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# Rearrange given Array such that each element raised to its index is odd

Given an array arr of length N, the task is to rearrange the elements of given array such that for each element, its bitwise XOR with its index is an odd value. If no rearrangement is possible return -1.

Example:

Input: arr[] = {1 2 4 3 5}
Output: 1 2 3 4 5
Explanation: In the above array:
for 1st element: value is 1 and index is 0 -> so 1 ^ 0 = 1, which is odd
for 2nd element: value is 2 and index is 1 -> so 2 ^ 1 = 3, which is odd
for 3rd element: value is 4 and index is 2 -> so 4 ^ 2 = 6, which is even -> rearranging will happen
for 4th element: value is 3 and index is 3 -> so 3 ^ 3 = 0, which is even -> rearranging will happen
for 5th element: value is 5 and index is 4 -> so 5 ^ 4 = 3, which is odd

So if we swap the positions of 4 and 3 as {1, 2, 3, 4, 5},
the XOR of 3^2 will become 1, and XOR of 4^3 will become 7, which are both odd.

Hence {1, 2, 3, 4, 5} is one of the possible rearrangements.

Input: arr[] = {1 2 7 3 5}
Output: -1

Approach: The idea to solve this problem is based on the properties of bitwise XOR operator, that:

• XOR of two odd elements is always even,
• XOR of two even elements is always even, and
• XOR of an odd and an even element is always odd.

So to rearrange the array as required, we will store all the even elements at odd indices, and odd elements at even indices.

Follow the below steps to understand how:

• First count how many odd and even index array have, which will be always n/2 and n-n/2 respectively
• Then Count how many odd and even elements array have
• Store the even and odd elements of the array separately
• Check if rearrangement is possible or not, i.e. the count of even elements is equal to odd indices and vice versa or not.
• If not possible, return -1.
• If the rearrangement is possible, Insert all odd elements at even indices, and even elements at odd indices.
• Return the rearranged array at the end.

Below is the implementation of the approach:

## C++

 `// C++ program to Rearrange the array``// Such that A[i]^ i is odd``#include ``using` `namespace` `std;` `// Function to rearrange given array``vector<``int``> rearrange(``int` `arr[], ``int` `n)``{``    ``vector<``int``> ans;``    ``int` `i;` `    ``// Count how many odd``    ``// and even index array have``    ``int` `oddIndex = n / 2,``  ``evenIndex = n - oddIndex;` `    ``// Count how many odd``    ``// and even elements array have``    ``int` `oddElement = 0, evenElement = 0;` `    ``// Store the even and odd elements``    ``// of the array separately``    ``vector<``int``> odd, even;` `    ``for` `(i = 0; i < n; i++)``        ``if` `(arr[i] % 2) {``            ``oddElement++;``            ``odd.push_back(arr[i]);``        ``}``        ``else` `{``            ``evenElement++;``            ``even.push_back(arr[i]);``        ``}` `    ``// To make XOR of each element``    ``// with its index as odd,``    ``// we have to place each even element``    ``// at an odd index and vice versa` `    ``// Therefore check if rearrangement``    ``// is possible or not``    ``if` `(oddElement != evenIndex``        ``|| oddIndex != evenElement) {``        ``ans.push_back(-1);``    ``}` `    ``// If the rearrangement is possible``    ``else` `{` `        ``// Insert odd elements at even indices``        ``// and even elements at odd indices``        ``int` `j = 0, k = 0;``        ``for` `(``int` `i = 0; i < n; i++)``            ``if` `(i % 2)``                ``ans.push_back(even[j++]);``            ``else``                ``ans.push_back(odd[k++]);``    ``}` `    ``// return the rearranged array``    ``return` `ans;``}` `// Driver Code``int` `main()``{``    ``int` `arr[] = { 1, 2, 4, 3, 5 };``    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr[0]);` `    ``vector<``int``> res = rearrange(arr, n);` `    ``for` `(``auto` `i : res)``        ``cout << i << ``" "``;``    ``return` `0;``}`

