，因为使用a不可能存在这样的对左对中的总和大于0。

下面是上述方法的实现：

C++
// C++ implementation to find the
// maximum sum subset having equal
// number of positive and negative 
// elements in the subset
  
#include 
  
using namespace std;
  
// Function to find maximum sum
// subset with equal number of 
// positive and negative elements
int findMaxSum(int* arr, int n)
{
    vector a;
    vector b;
      
    // Loop to store the positive 
    // and negative elements in
    // two different array
    for (int i = 0; i < n; i++) {
        if (arr[i] > 0) {
            a.push_back(arr[i]);
        }
        else if (arr[i] < 0) {
            b.push_back(arr[i]);
        }
    }
      
    // Sort both the array
    sort(a.begin(), a.end());
    sort(b.begin(), b.end());
      
    // Pointers starting from
    // the highest elements
    int p = a.size() - 1;
    int q = b.size() - 1;
    int s = 0;
      
    // Find pairs having sum 
    // greater than zero
    while (p >= 0 && q >= 0) {
        if (a[p] + b[q] > 0) {
            s = s + a[p] + b[q];
        }
        else {
            break;
        }
        p = p - 1;
        q = q - 1;
    }
    return s;
}
  
// Driver code
int main()
{
    int arr1[] = { 1, -2, 3, 4, -5, 8 };
    int n1 = sizeof(arr1) / sizeof(arr1[0]);
  
    cout << findMaxSum(arr1, n1) << endl;
    return 0;
}

Java
// Java implementation to find the
// maximum sum subset having equal
// number of positive and negative 
// elements in the subset
import java.util.*;
  
class GFG{
  
// Function to find maximum sum
// subset with equal number of 
// positive and negative elements
static int findMaxSum(int []arr, int n)
{
    Vector a = new Vector();
    Vector b = new Vector();
      
    // Loop to store the positive 
    // and negative elements in
    // two different array
    for(int i = 0; i < n; i++) 
    {
       if (arr[i] > 0) 
       {
           a.add(arr[i]);
       }
       else if (arr[i] < 0)
       {
           b.add(arr[i]);
       }
    }
      
    // Sort both the array
    Collections.sort(a);
    Collections.sort(b);
      
    // Pointers starting from
    // the highest elements
    int p = a.size() - 1;
    int q = b.size() - 1;
    int s = 0;
      
    // Find pairs having sum 
    // greater than zero
    while (p >= 0 && q >= 0)
    {
        if (a.get(p) + b.get(q) > 0)
        {
            s = s + a.get(p) + b.get(q);
        }
        else 
        {
            break;
        }
        p = p - 1;
        q = q - 1;
    }
      
    return s;
}
  
  
// Driver code
public static void main(String[] args)
{
    int arr1[] = { 1, -2, 3, 4, -5, 8 };
    int n1 = arr1.length;
  
    System.out.print(
           findMaxSum(arr1, n1) + "\n");
}
}
  
// This code is contributed by 29AjayKumar

Python3
# Python3 implementation to find the
# maximum sum subset having equal
# number of positive and negative 
# elements in the subset
  
# Function to find maximum sum
# subset with equal number of 
# positive and negative elements
def findMaxSum(arr, n):
      
    a = []
    b = []
      
    # Loop to store the positive 
    # and negative elements in
    # two different array
    for i in range(n):
        if (arr[i] > 0):
            a.append(arr[i])
              
        elif (arr[i] < 0):
            b.append(arr[i])
          
    # Sort both the array
    a.sort()
    b.sort()
      
    # Pointers starting from
    # the highest elements
    p = len(a) - 1
    q = len(b) - 1
    s = 0
      
    # Find pairs having sum 
    # greater than zero
    while (p >= 0 and q >= 0):
        if (a[p] + b[q] > 0):
            s = s + a[p] + b[q]
              
        else:
            break
        p = p - 1
        q = q - 1
          
    return s
      
# Driver code
arr1 = [ 1, -2, 3, 4, -5, 8 ]
n1 = len(arr1)
  
print(findMaxSum(arr1, n1))
  
# This code is contributed by shubhamsingh10

C#
// C# implementation to find the
// maximum sum subset having equal
// number of positive and negative 
// elements in the subset
using System;
using System.Collections.Generic;
  
class GFG{
  
// Function to find maximum sum
// subset with equal number of 
// positive and negative elements
static int findMaxSum(int []arr, int n)
{
      
    List a = new List();
    List b = new List();
      
    // Loop to store the positive 
    // and negative elements in
    // two different array
    for(int i = 0; i < n; i++) 
    {
        if (arr[i] > 0) 
        {
            a.Add(arr[i]);
        }
        else if (arr[i] < 0)
        {
            b.Add(arr[i]);
        }
    }
      
    // Sort both the array
    a.Sort();
    b.Sort();
      
    // Pointers starting from
    // the highest elements
    int p = a.Count - 1;
    int q = b.Count - 1;
    int s = 0;
      
    // Find pairs having sum 
    // greater than zero
    while (p >= 0 && q >= 0)
    {
        if (a[p] + b[q] > 0)
        {
            s = s + a[p] + b[q];
        }
        else
        {
            break;
        }
          
        p = p - 1;
        q = q - 1;
    }
    return s;
}
  
// Driver code
public static void Main(String[] args)
{
    int []arr1 = { 1, -2, 3, 4, -5, 8 };
    int n1 = arr1.Length;
  
    Console.Write(findMaxSum(arr1, n1) + "\n");
}
}
  
// This code is contributed by Amit Katiyar

输出： 
6

性能分析：

  • 时间复杂度： O(N * logN)
  • 辅助空间： O(N)