最多 X 次交换后的最大可能子数组和

给定一个由N 个整数组成的数组arr[]和一个整数X ，任务是在应用最多X 次交换后找到最大可能的子数组和。
例子：

Input: arr[] = {5, -1, 2, 3, 4, -2, 5}, X = 2 
Output: 19 
Swap (arr[0], arr[1]) and (arr[5], arr[6]). 
Now, the maximum sub-array sum will be (5 + 2 + 3 + 4 + 5) = 19
Input: arr[] = {-2, -3, -1, -10}, X = 10 
Output: -1 
 

方法：对于每个可能的子数组，将不属于该子数组的元素视为丢弃。现在，虽然有交换，并且当前考虑的子数组的总和可以最大化，即丢弃元素中最大的元素可以与子数组的最小元素交换，继续更新子数组的总和大批。当没有交换剩余或子阵列总和无法进一步最大化时，更新目前找到的当前最大子阵列总和，这将是最终所需的答案。
下面是上述方法的实现：

CPP

// C++ implementation of the approach
#include 
using namespace std;
 
// Function to return the maximum
// sub-array sum after at most x swaps
int SubarraySum(int a[], int n, int x)
{
    // To store the required answer
    int ans = -10000;
 
    // For all possible intervals
    for (int i = 0; i < n; i++) {
        for (int j = i; j < n; j++) {
 
            // Keep current ans as zero
            int curans = 0;
 
            // To store the integers which are
            // not part of the sub-array
            // currently under consideration
            priority_queue > pq;
 
            // To store elements which are
            // part of the sub-array
            // currently under consideration
            priority_queue, greater > pq2;
 
            // Create two sets
            for (int k = 0; k < n; k++) {
                if (k >= i && k <= j) {
                    curans += a[k];
                    pq2.push(a[k]);
                }
                else
                    pq.push(a[k]);
            }
            ans = max(ans, curans);
 
            // Swap at most X elements
            for (int k = 1; k <= x; k++) {
                if (pq.empty() || pq2.empty()
                    || pq2.top() >= pq.top())
                    break;
 
                // Remove the minimum of
                // the taken elements
                curans -= pq2.top();
                pq2.pop();
 
                // Add maximum of the
                // discarded elements
                curans += pq.top();
                pq.pop();
 
                // Update the answer
                ans = max(ans, curans);
            }
        }
    }
 
    // Return the maximized sub-array sum
    return ans;
}
 
// Driver code
int main()
{
    int a[] = { 5, -1, 2, 3, 4, -2, 5 }, x = 2;
    int n = sizeof(a) / sizeof(a[0]);
 
    cout << SubarraySum(a, n, x);
 
    return 0;
}

Java

// Java implementation of the approach
import java.io.*;
import java.util.*;
class GFG
{
 
  // Function to return the maximum
  // sub-array sum after at most x swaps
  static int SubarraySum(int[] a, int n, int x)
  {
 
    // To store the required answer
    int ans = -10000;
 
    // For all possible intervals
    for (int i = 0; i < n; i++)
    {
      for (int j = i; j < n; j++)
      {
 
        // Keep current ans as zero
        int curans = 0;
 
        // To store the integers which are
        // not part of the sub-array
        // currently under consideration
        ArrayList pq = new ArrayList();
 
        // To store elements which are
        // part of the sub-array
        // currently under consideration
        ArrayList pq2 = new ArrayList();
 
        // Create two sets
        for (int k = 0; k < n; k++) {
          if (k >= i && k <= j) {
            curans += a[k];
            pq2.add(a[k]);
          }
          else
            pq.add(a[k]);
        }
 
        Collections.sort(pq);
        Collections.reverse(pq);
        Collections.sort(pq2);
 
        ans = Math.max(ans, curans);
 
        // Swap at most X elements
        for (int k = 1; k <= x; k++) {
          if (pq.size() == 0 || pq2.size() == 0
              || pq2.get(0) >= pq.get(0))
            break;
 
          // Remove the minimum of
          // the taken elements
          curans -= pq2.get(0);
          pq2.remove(0);
 
          // Add maximum of the
          // discarded elements
          curans += pq.get(0);
          pq.remove(0);
 
          // Update the answer
          ans = Math.max(ans, curans);
        }
      }
    }
 
    // Return the maximized sub-array sum
    return ans;
  }
 
  // Driver code.
  public static void main (String[] args)
  {
 
    int[] a = { 5, -1, 2, 3, 4, -2, 5 };
    int x = 2;
    int n = a.length;
 
    System.out.println(SubarraySum(a, n, x));
  }
}
 
// This code is contributed by avanitrachhadiya2155

Python3

# Python3 implementation of the approach
 
# Function to return the maximum
# sub-array sum after at most x swaps
def SubarraySum(a, n, x) :
     
    # To store the required answer
    ans = -10000
     
    # For all possible intervals
    for i in range(n) :
     
      for j in range(i, n) :
     