## Java

 `// Java program to Rearrange the array``// Such that A[i]^ i is odd``import` `java.io.*;``import` `java.util.ArrayList; ` `class` `GFG {` `  ``// Function to rearrange given array``  ``static` `void`  `rearrange(``int` `arr[], ``int` `n)``  ``{``    ``ArrayList ans = ``new` `ArrayList<>(); ` `    ``// Count how many odd``    ``// and even index array have``    ``int` `oddIndex = n / ``2``, evenIndex = n - oddIndex;` `    ``// Count how many odd``    ``// and even elements array have``    ``int` `oddElement = ``0``, evenElement = ``0``;` `    ``// Store the even and odd elements``    ``// of the array separately``    ``ArrayList odd = ``new` `ArrayList<>(); ``    ``ArrayList even = ``new` `ArrayList<>(); ` `    ``for` `(``int` `i = ``0``; i < n; i++)``      ``if` `(arr[i] % ``2` `== ``1``) {``        ``oddElement++;``        ``odd.add(arr[i]);``      ``}``    ``else` `{``      ``evenElement++;``      ``even.add(arr[i]);``    ``}` `    ``// To make XOR of each element``    ``// with its index as odd,``    ``// we have to place each even element``    ``// at an odd index and vice versa` `    ``// Therefore check if rearrangement``    ``// is possible or not``    ``if` `(oddElement != evenIndex``        ``|| oddIndex != evenElement) {``      ``ans.add(-``1``);``    ``}` `    ``// If the rearrangement is possible``    ``else` `{` `      ``// Insert odd elements at even indices``      ``// and even elements at odd indices``      ``int` `j = ``0``, k = ``0``;``      ``for` `(``int` `i = ``0``; i < n; i++)``        ``if` `(i % ``2` `== ``1``)``          ``ans.add(even.get(j++));``      ``else``        ``ans.add(odd.get(k++));``    ``}` `    ``// print the rearranged array``    ``for``(``int` `i = ``0``; i < ans.size(); i++){``      ``System.out.print(ans.get(i) + ``" "``);``    ``}``  ``}` `  ``// Driver Code``  ``public` `static` `void` `main (String[] args) {``    ``int``[] arr = { ``1``, ``2``, ``4``, ``3``, ``5` `};``    ``int` `n = arr.length;` `    ``rearrange(arr, n);``  ``}``}` `// This code is contributed by hrithikgarg03188.`

## Python3

 `# python3 program to Rearrange the array``# Such that A[i]^ i is odd` `# Function to rearrange given array``def` `rearrange(arr, n):` `    ``ans ``=` `[]``    ``i ``=` `0` `    ``# Count how many odd``    ``# and even index array have``    ``oddIndex ``=` `n ``/``/` `2``    ``evenIndex ``=` `n ``-` `oddIndex` `    ``# Count how many odd``    ``# and even elements array have``    ``oddElement, evenElement ``=` `0``, ``0` `    ``# Store the even and odd elements``    ``# of the array separately``    ``odd, even ``=` `[], []` `    ``for` `i ``in` `range``(``0``, n):``        ``if` `(arr[i] ``%` `2``):``            ``oddElement ``+``=` `1``            ``odd.append(arr[i])` `        ``else``:``            ``evenElement ``+``=` `1``            ``even.append(arr[i])` `        ``# To make XOR of each element``        ``# with its index as odd,``        ``# we have to place each even element``        ``# at an odd index and vice versa` `        ``# Therefore check if rearrangement``        ``# is possible or not``    ``if` `(oddElement !``=` `evenIndex``            ``or` `oddIndex !``=` `evenElement):``        ``ans.append(``-``1``)` `        ``# If the rearrangement is possible``    ``else``:` `                ``# Insert odd elements at even indices``                ``# and even elements at odd indices``        ``j, k ``=` `0``, ``0``        ``for` `i ``in` `range``(``0``, n):``            ``if` `(i ``%` `2``):``                ``ans.append(even[j])``                ``j ``+``=` `1` `            ``else``:``                ``ans.append(odd[k])``                ``k ``+``=` `1` `        ``# return the rearranged array``    ``return` `ans` `# Driver Code``if` `__name__ ``=``=` `"__main__"``:` `    ``arr ``=` `[``1``, ``2``, ``4``, ``3``, ``5``]``    ``n ``=` `len``(arr)` `    ``res ``=` `rearrange(arr, n)` `    ``for` `i ``in` `res:``        ``print``(i, end``=``" "``)` `    ``# This code is contributed by rakeshsahni`

