Maximum sum in an array such that every element has exactly one adjacent element to it

Given an array arr[] of N integers, you can select some indexes such that every selected index has exactly one other selected index adjacent to it and the sum of elements at the chosen indexes should be maximum.

In other words, the task is to select elements from an array such that a single element alone is not selected and elements at three consecutive indices are not selected and the sum of selected elements should be maximum.

The task is to print the maximized sum.

Examples:

Input: arr[] = {1, 2, 3, 1, 4}
Output: 8
arr[0] + arr[1] + arr[3] + arr[4] = 1 + 2 + 1 + 4 = 8

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

Approach: Dynamic programming can be used to solve this problem. This problem can be translated to selecting pairs of adjacent integers such that no two pairs are adjacent or have an element in common
i.e. if (arr[i], arr[i + 1]) is a pair we selected then neither (arr[i + 2], arr[i + 3]) nor (arr[i + 1], arr[i + 2]) can be selected.
Let’s decide the states of the dp according to above statement.
For every index i, we will either select indexes i and i + 1 i.e. make a pair or not make it. In case, we make a pair, we won’t be able to select the index i + 2 as it will make 2 elements adjacent to i + 1. So, we will have to solve for i + 3 next. If we don’t make a pair, we will simply solve for i + 1.
So the recurrence relation will be.

dp[i] = max(arr[i] + arr[i + 1] + dp[i + 3], dp[i + 1])

There are N states in total and each state takes O(1) time to solve. Thus, the time complexity will be O(N).

Below is the implementation of the above approach:

C++

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// C++ implementation of the approach
#include <bits/stdc++.h>
#define arrSize 51
using namespace std;
  
// To store the states of dp
int dp[arrSize];
bool v[arrSize];
  
// Function to return the maximized sum
int sumMax(int i, int arr[], int n)
{
    // Base case
    if (i >= n - 1)
        return 0;
  
    // Checks if a state is
    // already solved
    if (v[i])
        return dp[i];
    v[i] = true;
  
    // Recurrence relation
    dp[i] = max(arr[i] + arr[i + 1]
                    + sumMax(i + 3, arr, n),
                sumMax(i + 1, arr, n));
  
    // Return the result
    return dp[i];
}
  
// Driver code
int main()
{
    int arr[] = { 1, 1, 1, 1 };
    int n = sizeof(arr) / sizeof(int);
  
    cout << sumMax(0, arr, n);
  
    return 0;
}

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Java

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// Java implementation of the approach
import java.io.*;
  
class GFG 
{
  
static int arrSize = 51;
  
// To store the states of dp
static int dp[] = new int[arrSize];
static boolean v[] = new boolean[arrSize];
  
// Function to return the maximized sum
static int sumMax(int i, int arr[], int n)
{
    // Base case
    if (i >= n - 1)
        return 0;
  
    // Checks if a state is
    // already solved
    if (v[i])
        return dp[i];
    v[i] = true;
  
    // Recurrence relation
    dp[i] = Math.max(arr[i] + arr[i + 1]
                    + sumMax(i + 3, arr, n),
                sumMax(i + 1, arr, n));
  
    // Return the result
    return dp[i];
}
  
// Driver code
public static void main (String[] args) 
{
    int arr[] = { 1, 1, 1, 1 };
    int n = arr.length;
  
    System.out.println(sumMax(0, arr, n));
}
}
  
// This code is contributed by anuj_67..

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Python3

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# Python 3 implementation of the approach
arrSize = 51
  
# To store the states of dp
dp = [0 for i in range(arrSize)]
v = [False for i in range(arrSize)]
  
# Function to return the maximized sum
def sumMax(i,arr,n):
    # Base case
    if (i >= n - 1):
        return 0
  
    # Checks if a state is
    # already solved
    if (v[i]):
        return dp[i]
    v[i] = True
    # Recurrence relation
    dp[i] = max(arr[i] + arr[i + 1] + sumMax(i + 3, arr, n),
                                        sumMax(i + 1, arr, n))
    # Return the result
    return dp[i]
  
# Driver code
if __name__ == '__main__':
    arr = [1, 1, 1, 1]
    n = len(arr)
    print(sumMax(0, arr, n))
  
# This code is contributed by
# Surendra_Gangwar

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C#

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// C# implementation of the approach
using System;
  
class GFG 
{
  
static int arrSize = 51;
  
// To store the states of dp
static int []dp = new int[arrSize];
static bool []v = new bool[arrSize];
  
// Function to return the maximized sum
static int sumMax(int i, int []arr, int n)
{
    // Base case
    if (i >= n - 1)
        return 0;
  
    // Checks if a state is
    // already solved
    if (v[i])
        return dp[i];
    v[i] = true;
  
    // Recurrence relation
    dp[i] = Math.Max(arr[i] + arr[i + 1]
                    + sumMax(i + 3, arr, n),
                sumMax(i + 1, arr, n));
  
    // Return the result
    return dp[i];
}
  
// Driver code
public static void Main () 
{
    int []arr = { 1, 1, 1, 1 };
    int n = arr.Length;
  
    Console.WriteLine(sumMax(0, arr, n));
}
}
  
// This code is contributed by anuj_67..

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Output:

2


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Improved By : vt_m, SURENDRA_GANGWAR