通过在选择第一个 Array 元素后否定整个总和，可能的最大子集总和

给定一个由N 个整数组成的数组A[] ，如果包含数组的第一个元素，所有元素的总和被否定，则任务是找到可能的最大子集总和。也可以考虑空子集(总和为 0)。

例子：

Input: N = 5, A[] = {1, 10, 4, -6, 3} 
Output: 17 
Explanation: 
On excluding A[0], subset with maximum sum = {10, 4, 3}, Sum = (10+4+3) = 17 
On including A[0], subset with maximum sum = {1, -6}, Sum = (1 + (-6)) = -5, on negating -(-5) = 5 
Maximum of the above two sum is max(17, 5) = 17
Input: N = 4, A[] = {3, -5, 1, -6} 
Output: 8 
Explanation: 
On excluding A[0], subset with maximum sum = {1}, Sum = 1 
On including A[0], subset with maximum sum = {3, -5, -6}, Sum = (3 + (-5) + (-6)) = -8, on negating -(-8) = 8 
Maximum of the above two sum is max(1, 8) = 8  

天真的方法：
解决问题的最简单方法是从数组中生成所有可能的子集，并通过将第一个元素作为每个子集的一部分来找到最大子集和maxSum和最小子集和minSum 。最后，打印maxSum和-(minSum)的最大值。
时间复杂度： O(2 ^N )
辅助空间： O(N)

有效的方法：
要优化上述方法，请考虑以下两种情况：

• 排除第一个元素，只需找到给定数组中所有正整数的总和(比如maxSum1) 。
• 对除 A[0] 之外的所有元素求和(比如 maxSum2)从数组中找出所有正整数，然后求反并将 A[0] 添加到总和中。
• 打印maxSum1和-maxSum2的最大值作为所需答案。

下面列出了每种情况的详细步骤：

情况 1：排除第一个元素 A[0]

• 从 A[1] 迭代到 A[N-1]。
• 维护一个变量maxSum1 ，以跟踪最大和，最初设置为 0。
• 如果遇到正元素(A[i] > 0) ，则将其包含在子集中，并且 maxSum1 将为(maxSum1 + A[i]) 。
• 最后，返回 maxSum1。

情况 2：包含第一个元素 A[0]

• 由于包含第一个元素的整个总和将被否定，因此找到不包括 A[0] 的最小总和。
• 可以使用在案例 1 中完成的相同步骤来实现最小和，但使用负值。
  • 取反从 A[1] 到 A[N-1] 的所有值。
  • 执行与案例 1 相同的步骤。
• 一旦获得总和(比如maxSum2 )，否定它 和 添加 A[0] 到它。
• 再次否定 maxSum2。

最后， maximum(maxSum1, maxsum2)将是必需的答案。

下面是上述方法的实现：

C++

// C++ Program to implement
// the above approach
#include 
using namespace std;
 
// Function returns maximum subset sum
// from the given array=
int maxSubset(vector& A, bool flag)
{
    int n = A.size();
    int sum = 0;
 
    // Case 2: Negate values
    // from A[1] to A[N-1]
    if (flag) {
        for (int i = 1; i < n; i++)
            A[i] = -A[i];
    }
 
    for (int i = 1; i < n; i++) {
 
        // Include only positives
        // for max subset sum
        if (A[i] > 0) {
            sum += A[i];
        }
    }
 
    // Return max sum obtained
    return sum;
}
 
// Function to return maximum of the
// maximum subset sum calculated
// for the two cases
int findBest(vector A)
{
    // Case 1
    int x = maxSubset(A, 0);
 
    // Case 2
    int y = maxSubset(A, 1);
 
    // Modifying the sum
    y = -y;
 
    // Including first element
    y += A[0];
 
    // Negating again
    y = -y;
 
    // Return the required answer
    return max(x, y);
}
 
// Driver Code
int main()
{
    vector A = { 1, 10, 4, -6, 3 };
 
    cout << findBest(A) << endl;
    return 0;
}

Java

// Java Program to implement
// the above approach
import java.util.*;
class GFG{
 
// Function returns maximum subset sum
// from the given array=
static int maxSubset(int []A, boolean flag)
{
    int n = A.length;
    int sum = 0;
 
    // Case 2: Negate values
    // from A[1] to A[N-1]
    if (flag)
    {
        for (int i = 1; i < n; i++)
            A[i] = -A[i];
    }
 
    for (int i = 1; i < n; i++)
    {
        // Include only positives
        // for max subset sum
        if (A[i] > 0)
        {
            sum += A[i];
        }
    }
 
