用它们的和重复替换相邻数组元素对后剩余对的最大乘积

给定一个大小为N的数组arr[] ，任务是在用它们的和重复替换一对相邻数组元素后，找到剩余对的最大乘积。
注意：将数组减少到大小为 2。

例子：

Input: arr[] = {2, 3, 5, 6, 7}
Output: 130
Explanation:
Replacing arr[1] and arr[2] with their sum (i.e. 3 + 5 = 8) modifies arr[] to {2, 8, 6, 7}
Replacing arr[2] and arr[3] with their sum (i.e. 6 + 7 = 13) modifies arr[] to {2, 8, 13}
Replacing arr[0] and arr[1] with their sum (2 + 8 = 10) modifies arr[] to {10, 13}
Maximum Product of the remaining pair = 10 * 13 = 130

Input: arr[] = {5, 6}
Output: 30

方法：给定的问题可以通过观察来解决。可以观察到，对于索引i ， X必须等于前i 个元素的总和，即 arr[1] + arr[2] + arr[3] + … + arr[i] 并且Y必须相等到其余元素的总和，即 arr[i + 1] + arr[i + 2] +…+ arr[N]。现在，这个问题可以通过使用前缀 sum 并找到它与每个索引处其余元素的总和的乘积来解决。请按照以下步骤解决问题：

• 将ans初始化为 INT_MIN 以存储所需的答案，将prefixSum 初始化为0以存储数组的前缀和。
• 将数组元素的总和存储在一个变量中，比如S 。
• 使用变量i在索引范围[0, N – 2] 上遍历数组并执行以下操作：
  • 将arr[i]的值添加到prefixSum 。
  • 将prefixSum的值存储在变量X 中，并将(sum – prefixSum)存储在变量Y 中。
  • 如果(X * Y)的值大于ans ，则将ans更新为(X * Y) 。
• 完成以上步骤后，打印ans的值作为结果。

下面是上述方法的实现：

C++

// C++ program for the above approach
 
#include 
using namespace std;
 
// Function to find the maximum product
// possible after repeatedly replacing
// pairs of adjacent array elements
// with their sum
void maxProduct(int arr[], int N)
{
    // Store the maximum product
    int max_product = INT_MIN;
 
    // Store the prefix sum
    int prefix_sum = 0;
 
    // Store the total sum of array
    int sum = 0;
 
    // Traverse the array to find
    // the total sum
    for (int i = 0; i < N; i++) {
        sum += arr[i];
    }
 
    // Iterate in the range [0, N-2]
    for (int i = 0; i < N - 1; i++) {
 
        // Add arr[i] to prefix_sum
        prefix_sum += arr[i];
 
        // Store the value of prefix_sum
        int X = prefix_sum;
 
        // Store the value of
        // (total sum - prefix sum)
        int Y = sum - prefix_sum;
 
        // Update the maximum product
        max_product = max(max_product,
                          X * Y);
    }
 
    // Print the answer
    cout << max_product;
}
 
// Driver Code
int main()
{
    int arr[] = { 2, 3, 5, 6, 7 };
    int N = sizeof(arr) / sizeof(arr[0]);
    maxProduct(arr, N);
 
    return 0;
}

Java

// Java program for the above approach
import java.io.*;
import java.util.*;
 
class GFG
{
  // Function to find the maximum product
  // possible after repeatedly replacing
  // pairs of adjacent array elements
  // with their sum
  static void maxProduct(int[] arr, int N)
  {
    // Store the maximum product
    int max_product = Integer.MIN_VALUE;
 
    // Store the prefix sum
    int prefix_sum = 0;
 
    // Store the total sum of array
    int sum = 0;
 
    // Traverse the array to find
    // the total sum
    for (int i = 0; i < N; i++)
    {
      sum += arr[i];
    }
 
    // Iterate in the range [0, N-2]
    for (int i = 0; i < N - 1; i++)
    {
 
      // Add arr[i] to prefix_sum
      prefix_sum += arr[i];
 
      // Store the value of prefix_sum
      int X = prefix_sum;
 
      // Store the value of
      // (total sum - prefix sum)
      int Y = sum - prefix_sum;
 
      // Update the maximum product
      max_product = Math.max(max_product, X * Y);
    }
 
    // Print the answer
    System.out.print(max_product);
  }
 
// Driver Code
public static void main(String[] args)
{
    int[] arr = { 2, 3, 5, 6, 7 };
    int N = arr.length;
    maxProduct(arr, N);
}
}
 
// This code is contributed by sanjoy_62.

