Find Maximum Sum by Replacing the Subarray in Given Range

In this tutorial, we will explore how to determine maximum sum by replacing the subarray.

We must first understand exactly what a Subarray is. A portion or subset of an array is the typical definition of a subarray. A group of variables that a programmer defines together is known as an array. The coder can specify a single array with numerous values labelled rather than establishing separate variables for each.

The syntax for using a subarray, such as delivering data to or from a method or function, will typically be similar to that of using the complete array in most programming languages, with a few minor exceptions. The exact subarray or section of the array is typically stated in brackets or parentheses after the name of the array.

Programmers must exercise the same level of consistency care when using a subarray as they would when using a full array or a single variable. This involves ensuring that all portions of the code that require the values can access them and that they aren't altered by a bug or mistake in the code as they are passed from one method or function to another. If there is an issue with the use of a subarray, it can be helpful to run through a code in debug mode to examine how values are evaluated line by line.

Many of the same operations that programmers can carry out on the complete array can also be carried out on a subarray. These tools are made more adaptable for storing numerous variables in a single defined set by enabling operations on a subarray.

By devising a method through observations for replacing the array with the greatest number of members, this problem can be solved greedily. If anyone else chooses max one) and we have more than two elements to the left or right, we may update the array with those elements by executing the following operational processes:

The application of the given strategy is shown below:

C++ Program:

#include <bits/stdc++.h>
using namespace std;

int find_maxsum(int arr[], int n)
{
	if (n == 2)
		return max(
			{ 2 * abs(arr[0] - arr[1]), arr[0] + arr[1] });
	else if (n == 3)
		return max({ 3 * (abs(arr[0] - arr[1])),
					3 * (abs(arr[2] - arr[1])), 3 * arr[0],
					3 * arr[2],
					arr[0] + arr[1] + arr[2] });
	int mx = 0;
	for (int i = 0; i < n; i++)
		mx = max(arr[i], mx);
	return n * mx;
}
int main()
{
	int arr[] = { 1, 9 };
	int n = sizeof(arr) / sizeof(arr[0]);
	int arr2[] = { 1, 1, 1 };
	int m = sizeof(arr2) / sizeof(arr2[0]);

	cout << find_maxsum(arr, n) << endl;
	cout << find_maxsum(arr2, m) << endl;
}

Output