Maximum possible sub-array sum after at most X swaps

Given an array arr[] of N integers and an integer X, the task is to find the maximum possible sub-array sum after applying at most X swaps.

Examples:

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

Approach: For every possible sub-array, consider the elements which are not part of this sub-array as discarded. Now, while there are swaps left and the sum of the sub-array currently under consideration can be maximized i.e. the greatest element among the discarded elements can be swapped with the minimum element of the sub-array, keep updating the sum of the sub-array. When there are no swaps left or the sub-array sum cannot be further maximized, update the current maximum sub-array sum found so far which will be the required answer in the end.



Below is the implementation of the above approach:

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// C++ implementation of the approach
#include <bits/stdc++.h>
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<int, vector<int> > pq;
  
            // To store elements which are
            // part of the sub-array
            // currently under consideration
            priority_queue<int, vector<int>, greater<int> > 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;
}
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Output:
19

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