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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 <iostream>
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[0] 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
int main()
{
    int arr[] = { 2, 3, 4, 5, 6 };
    int n = sizeof(arr) / sizeof(arr[0]);
 
    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[0] 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[0]
    # 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[0] 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() {
        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




<?php
// PHP program to update every array
// element with sum of previous and
// next numbers in array
 
function ReplaceElements($arr, $n)
{
    // Nothing to do when array
    // size is 1
    if ($n <= 1)
        return;
 
    // store current value of
    // arr[0] and update it
    $prev = $arr[0];
    $arr[0] = $arr[0] + $arr[1];
 
    // Update rest of the array elements
    for ($i = 1; $i < $n - 1; $i++)
    {
 
        // 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];
    return $arr;
}
 
// Driver Code
$arr = array(2, 3, 4, 5, 6);
$n = sizeof($arr);
 
$arr1 = ReplaceElements($arr, $n);
 
// Print the modified array
for ($i = 0; $i < $n; $i++)
    echo $arr1[$i] . " ";
 
// This code is contributed
// by Akanksha Rai
?>

Javascript




<script>
    // Javascript program to update every array element with
    // sum of previous and next numbers in array
     
    function ReplaceElements(arr, n) {
        // Nothing to do when array size is 1
        if (n <= 1) {
            return;
        }
   
        // store current value of arr[0] and update it
        let prev = arr[0];
        arr[0] = arr[0] + arr[1];
   
        // Update rest of the array elements
        for (let i = 1; i < n - 1; i++) {
   
            // Store current value of next iteration
            let 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];
    }
     
    let arr = [2, 3, 4, 5, 6];
    let n = arr.length;
 
    ReplaceElements(arr, n);
 
    // Print the modified array
    for (let i = 0; i < n; i++) {
      document.write(arr[i] + " ");
    }
 
</script>
Output: 
5 6 8 10 11

 

Time Complexity – O(N)
 




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