## C#

 `// C# program to Rearrange the array``// Such that A[i]^ i is odd``using` `System;``using` `System.Collections;` `class` `GFG {` `  ``// Function to rearrange given array``  ``static` `ArrayList rearrange(``int``[] arr, ``int` `n)``  ``{``    ``ArrayList ans = ``new` `ArrayList();` `    ``// Count how many odd``    ``// and even index array have``    ``int` `oddIndex = n / 2, evenIndex = n - oddIndex;` `    ``// Count how many odd``    ``// and even elements array have``    ``int` `oddElement = 0, evenElement = 0;` `    ``// Store the even and odd elements``    ``// of the array separately``    ``ArrayList odd = ``new` `ArrayList();``    ``ArrayList even = ``new` `ArrayList();` `    ``for` `(``int` `i = 0; i < n; i++)``      ``if` `(arr[i] % 2 == 1) {``        ``oddElement++;``        ``odd.Add(arr[i]);``      ``}``    ``else` `{``      ``evenElement++;``      ``even.Add(arr[i]);``    ``}` `    ``// To make XOR of each element``    ``// with its index as odd,``    ``// we have to place each even element``    ``// at an odd index and vice versa` `    ``// Therefore check if rearrangement``    ``// is possible or not``    ``if` `(oddElement != evenIndex``        ``|| oddIndex != evenElement) {``      ``ans.Add(-1);``    ``}` `    ``// If the rearrangement is possible``    ``else` `{` `      ``// Insert odd elements at even indices``      ``// and even elements at odd indices``      ``int` `j = 0, k = 0;``      ``for` `(``int` `i = 0; i < n; i++)``        ``if` `(i % 2 == 1)``          ``ans.Add(even[j++]);``      ``else``        ``ans.Add(odd[k++]);``    ``}` `    ``// return the rearranged array``    ``return` `ans;``  ``}` `  ``// Driver Code``  ``public` `static` `void` `Main()``  ``{``    ``int``[] arr = { 1, 2, 4, 3, 5 };``    ``int` `n = arr.Length;` `    ``ArrayList res = rearrange(arr, n);` `    ``for` `(``int` `i = 0; i < res.Count; i++) {``      ``Console.Write(res[i] + ``" "``);``    ``}``  ``}``}` `// This code is contributed by Samim Hossain Mondal.`

## Javascript

 ``

Output

```1 2 3 4 5

```

Time complexity: O(N).
Auxiliary Space: O(N).

### Approach: Recursive Backtracking

We can solve this problem using recursive backtracking. The basic idea is to generate all possible permutations of the given array, and check if the condition is satisfied for each permutation. We can use a recursive function to generate all permutations.

Steps:

1. Define a recursive function permute(arr, l, r) to generate all permutations of the given array. The function takes the array, the left index and the right index as input.
2. If l==r, it means we have generated a permutation. Check if the condition is satisfied for this permutation. If yes, print the permutation and return True. Otherwise, return False.
3. Otherwise, for each index i from l to r, swap the elements at l and i, and recursively call the function permute(arr, l+1, r). After the recursive call, swap the elements back to restore the original array.

## C++

 `#include ``using` `namespace` `std;` `bool` `permute(vector<``int``>& arr, ``int` `l, ``int` `r) {``    ``if` `(l == r) {``        ``// check if condition is satisfied``        ``bool` `flag = ``true``;``        ``for` `(``int` `i = 0; i < arr.size(); i++) {``            ``if` `((arr[i] ^ i) % 2 == 0) {``                ``flag = ``false``;``                ``break``;``            ``}``        ``}``        ``if` `(flag) {``            ``for` `(``int` `i = 0; i < arr.size(); i++)``                ``cout << arr[i] << ``" "``;``            ``cout << endl;``            ``return` `true``;``        ``} ``else` `{``            ``return` `false``;``        ``}``    ``} ``else` `{``        ``for` `(``int` `i = l; i <= r; i++) {``            ``// swap elements at index l and i``            ``swap(arr[l], arr[i]);``            ``// recursive call``            ``if` `(permute(arr, l + 1, r))``                ``return` `true``;``            ``// restore original array``            ``swap(arr[l], arr[i]);``        ``}``        ``return` `false``;``    ``}``}` `int` `main() {``    ``vector<``int``> arr = {1, 2, 4, 3, 5};``    ``if` `(!permute(arr, 0, arr.size() - 1))``        ``cout << ``"-1"` `<< endl;``    ``return` `0;``}`

## Java

 `import` `java.util.*;` `public` `class` `GFG {``    ``public` `static` `boolean` `permute(ArrayList arr, ``int` `l, ``int` `r) {``        ``if` `(l == r) {``          ``// check if condition is satisfied``            ``boolean` `flag = ``true``;``            ``for` `(``int` `i = ``0``; i < arr.size(); i++) {``                ``if` `((arr.get(i) ^ i) % ``2` `== ``0``) {``                    ``flag = ``false``;``                    ``break``;``                ``}``            ``}``            ``if` `(flag) {``                ``for` `(``int` `i = ``0``; i < arr.size(); i++)``                    ``System.out.print(arr.get(i) + ``" "``);``                ``System.out.println();``                ``return` `true``;``            ``} ``else` `{``                ``return` `false``;``            ``}``        ``} ``else` `{``            ``for` `(``int` `i = l; i <= r; i++) {``               ``// swap elements at index l and i``                ``Collections.swap(arr, l, i);``              ``// recursive call``                ``if` `(permute(arr, l + ``1``, r))``                    ``return` `true``;``              ``// restore original array``                ``Collections.swap(arr, l, i);``            ``}``            ``return` `false``;``        ``}``    ``}` `    ``public` `static` `void` `main(String[] args) {``        ``ArrayList arr = ``new` `ArrayList<>(Arrays.asList(``1``, ``2``, ``4``, ``3``, ``5``));``        ``if` `(!permute(arr, ``0``, arr.size() - ``1``))``            ``System.out.println(``"-1"``);``    ``}``}`

