Replace every array element by sum of previous and next
Last Updated :
14 Sep, 2022
Given an array of integers, update every element with sum of previous and next elements with following exceptions.
- First element is replaced by sum of first and second.
- 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
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 that is as follows:
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 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 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# 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 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
?>
<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>
Time Complexity: O(N)
Auxiliary Space: O(1) because it is using constant space for variables
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