Python3

# Python program for the above approach
import sys
 
# Function to find the maximum product
# possible after repeatedly replacing
# pairs of adjacent array elements
# with their sum
def maxProduct(arr, N):
   
    # Store the maximum product
    max_product = -sys.maxsize;
 
    # Store the prefix sum
    prefix_sum = 0;
 
    # Store the total sum of array
    sum = 0;
 
    # Traverse the array to find
    # the total sum
    for i in range(N):
        sum += arr[i];
 
    # Iterate in the range [0, N-2]
    for i in range(N - 1):
       
        # Add arr[i] to prefix_sum
        prefix_sum += arr[i];
 
        # Store the value of prefix_sum
        X = prefix_sum;
 
        # Store the value of
        # (total sum - prefix sum)
        Y = sum - prefix_sum;
 
        # Update the maximum product
        max_product = max(max_product, X * Y);
 
    # Prthe answer
    print(max_product);
 
# Driver Code
if __name__ == '__main__':
    arr = [2, 3, 5, 6, 7];
    N = len(arr);
    maxProduct(arr, N);
 
# This code is contributed by shikhasingrajput

C#

// C# program for the above approach
using System;
class GFG
{
 
  // Function to find the maximum product
  // possible after repeatedly replacing
  // pairs of adjacent array elements
  // with their sum
  static void maxProduct(int[] arr, int N)
  {
    // Store the maximum product
    int max_product = Int32.MinValue;
 
    // Store the prefix sum
    int prefix_sum = 0;
 
    // Store the total sum of array
    int sum = 0;
 
    // Traverse the array to find
    // the total sum
    for (int i = 0; i < N; i++)
    {
      sum += arr[i];
    }
 
    // Iterate in the range [0, N-2]
    for (int i = 0; i < N - 1; i++)
    {
 
      // Add arr[i] to prefix_sum
      prefix_sum += arr[i];
 
      // Store the value of prefix_sum
      int X = prefix_sum;
 
      // Store the value of
      // (total sum - prefix sum)
      int Y = sum - prefix_sum;
 
      // Update the maximum product
      max_product = Math.Max(max_product, X * Y);
    }
 
    // Print the answer
    Console.WriteLine(max_product);
  } 
 
  // Driver code
  static void Main()
  {
    int[] arr = { 2, 3, 5, 6, 7 };
    int N = arr.Length;
    maxProduct(arr, N);
  }
}
 
// This code is contributed by divyeshrabadiya07.

Javascript

// javascript program for the above approach
 
    // Function to find the maximum product
    // possible after repeatedly replacing
    // pairs of adjacent array elements
    // with their sum
    function maxProduct(arr , N) {
        // Store the maximum product
        var max_product = Number.MIN_VALUE;
 
        // Store the prefix sum
        var prefix_sum = 0;
 
        // Store the total sum of array
        var sum = 0;
 
        // Traverse the array to find
        // the total sum
        for (i = 0; i < N; i++) {
            sum += arr[i];
        }
 
        // Iterate in the range [0, N-2]
        for (i = 0; i < N - 1; i++) {
 
            // Add arr[i] to prefix_sum
            prefix_sum += arr[i];
 
            // Store the value of prefix_sum
            var X = prefix_sum;
 
            // Store the value of
            // (total sum - prefix sum)
            var Y = sum - prefix_sum;
 
            // Update the maximum product
            max_product = Math.max(max_product, X * Y);
        }
 
        // Prvar the answer
        document.write(max_product);
    }
 
    // Driver Code
     
        var arr = [ 2, 3, 5, 6, 7 ];
        var N = arr.length;
        maxProduct(arr, N);
 
// This code contributed by umadevi9616

输出：

130

时间复杂度： O(N)
辅助空间： O(1)

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