# Replace array elements by sum of next two consecutive elements

• Difficulty Level : Basic
• Last Updated : 23 Apr, 2021

Given an array arr[] of size n, The task is to replace every element of the array by the sum of next two consecutive elements in a circular manner i.e. arr[0] = arr[1] + arr[2], arr[1] = arr[2] + arr[3], … arr[n – 1] = arr[0] + arr[1].
Examples:

Input: arr[] = {3, 4, 2, 1, 6}
Output: 6 3 7 9 7
Input: arr[] = {5, 2, 1, 3, 8}
Output: 3 4 11 13 7

Approach: Store the first and second element of the array in variables first and second. Now for every element except the last and the second last element of the array, update arr[i] = arr[i + 1] + arr[i + 2]. Then update the last and the second last element as arr[n – 2] = arr[n – 1] + first and arr[n – 1] = first + second.
Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach``#include ``using` `namespace` `std;` `// Utility function to print the``// contents of an array``void` `printArr(``int` `arr[], ``int` `n)``{``    ``for` `(``int` `i = 0; i < n; i++)``        ``cout << arr[i] << ``" "``;``}` `// Function to update every element of``// the array as the sum of next two elements``void` `updateArr(``int` `arr[], ``int` `n)``{` `    ``// Invalid array``    ``if` `(n < 3)``        ``return``;` `    ``// First and second elements of the array``    ``int` `first = arr[0];``    ``int` `second = arr[1];` `    ``// Update every element as required``    ``// except the last and the``    ``// second last element``    ``for` `(``int` `i = 0; i < n - 2; i++)``        ``arr[i] = arr[i + 1] + arr[i + 2];` `    ``// Update the last and the second``    ``// last element of the array``    ``arr[n - 2] = arr[n - 1] + first;``    ``arr[n - 1] = first + second;` `    ``// Print the updated array``    ``printArr(arr, n);``}` `// Driver code``int` `main()``{``    ``int` `arr[] = { 3, 4, 2, 1, 6 };``    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr[0]);``    ``updateArr(arr, n);` `    ``return` `0;``}`

## Java

 `// Java implementation of the approach``class` `GFG``{` `    ``// Utility function to print the``    ``// contents of an array``    ``static` `void` `printArr(``int``[] arr, ``int` `n)``    ``{``        ``for` `(``int` `i = ``0``; i < n; i++)``        ``{``            ``System.out.print(arr[i] + ``" "``);``        ``}``    ``}` `    ``// Function to update every element of``    ``// the array as the sum of next two elements``    ``static` `void` `updateArr(``int``[] arr, ``int` `n)``    ``{` `        ``// Invalid array``        ``if` `(n < ``3``)``        ``{``            ``return``;``        ``}` `        ``// First and second elements of the array``        ``int` `first = arr[``0``];``        ``int` `second = arr[``1``];` `        ``// Update every element as required``        ``// except the last and the``        ``// second last element``        ``for` `(``int` `i = ``0``; i < n - ``2``; i++)``        ``{``            ``arr[i] = arr[i + ``1``] + arr[i + ``2``];``        ``}` `        ``// Update the last and the second``        ``// last element of the array``        ``arr[n - ``2``] = arr[n - ``1``] + first;``        ``arr[n - ``1``] = first + second;` `        ``// Print the updated array``        ``printArr(arr, n);``    ``}` `    ``// Driver code``    ``public` `static` `void` `main(String[] args)``    ``{``        ``int``[] arr = {``3``, ``4``, ``2``, ``1``, ``6``};``        ``int` `n = arr.length;``        ``updateArr(arr, n);``    ``}``}` `// This code is contributed by 29AjayKumar`

## Python3

 `# Python 3 implementation of the approach` `# Utility function to print the``# contents of an array``def` `printArr(arr, n):``    ``for` `i ``in` `range``(n):``        ``print``(arr[i], end ``=` `" "``)` `# Function to update every element of``# the array as the sum of next two elements``def` `updateArr(arr, n):``    ` `    ``# Invalid array``    ``if` `(n < ``3``):``        ``return` `    ``# First and second elements of the array``    ``first ``=` `arr[``0``]``    ``second ``=` `arr[``1``]` `    ``# Update every element as required``    ``# except the last and the``    ``# second last element``    ``for` `i ``in` `range``(n ``-` `2``):``        ``arr[i] ``=` `arr[i ``+` `1``] ``+` `arr[i ``+` `2``]` `    ``# Update the last and the second``    ``# last element of the array``    ``arr[n ``-` `2``] ``=` `arr[n ``-` `1``] ``+` `first``    ``arr[n ``-` `1``] ``=` `first ``+` `second` `    ``# Print the updated array``    ``printArr(arr, n)` `# Driver code``if` `__name__ ``=``=` `'__main__'``:``    ``arr ``=` `[``3``, ``4``, ``2``, ``1``, ``6``]``    ``n ``=` `len``(arr)``    ``updateArr(arr, n)``    ` `# This code is contributed by``# Surendra_Gangwar`

## C#

 `// C# implementation of the approach``using` `System;` `class` `GFG``{``    ` `// Utility function to print the``// contents of an array``static` `void` `printArr(``int` `[]arr, ``int` `n)``{``    ``for` `(``int` `i = 0; i < n; i++)``        ``Console.Write(arr[i] + ``" "``);``}` `// Function to update every element of``// the array as the sum of next two elements``static` `void` `updateArr(``int` `[]arr, ``int` `n)``{` `    ``// Invalid array``    ``if` `(n < 3)``        ``return``;` `    ``// First and second elements of the array``    ``int` `first = arr[0];``    ``int` `second = arr[1];` `    ``// Update every element as required``    ``// except the last and the``    ``// second last element``    ``for` `(``int` `i = 0; i < n - 2; i++)``        ``arr[i] = arr[i + 1] + arr[i + 2];` `    ``// Update the last and the second``    ``// last element of the array``    ``arr[n - 2] = arr[n - 1] + first;``    ``arr[n - 1] = first + second;` `    ``// Print the updated array``    ``printArr(arr, n);``}` `// Driver code``public` `static` `void` `Main()``{``    ``int` `[]arr = { 3, 4, 2, 1, 6 };``    ``int` `n = arr.Length;``    ``updateArr(arr, n);``}``}` `// This code is contributed``// by Akanksha Rai`

## PHP

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## Javascript

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Output:
`6 3 7 9 7`

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