通过将任何数组元素 arr[i] 更改为 (-1)*arr[i] – 1 任意次数来最大化数组乘积

给定一个由N 个整数组成的数组arr[] ，任务是通过将每个数组元素arr[i]更改为(-1)*arr[i] – 1任意次数来修改数组元素，使得数组最大化。

例子：

Input: arr[] = {-3, 0, 1}
Output: 2 -1 -2
Explanation:
Performing the following operations on the array elements maximizes the product:
Applying the operation on arr[0] once, arr[0] = (- 1) * (- 3) – 1 = 2.
Applying the operation on arr[1] once, arr[1] = 0 – 1 = -1.
Applying the operation on arr[2] once, arr[2] = -1 – 1 = -2.
The modified array is {2, -1, -2} with product 4, which is maximum possible.

Input: arr[] = {-9, 2, 0, 1, -5, 3, 16}
Output: -9 -3 -1 -2 -5 -4 16

方法：根据以下观察可以解决给定的问题：

• 为了使乘积最大，元素的绝对值应该随着运算减少负元素的绝对值而变高。因此，给定的操作应该只应用于正值。
• 如果元素个数为奇数，则对绝对值最高的元素进行运算，因为奇数个负数的乘积为负。因此，将乘积的一个正数更改为正和最大值。
• 对负数执行操作会减小其绝对值。因此，找到绝对值最大的元素并对其进行一次操作，因为这样会使乘积减少最少。

请按照以下步骤解决问题：

• 遍历给定的数组arr[]并对每个元素arr[i] ，检查arr[i] ≥ 0是否。如果发现为真，则更新arr[i] = (-1)*arr[i] – 1 。否则，继续。
• 遍历数组后，检查N的值是否为奇数。如果发现为真，则找到绝对值最大的元素并对其进行一次操作。
• 完成上述步骤后，打印修改后的数组arr[] 。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Function to find array with maximum
// product by changing array
// elements to (-1)arr[i] - 1
void findArrayWithMaxProduct(int arr[],
                             int N)
{
    // Traverse the array
    for (int i = 0; i < N; i++)
 
        // Applying the operation on
        // all the positive elements
        if (arr[i] >= 0) {
            arr[i] = -arr[i] - 1;
        }
 
    if (N % 2 == 1) {
 
        // Stores maximum element in array
        int max_element = -1;
 
        // Stores index of maximum element
        int index = -1;
 
        for (int i = 0; i < N; i++)
 
            // Check if current element
            // is greater than the
            // maximum element
            if (abs(arr[i]) > max_element) {
 
                max_element = abs(arr[i]);
 
                // Find index of the
                // maximum element
                index = i;
            }
 
        // Perform the operation on the
        // maximum element
        arr[index] = -arr[index] - 1;
    }
 
    // Print the elements of the array
    for (int i = 0; i < N; i++)
        printf("%d ", arr[i]);
}
 
// Driver Code
int main()
{
    int arr[] = { -3, 0, 1 };
    int N = sizeof(arr) / sizeof(arr[0]);
 
    // Function Call
    findArrayWithMaxProduct(arr, N);
 
    return 0;
}

Java

// Java program for the above approach
import java.io.*;
class GFG
{
 
  // Function to find array with maximum
  // product by changing array
  // elements to (-1)arr[i] - 1
  static void findArrayWithMaxProduct(int arr[], int N)
  {
 
    // Traverse the array
    for (int i = 0; i < N; i++)
 
      // Applying the operation on
      // all the positive elements
      if (arr[i] >= 0)
      {
        arr[i] = -arr[i] - 1;
      }
 
    if (N % 2 == 1)
    {
 
      // Stores maximum element in array
      int max_element = -1;
 
      // Stores index of maximum element
      int index = -1;
 
      for (int i = 0; i < N; i++)
 
        // Check if current element
        // is greater than the
        // maximum element
        if (Math.abs(arr[i]) > max_element) {
 
          max_element = Math.abs(arr[i]);
 
          // Find index of the
          // maximum element
          index = i;
        }
 
      // Perform the operation on the
      // maximum element
      arr[index] = -arr[index] - 1;
    }
 
    // Print the elements of the array
    for (int i = 0; i < N; i++)
      System.out.print(arr[i] + " ");
  }
 
  // Driver Code
  public static void main(String[] args)
  {
    int arr[] = { -3, 0, 1 };
    int N = arr.length;
 
    // Function Call
    findArrayWithMaxProduct(arr, N);
  }
}
 
// This code is contributed by jithin.

