# Maximize difference between the Sum of the two halves of the Array after removal of N elements

Given an integer N and array arr[] consisting of 3 * N integers, the task is to find the maximum difference between first half and second half of the array after the removal of exactly N elements from the array.

Examples:

Input: N = 2, arr[] = {3, 1, 4, 1, 5, 9}
Output: 1
Explanation:
Removal of arr[1] and arr[5] from the array maximizes the difference = (3 + 4) – (1 + 5) = 7 – 6 = 1.

Input: N = 1, arr[] = {1, 2, 3}
Output: -1

Approach:
Follow the steps given below to solve the problem

• Traverse the array from the beginning and keep updating the sum of the largest N elements from the beginning of the array.
• Similarly, keep updating the sum of the smallest N elements from the end of the array.
• Traverse these sums and calculate the differences at each point and update the maximum difference obtained.
• Print the maximum difference obtained.

Below is the implementation of the above approach:

 `// C++ Program to implement ` `// the above approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function to print the maximum difference ` `// possible between the two halves of the array ` `long` `long` `FindMaxDif(vector<``long` `long``> a, ``int` `m) ` `{ ` `    ``int` `n = m / 3; ` ` `  `    ``vector<``long` `long``> l(m + 5), r(m + 5); ` ` `  `    ``// Stores n maximum values from the start ` `    ``multiset<``long` `long``> s; ` ` `  `    ``for` `(``int` `i = 1; i <= m; i++) { ` ` `  `        ``// Insert first n elements ` `        ``if` `(i <= n) { ` ` `  `            ``// Update sum of largest n ` `            ``// elements from left ` `            ``l[i] = a[i - 1] + l[i - 1]; ` `            ``s.insert(a[i - 1]); ` `        ``} ` ` `  `        ``// For the remaining elements ` `        ``else` `{ ` `            ``l[i] = l[i - 1]; ` ` `  `            ``// Obtain minimum value ` `            ``// in the set ` `            ``long` `long` `d = *s.begin(); ` ` `  `            ``// Insert only if it is greater ` `            ``// than minimum value ` `            ``if` `(a[i - 1] > d) { ` ` `  `                ``// Update sum from left ` `                ``l[i] -= d; ` `                ``l[i] += a[i - 1]; ` ` `  `                ``// Remove the minimum ` `                ``s.erase(s.find(d)); ` ` `  `                ``// Insert the current element ` `                ``s.insert(a[i - 1]); ` `            ``} ` `        ``} ` `    ``} ` ` `  `    ``// Clear the set ` `    ``s.clear(); ` ` `  `    ``// Store n minimum elements from the end ` `    ``for` `(``int` `i = m; i >= 1; i--) { ` ` `  `        ``// Insert the last n elements ` `        ``if` `(i >= m - n + 1) { ` ` `  `            ``// Update sum of smallest n ` `            ``// elements from right ` `            ``r[i] = a[i - 1] + r[i + 1]; ` `            ``s.insert(a[i - 1]); ` `        ``} ` ` `  `        ``// For the remaining elements ` `        ``else` `{ ` ` `  `            ``r[i] = r[i + 1]; ` ` `  `            ``// Obtain the minimum ` `            ``long` `long` `d = *s.rbegin(); ` ` `  `            ``// Insert only if it is smaller ` `            ``// than maximum value ` `            ``if` `(a[i - 1] < d) { ` ` `  `                ``// Update sum from right ` `                ``r[i] -= d; ` `                ``r[i] += a[i - 1]; ` ` `  `                ``// Remove the minimum ` `                ``s.erase(s.find(d)); ` ` `  `                ``// Insert the new element ` `                ``s.insert(a[i - 1]); ` `            ``} ` `        ``} ` `    ``} ` ` `  `    ``long` `long` `ans = -9e18L; ` ` `  `    ``for` `(``int` `i = n; i <= m - n; i++) { ` ` `  `        ``// Compare the difference and ` `        ``// store the maximum ` `        ``ans = max(ans, l[i] - r[i + 1]); ` `    ``} ` ` `  `    ``// Return the maximum ` `    ``// possible difference ` `    ``return` `ans; ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` ` `  `    ``vector<``long` `long``> vtr = { 3, 1, 4, 1, 5, 9 }; ` `    ``int` `n = vtr.size(); ` ` `  `    ``cout << FindMaxDif(vtr, n); ` ` `  `    ``return` `0; ` `} `

Output:
```1
```

Time Complexity: O(NlogN)
Auxiliary Space: O(N)

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