Given an array of integers, replace every element with xor of previous and next elements with following exceptions.
- First element is replaced by xor of first and second.
- Last element is replaced by xor of last and second last.
Examples:
Input: arr[] = { 2, 3, 4, 5, 6}
Output: 1 6 6 2 3
We get the following array as {2^3, 2^4, 3^5, 4^6, 5^6}Input: arr[] = { 1, 2, 1, 5}
Output: 3, 0, 7, 4
We get the following array as {1^2, 1^1, 2^5, 1^5}
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.
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){ int newarr[n];//created a new array to store the elements. for(int i=0;i<n;i++){ newarr[i]=arr[i];} arr[0]=arr[0]^arr[1];// changed the first element for(int i=1;i<n-1;i++){// iterated from second element to last second arr[i]=newarr[i-1]^newarr[i+1];}// element and changed every element according to question. arr[n-1]=newarr[n-1]^newarr[n-2];// changed the last element of the array.}// Driver programint 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;}//This code is contributed by Naveen Gujjar from Haryana |
Java
public class Main { public static void replaceElements(int[] arr, int n) { int[] newarr = new int[n]; // created a new array to store the elements. for (int i = 0; i < n; i++) { newarr[i] = arr[i]; } arr[0] = arr[0] ^ arr[1]; // changed the first element for (int i = 1; i < n - 1; i++) { // iterated from second element to last second arr[i] = newarr[i - 1] ^ newarr[i + 1]; // element and changed every element according to the question. } arr[n - 1] = newarr[n - 1] ^ newarr[n - 2]; // changed the last element of the array. } // 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] + " "); } }} |
Python3
# Function to update every array element with sum of previous and next numbers in arraydef replaceElements(arr, n): newarr = [0] * n # created a new array to store the elements for i in range(n): newarr[i] = arr[i] arr[0] = arr[0] ^ arr[1] # changed the first element for i in range(1, n-1): # iterated from second element to last second element and changed every element according to question arr[i] = newarr[i-1] ^ newarr[i+1] # changed the last element of the array arr[n-1] = newarr[n-1] ^ newarr[n-2]# Driver programarr = [2, 3, 4, 5, 6]n = len(arr)replaceElements(arr, n)# Print the modified arrayfor i in range(n): print(arr[i], end=" ") |
Javascript
// JS program to update every array element with// sum of previous and next numbers in arrayfunction ReplaceElements(arr, n) { let newarr = [...arr]; // created a new array to store the elements. arr[0] = arr[0] ^ arr[1]; // changed the first element for (let i = 1; i < n - 1; i++) { // iterated from second element to last second element arr[i] = newarr[i - 1] ^ newarr[i + 1]; // element and changed every element according to question. } arr[n - 1] = newarr[n - 1] ^ newarr[n - 2]; // changed the last element of the array.}// Driver programlet arr = [2, 3, 4, 5, 6];let n = arr.length;ReplaceElements(arr, n);// Print the modified arrayfor (let i = 0; i < n; i++) { console.log(arr[i] + " ");} |
C#
// C# program to update every array element with// sum of previous and next numbers in arrayusing System;class MainClass { static void ReplaceElements(int[] arr, int n) { int[] newarr = new int[n]; // created a new array to // store the elements. for (int i = 0; i < n; i++) { newarr[i] = arr[i]; } arr[0] = arr[0] ^ arr[1]; // changed the first element for (int i = 1; i < n - 1; i++) { // iterated from second element to last // second arr[i] = newarr[i - 1] ^ newarr[i + 1]; // element and changed // every element according // to question. } arr[n - 1] = newarr[n - 1] ^ newarr[n - 2]; // changed the last element // of the array. } 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] + " "); }} |
Output
1 6 6 2 3
Time Complexity: O(n)
Auxiliary Space: O(n)
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. Xor the previous element using the extra variable and the next element to get each element.
Below is the implementation of the above approach:
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 interaction int curr = arr[i]; // Update current value using previous 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 programint 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 arrayimport 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 interaction int curr = arr[i]; // Update current value using // previous 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 Codepublic 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 anuj_67.. |
Python3
# Python3 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 interaction curr = arr[i] # Update current value using # previous 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 Codearr = [2, 3, 4, 5, 6]n = len(arr)ReplaceElements(arr, n)for i in range(n): print(arr[i], end = " ")# This code is contributed# by Shrikant13 |
C#
// C# program to update every array // element with sum of previous and// next numbers in arrayusing System;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 interaction int curr = arr[i]; // Update current value using // previous 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 Codepublic 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 Akanskha Rai(Abby_akku) |
PHP
<?php // PHP program to update every array // element with sum of previous and // next numbers in arrayfunction 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 // interaction $curr = $arr[$i]; // Update current value using // previous 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$arr = array( 2, 3, 4, 5, 6 );$n = sizeof($arr);ReplaceElements($arr, $n);// Print the modified arrayfor ($i = 0; $i < $n; $i++) echo $arr[$i] . " ";// This code is contributed by ita_c?> |
Javascript
<script>// Javascript program to update every array// element with sum of previous and next // numbers in arrayfunction 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 interaction let curr = arr[i]; // Update current value using previous 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 codelet arr = [ 2, 3, 4, 5, 6 ];let n = arr.length;ReplaceElements(arr, n);// Print the modified arrayfor(let i = 0; i < n; i++) document.write(arr[i] + " "); // This code is contributed by subhammahato348</script> |
Output
1 6 6 2 3
Time complexity: O(N), where N is the number of elements in the given array.
Auxiliary space: O(1)
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