Python3

# Python3 program for the above approach
 
# Function to find array with maximum
# product by changing array
# elements to (-1)arr[i] - 1
def findArrayWithMaxProduct(arr, N):
   
    # Traverse the array
    for i in range(N):
 
        # Applying the operation on
        # all the positive elements
        if (arr[i] >= 0):
            arr[i] = -arr[i] - 1
    if (N % 2 == 1):
 
        # Stores maximum element in array
        max_element = -1
 
        # Stores index of maximum element
        index = -1
        for i in range(N):
 
            # Check if current element
            # is greater than the
            # maximum element
            if (abs(arr[i]) > max_element):
                max_element = abs(arr[i])
 
                # Find index of the
                # maximum element
                index = i
 
        # Perform the operation on the
        # maximum element
        arr[index] = -arr[index] - 1
         
    # Prthe elements of the array
    for i in arr:
        print(i, end = " ")
 
# Driver Code
if __name__ == '__main__':
    arr = [-3, 0, 1]
    N = len(arr)
 
    # Function Call
    findArrayWithMaxProduct(arr, N)
 
    # This code is contributed by mohit kumar 29.

C#

// C# program for the above approach
using System;
class GFG
{
 
  // Function to find array with maximum
  // product by changing array
  // elements to (-1)arr[i] - 1
  static void findArrayWithMaxProduct(int[] arr, int N)
  {
 
    // Traverse the array
    for (int i = 0; i < N; i++)
 
      // Applying the operation on
      // all the positive elements
      if (arr[i] >= 0)
      {
        arr[i] = -arr[i] - 1;
      }
 
    if (N % 2 == 1)
    {
 
      // Stores maximum element in array
      int max_element = -1;
 
      // Stores index of maximum element
      int index = -1;
      for (int i = 0; i < N; i++)
 
        // Check if current element
        // is greater than the
        // maximum element
        if (Math.Abs(arr[i]) > max_element) {
 
          max_element = Math.Abs(arr[i]);
 
          // Find index of the
          // maximum element
          index = i;
        }
 
      // Perform the operation on the
      // maximum element
      arr[index] = -arr[index] - 1;
    }
 
    // Print the elements of the array
    for (int i = 0; i < N; i++)
      Console.Write(arr[i] + " ");
  }
 
// Driver Code
static public void Main()
{
    int[] arr = { -3, 0, 1 };
    int N = arr.Length;
 
    // Function Call
    findArrayWithMaxProduct(arr, N);
}
}
 
// This code is contributed by sanjoy_62.

Javascript

      // JavaScript program for the above approach
      // Function to find array with maximum
      // product by changing array
      // elements to (-1)arr[i] - 1
      function findArrayWithMaxProduct(arr, N)
      {
       
        // Traverse the array
        for (var i = 0; i < N; i++)
         
          // Applying the operation on
          // all the positive elements
          if (arr[i] >= 0) {
            arr[i] = -arr[i] - 1;
          }
 
        if (N % 2 === 1)
        {
         
          // Stores maximum element in array
          var max_element = -1;
 
          // Stores index of maximum element
          var index = -1;
          for (var i = 0; i < N; i++)
           
            // Check if current element
            // is greater than the
            // maximum element
            if (Math.abs(arr[i]) > max_element)
            {
              max_element = Math.abs(arr[i]);
 
              // Find index of the
              // maximum element
              index = i;
            }
 
          // Perform the operation on the
          // maximum element
          arr[index] = -arr[index] - 1;
        }
 
        // Print the elements of the array
        for (var i = 0; i < N; i++) document.write(arr[i] + " ");
      }
 
      // Driver Code
      var arr = [-3, 0, 1];
      var N = arr.length;
 
      // Function Call
      findArrayWithMaxProduct(arr, N);
       
      // This code is contributed by rdtank.
    

输出：

2 -1 -2

时间复杂度： O(N)
辅助空间： O(1)

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