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++ 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 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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`# 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# 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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