# Replace every array element by sum of previous and next

• Difficulty Level : Easy
• Last Updated : 12 May, 2021

Given an array of integers, update every element with sum of previous and next elements with following exceptions.
a) First element is replaced by sum of first and second.
b) Last element is replaced by sum of last and second last.
Examples:

```Input : arr[] = { 2, 3, 4, 5, 6}
Output : 5 6 8 10 11
Explanation: We get the following array as {2+3, 2+4, 3+5, 4+6, 5+6}

Input : arr[] = { 3, 2, 1}
Output : 5 4 3```

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A Simple Solution is to create an auxiliary array, copy contents of given array to auxiliary array. Finally traverse the auxiliary array and update given array using copied values. Time complexity of this solution is O(n), but it requires O(n) extra space.
An efficient solution can solve the problem in O(n) time and O(1) space. The idea is to keep track of previous element in loop. Add the previous element using the extra variable and the next element to get each element.
Below is the implementation of this idea.

## C++

 `// C++ program to update every array element with``// sum of previous and next numbers in array``#include ``using` `namespace` `std;` `void` `ReplaceElements(``int` `arr[], ``int` `n)``{``    ``// Nothing to do when array size is 1``    ``if` `(n <= 1)``        ``return``;` `    ``// store current value of arr and update it``    ``int` `prev = arr;``    ``arr = arr + arr;` `    ``// Update rest of the array elements``    ``for` `(``int` `i = 1; i < n - 1; i++) {` `        ``// Store current value of next iteration``        ``int` `curr = arr[i];` `        ``// Update current value using previews value``        ``arr[i] = prev + arr[i + 1];` `        ``// Update previous value``        ``prev = curr;``    ``}` `    ``// Update last array element separately``    ``arr[n - 1] = prev + arr[n - 1];``}` `// Driver program``int` `main()``{``    ``int` `arr[] = { 2, 3, 4, 5, 6 };``    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr);` `    ``ReplaceElements(arr, n);` `    ``// Print the modified array``    ``for` `(``int` `i = 0; i < n; i++)``        ``cout << arr[i] << ``" "``;``    ``return` `0;``}`

## Java

 `// Java program to update every array element with``// sum of previous and next numbers in array` `import` `java.io.*;` `class` `GFG {``        ``static` `void` `ReplaceElements(``int` `arr[], ``int` `n) {``        ``// Nothing to do when array size is 1``        ``if` `(n <= ``1``) {``            ``return``;``        ``}` `        ``// store current value of arr and update it``        ``int` `prev = arr[``0``];``        ``arr[``0``] = arr[``0``] + arr[``1``];` `        ``// Update rest of the array elements``        ``for` `(``int` `i = ``1``; i < n - ``1``; i++) {` `            ``// Store current value of next iteration``            ``int` `curr = arr[i];` `            ``// Update current value using previews value``            ``arr[i] = prev + arr[i + ``1``];` `            ``// Update previous value``            ``prev = curr;``        ``}` `        ``// Update last array element separately``        ``arr[n - ``1``] = prev + arr[n - ``1``];``    ``}` `// Driver program``    ` `    ``public` `static` `void` `main (String[] args) {` `        ``int` `arr[] = {``2``, ``3``, ``4``, ``5``, ``6``};``        ``int` `n = arr.length;``        ``ReplaceElements(arr, n);``        ``// Print the modified array``        ``for` `(``int` `i = ``0``; i < n; i++) {``            ``System.out.print(arr[i] + ``" "``);``        ``}``    ``}``}``// This code is contributed by akt_mit`

## Python 3

 `# Python 3 program to update every array``# element with sum of previous and next``# numbers in array` `def` `ReplaceElements(arr, n):` `    ``# Nothing to do when array size is 1``    ``if` `(n <``=` `1``):``        ``return` `    ``# store current value of arr``    ``# and update it``    ``prev ``=` `arr[``0``]``    ``arr[``0``] ``=` `arr[``0``] ``+` `arr[``1``]` `    ``# Update rest of the array elements``    ``for` `i ``in` `range``(``1``, n ``-` `1``):` `        ``# Store current value of``        ``# next iteration``        ``curr ``=` `arr[i]` `        ``# Update current value using``        ``# previews value``        ``arr[i] ``=` `prev ``+` `arr[i ``+` `1``]` `        ``# Update previous value``        ``prev ``=` `curr` `    ``# Update last array element separately``    ``arr[n ``-` `1``] ``=` `prev ``+` `arr[n ``-` `1``]` `# Driver Code``if` `__name__ ``=``=` `"__main__"``:` `    ``arr ``=` `[ ``2``, ``3``, ``4``, ``5``, ``6` `]``    ``n ``=` `len``(arr)` `    ``ReplaceElements(arr, n)` `    ``# Print the modified array``    ``for` `i ``in` `range``(n):``        ``print` `(arr[i], end ``=` `" "``)` `# This code is contributed``# by ChitraNayal`

## C#

 `// C# program to update every array element with``// sum of previous and next numbers in array``using` `System;``public` `class` `GFG {` `    ``static` `void` `ReplaceElements(``int` `[]arr, ``int` `n) {``        ``// Nothing to do when array size is 1``        ``if` `(n <= 1) {``            ``return``;``        ``}` `        ``// store current value of arr and update it``        ``int` `prev = arr;``        ``arr = arr + arr;` `        ``// Update rest of the array elements``        ``for` `(``int` `i = 1; i < n - 1; i++) {` `            ``// Store current value of next iteration``            ``int` `curr = arr[i];` `            ``// Update current value using previews value``            ``arr[i] = prev + arr[i + 1];` `            ``// Update previous value``            ``prev = curr;``        ``}` `        ``// Update last array element separately``        ``arr[n - 1] = prev + arr[n - 1];``    ``}` `// Driver program``    ``public` `static` `void` `Main() {``        ``int` `[]arr = {2, 3, 4, 5, 6};``        ``int` `n = arr.Length;` `        ``ReplaceElements(arr, n);` `        ``// Print the modified array``        ``for` `(``int` `i = 0; i < n; i++) {``            ``Console.Write(arr[i] + ``" "``);``        ``}``    ``}``}` `// This code is contributed by Rajput-JI`

## PHP

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## Javascript

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Output:
`5 6 8 10 11`

Time Complexity – O(N)

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