    // Return max sum obtained
    return sum;
}
 
// Function to return maximum of the
// maximum subset sum calculated
// for the two cases
static int findBest(int []A)
{
    // Case 1
    int x = maxSubset(A, false);
 
    // Case 2
    int y = maxSubset(A, true);
 
    // Modifying the sum
    y = -y;
 
    // Including first element
    y += A[0];
 
    // Negating again
    y = -y;
 
    // Return the required answer
    return Math.max(x, y);
}
 
// Driver Code
public static void main(String[] args)
{
    int []A = {1, 10, 4, -6, 3};
    System.out.print(findBest(A) + "\n");
}
}
 
// This code is contributed by 29AjayKumar

Python3

# Python3 program to implement
# the above approach
 
# Function returns maximum subset
# sum from the given array=
def maxSubset(A, flag):
 
    n = len(A)
    sum = 0;
 
    # Case 2: Negate values
    # from A[1] to A[N-1]
    if (flag):
        for i in range(1, n):
            A[i] = -A[i]
 
    for i in range(1, n):
 
        # Include only positives
        # for max subset sum
        if (A[i] > 0):
            sum += A[i]
 
    # Return max sum obtained
    return sum
 
# Function to return maximum of the
# maximum subset sum calculated
# for the two cases
def findBest(A):
 
    # Case 1
    x = maxSubset(A, 0)
 
    # Case 2
    y = maxSubset(A, 1)
 
    # Modifying the sum
    y = -y
 
    # Including first element
    y += A[0]
 
    # Negating again
    y = -y
 
    # Return the required answer
    return max(x, y)
 
# Driver Code
if __name__ == "__main__":
 
    A = [ 1, 10, 4, -6, 3 ]
 
    print (findBest(A))
 
# This code is contributed by chitranayal

C#

// C# Program to implement
// the above approach
using System;
class GFG{
 
// Function returns maximum
// subset sum from the given array
static int maxSubset(int []A,
                     bool flag)
{
  int n = A.Length;
  int sum = 0;
 
  // Case 2: Negate values
  // from A[1] to A[N-1]
  if (flag)
  {
    for (int i = 1; i < n; i++)
      A[i] = -A[i];
  }
 
  for (int i = 1; i < n; i++)
  {
    // Include only positives
    // for max subset sum
    if (A[i] > 0)
    {
      sum += A[i];
    }
  }
 
  // Return max sum obtained
  return sum;
}
 
  // Function to return maximum of the
  // maximum subset sum calculated
  // for the two cases
  static int findBest(int []A)
  {
    // Case 1
    int x = maxSubset(A, false);
 
    // Case 2
    int y = maxSubset(A, true);
 
    // Modifying the sum
    y = -y;
 
    // Including first element
    y += A[0];
 
    // Negating again
    y = -y;
 
    // Return the required answer
    return Math.Max(x, y);
}
 
// Driver Code
public static void Main(String[] args)
{
  int []A = {1, 10, 4, -6, 3};
  Console.Write(findBest(A) + "\n");
}
}
 
// This code is contributed by 29AjayKumar

Javascript

// JavaScript program for the above approach
 
// Function returns maximum subset sum
// from the given array=
function maxSubset(A, flag)
{
    let n = A.length;
    let sum = 0;
  
    // Case 2: Negate values
    // from A[1] to A[N-1]
    if (flag)
    {
        for (let i = 1; i < n; i++)
            A[i] = -A[i];
    }
  
    for (let i = 1; i < n; i++)
    {
        // Include only positives
        // for max subset sum
        if (A[i] > 0)
        {
            sum += A[i];
        }
    }
  
    // Return max sum obtained
    return sum;
}
  
// Function to return maximum of the
// maximum subset sum calculated
// for the two cases
function findBest(A)
{
    // Case 1
    let x = maxSubset(A, false);
  
    // Case 2
    let y = maxSubset(A, true);
  
    // Modifying the sum
    y = -y;
  
    // Including first element
    y += A[0];
  
    // Negating again
    y = -y;
  
    // Return the required answer
    return Math.max(x, y);
}
  
     
// Driver Code
 
    let A = [ 1, 10, 4, -6, 3 ];
    document.write(findBest(A) + "\n");
              

输出：

17

时间复杂度： O(N)
辅助空间： O(1)

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