Given an array arr[] of size N, find the maximum value of adjacent sum by rearranging array in any order.

Examples:

Input: arr[] = {1, 2, 3, 4}
Output: 17
Explanation: The arrangement is {1, 3, 4, 2} where sum of all the adjacent element is (1 + 3) + (3 + 4) + (4 + 2) = 17 which is maximum.

Input: arr[] = {5, 2, 3, 4, 6}
Output: 35
Explanation: The arrangement is {2, 5, 4, 6, 3} where sum of all adjacent element is (2 + 5) + (5 + 4) + (4 + 6) + (6 + 3) = 35 which is maximum.

Approach: To solve the problem, follow the below idea:

While calculating sum, each element comes 2 times except 2 elements(the first element and the last element) so to maximize sum we just keep the smallest 2 elements on the ends so that they don’t contribute to sum as much and since each elements comes 2 times(besides the end elements) so the final sum will be (2 * sum of all elements) – smallest element- second smallest element

Step-by-step algorithm:

• Find the smallest and the second smallest element in the array.
• Find the sum of all elements as sum.
• Return the maximum sum as (2 * sum) – smallest element – second smallest element.

Below is the implementation of the algorithm:

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

// Function to output max adj sum
void solve(int arr[], int N)
{
    int smallestIdx = 0;
    // traversing the array to find
    // smallest element.
    for (int i = 0; i < N; i++) {
        if (arr[i] < arr[smallestIdx]) {
            smallestIdx = i;
        }
    }

    int secondSmallestIdx = 0;
    if (smallestIdx == 0)
        secondSmallestIdx = 1;

    // traversing the array to find second smallest element
    for (int i = 0; i < N; i++) {
        if (arr[i] < arr[secondSmallestIdx]
            && smallestIdx != i) {
            secondSmallestIdx = i;
        }
    }

    // the maximum sum will be 2*sum of all the element
    // minus the smallest and the second smallest element.
    int sum = 0;
    for (int i = 0; i < N; i++) {
        sum += arr[i];
    }
    cout << "maximum adjacent sum is: "
         << 2 * sum - arr[smallestIdx]
                - arr[secondSmallestIdx]
         << endl;
}
int main()
{
    // Sample Input
    int N = 5;
    int arr[] = { 5, 2, 3, 4, 6 };

    solve(arr, N);
    return 0;
}

import java.util.Arrays;

public class Main {
    // Function to output max adjacent sum
    static void solve(int[] arr, int N) {
        int smallestIdx = 0;

        // Traversing the array to find the smallest element
        for (int i = 0; i < N; i++) {
            if (arr[i] < arr[smallestIdx]) {
                smallestIdx = i;
            }
        }

        int secondSmallestIdx = 0;
        if (smallestIdx == 0) {
            secondSmallestIdx = 1;
        }

        // Traversing the array to find the second smallest element
        for (int i = 0; i < N; i++) {
            if (arr[i] < arr[secondSmallestIdx] && smallestIdx != i) {
                secondSmallestIdx = i;
            }
        }

        // The maximum sum will be 2 * sum of all the elements
        // minus the smallest and the second smallest element.
        int sum = Arrays.stream(arr).sum();
        System.out.println("maximum adjacent sum is: " +
                (2 * sum - arr[smallestIdx] - arr[secondSmallestIdx]));
    }

    public static void main(String[] args) {
        // Sample Input
        int N = 5;
        int[] arr = {5, 2, 3, 4, 6};

        solve(arr, N);
    }
}

// This code is contributed by shivamgupta0987654321

# Function to output max adjacent sum
def solve(arr):
    N = len(arr)
    smallestIdx = 0

    # Traversing the array to find the smallest element
    for i in range(N):
        if arr[i] < arr[smallestIdx]:
            smallestIdx = i

    secondSmallestIdx = 0
    if smallestIdx == 0:
        secondSmallestIdx = 1

    # Traversing the array to find the second smallest element
    for i in range(N):
        if arr[i] < arr[secondSmallestIdx] and smallestIdx != i:
            secondSmallestIdx = i

    # The maximum sum will be 2 times the sum of all elements
    # minus the smallest and the second smallest element.
    sum_of_elements = sum(arr)
    max_adj_sum = 2 * sum_of_elements - arr[smallestIdx] - arr[secondSmallestIdx]
    print("Maximum adjacent sum is:", max_adj_sum)

arr = [5, 2, 3, 4, 6]
solve(arr)

function solve(arr) {
    let N = arr.length;
    let smallestIdx = 0;

    // Traversing the array to find the smallest element
    for (let i = 0; i < N; i++) {
        if (arr[i] < arr[smallestIdx]) {
            smallestIdx = i;
        }
    }

    let secondSmallestIdx = 0;
    if (smallestIdx === 0) {
        secondSmallestIdx = 1;
    }

    // Traversing the array to find the second smallest element
    for (let i = 0; i < N; i++) {
        if (arr[i] < arr[secondSmallestIdx] && smallestIdx !== i) {
            secondSmallestIdx = i;
        }
    }

    // The maximum sum will be 2 times the sum of all elements
    // minus the smallest and the second smallest element.
    let sum_of_elements = arr.reduce((acc, curr) => acc + curr, 0);
    let max_adj_sum = 2 * sum_of_elements - arr[smallestIdx] - arr[secondSmallestIdx];
    console.log("Maximum adjacent sum is:", max_adj_sum);
}

let arr = [5, 2, 3, 4, 6];
solve(arr);

maximum adjacent sum is :35

Time Complexity: O(N), where N is the size of input array arr[].
Auxiliary Space: O(1)