Given an array of integers, replace every element with xor 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: 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.
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 interation
int curr = arr[i];
// Update current value using previos 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;
} |
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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 interation
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 Code 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 anuj_67.. |
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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 interation
curr = arr[i]
# Update current value using
# previos 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 = [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 |
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C#
// C# program to update every array // element with sum of previous and // next numbers in array using 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 interation
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 Code 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 Akanskha Rai(Abby_akku) |
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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
// interation
$curr = $arr[$i];
// Update current value using
// previos 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, $n);
// Print the modified array for ($i = 0; $i < $n; $i++)
echo $arr[$i] . " ";
// This code is contributed by ita_c ?> |
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Output:
1 6 6 2 3
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