使用给定操作最大化数组总和

有一个由 (2 * n – 1) 个整数组成的数组。我们可以更改数组中恰好 n 个元素的符号。换句话说，我们可以精确地选择 n 个数组元素，并将它们中的每一个乘以 -1。找到数组的最大和。
例子：

Input : arr[] = 50 50 50
Output : 150
There is no need to change anything.
The sum of elements equals 150 
which is maximum.

Input :  arr[] = -1 -100 -1
Output : 100
Change the sign of the first two 
elements. Sum of the elements 
equal to 100.

方法：
第 1 步：-将循环迭代 2*n-1 次并重复第 2、3 和 4 步。
第2步： -计算没有。负数 (neg)。
第 3 步：-通过取数字的绝对值来计算数组的总和（总和）。
第 4 步：-通过取数字的绝对值 (min) 找到数组的最小数字。
第5步： -检查是否没有。负数是奇数，n 的值（给定）是偶数，然后从总和中减去 m 的两倍，这将是数组的 max_sum，否则 sum 的值将是数组的 max_sum。
下面是上述方法的实现：

C++

// CPP program to get the maximum
// sum of the array.
#include 
using namespace std;
 
// function to find maximum sum
int maxSum(int arr[], int n)
{
    int neg = 0, sum = 0,
     m = INT_MAX, max_sum;
     
    // step 1
    for (int i = 0; i < 2 * n - 1; i++) {
     
        // step 2
        neg += (arr[i] < 0);
     
        // step 3
        sum += abs(arr[i]);
     
        // step 4
        m = min(m, abs(arr[i]));
    }
     
    // step 5
    if (neg % 2 && n % 2 == 0) {
        max_sum = sum -= 2 * m;
        return (max_sum);
    }
     
    max_sum = sum;
    return (max_sum);
}
 
// Driver Function
int main()
{
    int arr[] = { -1, -100, -1 };
    int n = 2;
    cout << maxSum(arr, n) << endl;
    return 0;
}

Java

// Java program to get the maximum
// sum of the array.
import java.io.*;
import java.math.*;
 
class GFG {
 
// function to find maximum sum
static int maxSum(int arr[], int n) {
    int neg = 0, sum = 0, m = 100000000, max_sum;
 
    // step 1
    for (int i = 0; i < 2 * n - 1; i++) {
 
    // step 2
    if (arr[i] < 0)
        neg += 1;
 
    // step 3
    sum += Math.abs(arr[i]);
 
    // step 4
    m = Math.min(m, Math.abs(arr[i]));
    }
 
    // step 5
    if (neg % 2 == 1 && n % 2 == 0) {
    max_sum = sum -= 2 * m;
    return (max_sum);
    }
 
    max_sum = sum;
    return (max_sum);
}
 
// Driver Function
public static void main(String args[]) {
    int arr[] = {-1, -100, -1};
    int n = 2;
    System.out.println(maxSum(arr, n));
}
}
 
/*This code is contributed by Nikita Tiwari.*/

Python3

# Python3 code to get the maximum
# sum of the array.
import sys
 
# function to find maximum sum
def maxSum (arr, n):
    neg = 0
    sum = 0
    m = sys.maxsize
     
    # step 1
    for i in range(2 * n - 1):
         
        # step 2
        neg += (arr[i] < 0)
         
        # step 3
        sum += abs(arr[i])
         
        # step 4
        m = min(m, abs(arr[i]))
     
    # step 5
    if neg % 2 and n % 2 == 0:
        max_sum = sum - 2 * m
        return (max_sum)
     
    max_sum = sum
    return max_sum
     
# Driver Code
arr = [ -1, -100, -1 ]
n = 2
print( maxSum(arr, n))
 
# This code is contributed by "Sharad_Bhardwaj".

C#

// C# program to get the maximum
// sum of the array.
using System;
 
class GFG
{
 
// function to find maximum sum
static int maxSum(int []arr, int n)
{
    int neg = 0, sum = 0;
    int m = 100000000, max_sum;
 
    // step 1
    for (int i = 0; i < 2 * n - 1; i++)
    {
 
        // step 2
        if (arr[i] < 0)
        neg += 1;
 
        // step 3
        sum += Math.Abs(arr[i]);
 
        // step 4
        m = Math.Min(m, Math.Abs(arr[i]));
    }
 
    // step 5
    if (neg % 2 == 1 && n % 2 == 0)
    {
        max_sum = sum -= 2 * m;
        return (max_sum);
    }
 
    max_sum = sum;
    return (max_sum);
}
 
// Driver Code
public static void Main()
{
    int []arr = {-1, -100, -1};
    int n = 2;
    Console.WriteLine(maxSum(arr, n));
}
}
 
// This code is contributed by vt_m.

PHP

Javascript

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