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:

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<``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;` `}`

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 code 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<``int``> pq = ``new` `List<``int``>(); `   `        ``// To store elements which are ` `        ``// part of the sub-array ` `        ``// currently under consideration ` `        ``List<``int``> pq2 = ``new` `List<``int``>(); `   `        ``// 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

 ``

Output:

`19`

Time Complexity: O(n3 logn)

Auxiliary Space: O(n)

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