        # Keep current ans as zero
        curans = 0
     
        # To store the integers which are
        # not part of the sub-array
        # currently under consideration
        pq = []
     
        # To store elements which are
        # part of the sub-array
        # currently under consideration
        pq2 = []
     
        # Create two sets
        for k in range(n) :
          if (k >= i and k <= j) :
            curans += a[k]
            pq2.append(a[k])
           
          else :
            pq.append(a[k])
     
        pq.sort()
        pq.reverse()
        pq2.sort()
     
        ans = max(ans, curans)
     
        # Swap at most X elements
        for k in range(1, x + 1) :
          if (len(pq) == 0 or len(pq2) == 0 or pq2[0] >= pq[0]) :
            break
     
          # Remove the minimum of
          # the taken elements
          curans -= pq2[0]
          pq2.pop(0)
     
          # Add maximum of the
          # discarded elements
          curans += pq[0]
          pq.pop(0)
     
          # Update the answer
          ans = max(ans, curans)
     
    # Return the maximized sub-array sum
    return ans
     
    # Driver code
a = [ 5, -1, 2, 3, 4, -2, 5 ]
x = 2;
n = len(a)
print(SubarraySum(a, n, x))
 
# This ccode is contributed by divyesh072019.

C#

// C# implementation of the approach
using System;
using System.Collections.Generic;
class GFG
{
 
  // Function to return the maximum
  // sub-array sum after at most x swaps
  static int SubarraySum(int[] a, int n, int x)
  {
 
    // To store the required answer
    int ans = -10000;
 
    // For all possible intervals
    for (int i = 0; i < n; i++)
    {
      for (int j = i; j < n; j++)
      {
 
        // Keep current ans as zero
        int curans = 0;
 
        // To store the integers which are
        // not part of the sub-array
        // currently under consideration
        List pq = new List();
 
        // To store elements which are
        // part of the sub-array
        // currently under consideration
        List pq2 = new List();
 
        // Create two sets
        for (int k = 0; k < n; k++) {
          if (k >= i && k <= j) {
            curans += a[k];
            pq2.Add(a[k]);
          }
          else
            pq.Add(a[k]);
        }
 
        pq.Sort();
        pq.Reverse();
        pq2.Sort();
 
        ans = Math.Max(ans, curans);
 
        // Swap at most X elements
        for (int k = 1; k <= x; k++) {
          if (pq.Count == 0 || pq2.Count == 0
              || pq2[0] >= pq[0])
            break;
 
          // Remove the minimum of
          // the taken elements
          curans -= pq2[0];
          pq2.RemoveAt(0);
 
          // Add maximum of the
          // discarded elements
          curans += pq[0];
          pq.RemoveAt(0);
 
          // Update the answer
          ans = Math.Max(ans, curans);
        }
      }
    }
 
    // Return the maximized sub-array sum
    return ans;
  }
 
  // Driver code.
  static void Main() {
    int[] a = { 5, -1, 2, 3, 4, -2, 5 };
    int x = 2;
    int n = a.Length;
    Console.WriteLine(SubarraySum(a, n, x));
  }
}
 
// This code is contributed by divyeshrabaiya07.

Javascript

    // Javascript implementation of the approach
       
    // Function to return the maximum
    // sub-array sum after at most x swaps
    function SubarraySum(a, n, x)
    {
 
      // To store the required answer
      let ans = -10000;
 
      // For all possible intervals
      for (let i = 0; i < n; i++)
      {
        for (let j = i; j < n; j++)
        {
 
          // Keep current ans as zero
          let curans = 0;
 
          // To store the integers which are
          // not part of the sub-array
          // currently under consideration
          let pq = [];
 
          // To store elements which are
          // part of the sub-array
          // currently under consideration
          let pq2 = [];
 
          // Create two sets
          for (let k = 0; k < n; k++) {
            if (k >= i && k <= j) {
              curans += a[k];
              pq2.push(a[k]);
            }
            else
              pq.push(a[k]);
          }
 
          pq.sort();
          pq.reverse();
          pq2.sort();
 
          ans = Math.max(ans, curans);
 
          // Swap at most X elements
          for (let k = 1; k <= x; k++) {
            if (pq.length == 0 || pq2.length == 0
                || pq2[0] >= pq[0])
              break;
 
            // Remove the minimum of
            // the taken elements
            curans -= pq2[0];
            pq2.shift();
 
            // Add maximum of the
            // discarded elements
            curans += pq[0];
            pq.shift();
 
            // Update the answer
            ans = Math.max(ans, curans);
          }
        }
      }
 
      // Return the maximized sub-array sum
      return ans;
    }
       
    let a = [ 5, -1, 2, 3, 4, -2, 5 ];
    let x = 2;
    let n = a.length;
    document.write(SubarraySum(a, n, x));
         
        // This code is contributed by suresh07.

输出：

19

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