## Python3

 `def` `permute(arr, l, r):``    ``if` `l ``=``=` `r:``        ``# check if condition is satisfied``        ``flag ``=` `True``        ``for` `i ``in` `range``(``len``(arr)):``            ``if` `(arr[i] ^ i) ``%` `2` `=``=` `0``:``                ``flag ``=` `False``                ``break``        ``if` `flag:``            ``print``(arr)``            ``return` `True``        ``else``:``            ``return` `False``    ``else``:``        ``for` `i ``in` `range``(l, r``+``1``):``            ``# swap elements at index l and i``            ``arr[l], arr[i] ``=` `arr[i], arr[l]``            ``# recursive call``            ``if` `permute(arr, l``+``1``, r):``                ``return` `True``            ``# restore original array``            ``arr[l], arr[i] ``=` `arr[i], arr[l]``        ``return` `False` `arr ``=` `[``1``, ``2``, ``4``, ``3``, ``5``]``if` `not` `permute(arr, ``0``, ``len``(arr)``-``1``):``    ``print``(``"-1"``)`

## C#

 `using` `System;``using` `System.Collections.Generic;` `public` `class` `GFG``{``    ``public` `static` `bool` `Permute(List<``int``> arr, ``int` `l, ``int` `r)``    ``{``        ``if` `(l == r)``        ``{``            ``// check if condition is satisfied``            ``bool` `flag = ``true``;``            ``for` `(``int` `i = 0; i < arr.Count; i++)``            ``{``                ``if` `((arr[i] ^ i) % 2 == 0)``                ``{``                    ``flag = ``false``;``                    ``break``;``                ``}``            ``}``            ``if` `(flag)``            ``{``                ``foreach` `(``int` `num ``in` `arr)``                    ``Console.Write(num + ``" "``);``                ``Console.WriteLine();``                ``return` `true``;``            ``}``            ``else``            ``{``                ``return` `false``;``            ``}``        ``}``        ``else``        ``{``            ``for` `(``int` `i = l; i <= r; i++)``            ``{``                ``// swap elements at index l and i``                ``int` `temp = arr[l];``                ``arr[l] = arr[i];``                ``arr[i] = temp;``                ``// recursive call``                ``if` `(Permute(arr, l + 1, r))``                    ``return` `true``;``                ``// restore original array``                ``temp = arr[l];``                ``arr[l] = arr[i];``                ``arr[i] = temp;``            ``}``            ``return` `false``;``        ``}``    ``}` `    ``public` `static` `void` `Main(``string``[] args)``    ``{``        ``List<``int``> arr = ``new` `List<``int``>(``new` `int``[] { 1, 2, 4, 3, 5 });``        ``if` `(!Permute(arr, 0, arr.Count - 1))``            ``Console.WriteLine(``"-1"``);``    ``}``}`

## Javascript

 `function` `permute(arr, l, r) {``    ``if` `(l == r) {``        ``// Check if condition is satisfied``        ``let flag = ``true``;``        ``for` `(let i = 0; i < arr.length; i++) {``            ``if` `((arr[i] ^ i) % 2 == 0) {``                ``flag = ``false``;``                ``break``;``            ``}``        ``}``        ``if` `(flag) {``            ``console.log(arr);``            ``return` `true``;``        ``} ``else` `{``            ``return` `false``;``        ``}``    ``} ``else` `{``        ``for` `(let i = l; i <= r; i++) {``            ``// Swap elements at index l and i``            ``[arr[l], arr[i]] = [arr[i], arr[l]];``            ``// Recursive call``            ``if` `(permute(arr, l + 1, r)) {``                ``return` `true``;``            ``}``            ``// Restore original array``            ``[arr[l], arr[i]] = [arr[i], arr[l]];``        ``}``        ``return` `false``;``    ``}``}` `let arr = [1, 2, 4, 3, 5];``if` `(!permute(arr, 0, arr.length - 1)) {``    ``console.log(``"-1"``);``}`

Output

```[1, 2, 3, 4, 5]

```

The time complexity of this approach is O(N!) since we are generating all possible permutations of the array.

The auxiliary space used is O(N) since we are using a single array to store the permutations.