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 the 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++ 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;
}

// 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..

# 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

// 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..

<script>

// Javascript implementation of the approach
var arrSize = 51;

// To store the states of dp
var dp = Array(arrSize);
var v = Array(arrSize);

// Function to return the maximized sum
function 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] = 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
var arr = [1, 1, 1, 1];
var n = arr.length;
document.write( sumMax(0, arr, n));

</script>   

2

Time Complexity: O(N), where N is the given size of the array.

Auxiliary Space: O(51), no extra space is required, so it is a constant.

Efficient approach : Using DP Tabulation method ( Iterative approach )

The approach to solve this problem is same but DP tabulation(bottom-up) method is better then Dp + memoization(top-down) because memoization method needs extra stack space of recursion calls.

Steps to solve this problem :

• Create a DP to store the solution of the subproblems.
• Initialize the DP  with base cases
• Now Iterate over subproblems to get the value of current problem form previous computation of subproblems stored in DP
• Return the final solution stored in dp[0].

Implementation :

// C++ implementation of the approach
#include <bits/stdc++.h>
#define arrSize 51
using namespace std;

// To store the states of dp
int dp[arrSize];

// Function to return the maximized sum
int sumMax(int arr[], int n) {
    // Initializing the base cases
    dp[n-1] = dp[n] = dp[n+1] = 0;

    // Calculating the dp table
    for(int i=n-2; i>=0; i--) {
        dp[i] = max(arr[i] + arr[i+1] + dp[i+3], dp[i+1]);
    }

    // Return the result
    return dp[0];
}

// Driver Code
int main()
{
    int arr[] = {1,1,1,1};
    int n = sizeof(arr) / sizeof(int);
    
      // function call
    cout << sumMax(arr, n);

    return 0;
}

import java.util.Arrays;

public class Main {

    // To store the states of dp
    private static final int arrSize = 51;
    private static int[] dp = new int[arrSize];

    // Function to return the maximized sum
    private static int sumMax(int[] arr, int n) {
        // Initializing the base cases
        dp[n - 1] = dp[n] = dp[n + 1] = 0;

        // Calculating the dp table
        for (int i = n - 2; i >= 0; i--) {
            dp[i] = Math.max(arr[i] + arr[i + 1] + dp[i + 3], dp[i + 1]);
        }

        // Return the result
        return dp[0];
    }

    // Driver Code
    public static void main(String[] args) {
        int[] arr = {1, 1, 1, 1};
        int n = arr.length;

        // function call
        System.out.println(sumMax(arr, n));
    }
}

# Python implementation of the approach
import sys
sys.setrecursionlimit(10**7)

# To store the states of dp
dp = [0] * 51

# Function to return the maximized sum
def sumMax(arr, n):
    # Initializing the base cases
    dp[n-1] = dp[n] = dp[n+1] = 0

    # Calculating the dp table
    for i in range(n-2, -1, -1):
        dp[i] = max(arr[i] + arr[i+1] + dp[i+3], dp[i+1])

    # Return the result
    return dp[0]

# Driver Code
if __name__ == '__main__':
    arr = [1,1,1,1]
    n = len(arr)
    
    # function call
    print(sumMax(arr, n))

using System;

public class Program 
{

  // To store the states of dp
  const int arrSize = 51;
  static int[] dp = new int[arrSize];

  // Function to return the maximized sum
  static int sumMax(int[] arr, int n)
  {
    // Initializing the base cases
    dp[n - 1] = dp[n] = dp[n + 1] = 0;

    // Calculating the dp table
    for (int i = n - 2; i >= 0; i--) {
      dp[i] = Math.Max(
        arr[i] + arr[i + 1] + dp[i + 3], dp[i + 1]);
    }

    // Return the result
    return dp[0];
  }

  public static void Main()
  {
    int[] arr = { 1, 1, 1, 1 };
    int n = arr.Length;

    // function call
    Console.WriteLine(sumMax(arr, n));
  }
}

// This code is contributed by user_dtewbxkn77n

function sumMax(arr, n) {
  // Initializing the base cases
  const dp = Array(n + 2).fill(0);

  // Calculating the dp table
  for (let i = n - 2; i >= 0; i--) {
    dp[i] = Math.max(arr[i] + arr[i + 1] + dp[i + 3], dp[i + 1]);
  }

  // Return the result
  return dp[0];
}

// Driver Code
const arr = [1, 1, 1, 1];
const n = arr.length;

// Function call
console.log(sumMax(arr, n));

Output: 

2

Time Complexity: O(N), where N is the given size of the array.

Auxiliary Space: O(51), no extra space is required, so it is a constant.