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 elements 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 <bits/stdc++.h>using namespace std;// Utility function to print the// contents of an arrayvoid 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 elementsvoid 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 codeint 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 approachclass 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 arraydef 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 elementsdef 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 codeif __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 approachusing System;class GFG{ // Utility function to print the// contents of an arraystatic 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 elementsstatic 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 codepublic 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
<?php// PHP implementation of the approach// Utility function to print the// contents of an arrayfunction printArr($arr, $n){ for ($i = 0; $i < $n; $i++) echo $arr[$i], " ";}// Function to update every element // of the array as the sum of next // two elementsfunction 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 = 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, $n);}// Driver code$arr = array (3, 4, 2, 1, 6 );$n = sizeof($arr);updateArr($arr, $n);// This code is contributed by ajit.?> |
Javascript
<script>// Javascript implementation of the approach// Utility function to print the// contents of an arrayfunction printArr( arr, n){ for (let i = 0; i < n; i++) document.write(arr[i] + " ");}// Function to update every element of// the array as the sum of next two elementsfunction updateArr( arr, n){ // Invalid array if (n < 3) return; // First and second elements of the array let first = arr[0]; let second = arr[1]; // Update every element as required // except the last and the // second last element for (let 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 let arr = [ 3, 4, 2, 1, 6 ]; let n = arr.length; updateArr(arr, n);// This code is contributed by jana_sayantan. </script> |
Output:
6 3 7 9 7
Time complexity: O(n) where n is length of given array
Auxiliary Space: O(1)
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