Minimum value of N such that xor from 1 to N is equal to K

Given a value K which is the XOR of all the values from 1 to N, the task is to find the minimum value of N such that XOR from 1 to N is equal to K.
Examples: 

Input: K = 7
Output: 6
1 ^ 2 ^ 3 ^ 4 ^ 5 ^ 6 = 7

Input: K = 10
Output: Not Possible

Approach: This problem is similar to the Calculate XOR from 1 to n. Below are the conditions to be checked: 

1. If k = 0, then N = 3.
2. If k = 1, then N = 1.
3. If k % 4 = 0, then N = k.
4. If k % 4 = 3, then N = k-1.

Below is the implementation of above approach: 

C++

// C++ implementation of above approach
#include <bits/stdc++.h>
using namespace std;

// Function to find the value of N
int findN(int k)
{

    // variable to store the result
    int ans;

    // handling case for '0'
    if (k == 0)
        ans = 3;

    // handling case for '1'
    if (k == 1)
        ans = 1;

    // when number is completely divided by
    // 4 then minimum 'x' will be 'k'
    else if (k % 4 == 0)
        ans = k;

    // when number divided by 4 gives 3 as
    // remainder then minimum 'x' will be 'k-1'
    else if (k % 4 == 3)
        ans = k - 1;

    // else it is not possible to get
    // k for any value of x
    else
        ans = -1;

    return ans;
}

// Driver code
int main()
{
    // let the given number be 7
    int k = 7;

    int res = findN(k);
    if (res == -1)
        cout << "Not possible";
    else
        cout << res;

    return 0;
}

Java

// Java implementation of
// above approach
import java.io.*;

class GFG
{

// Function to find the
// value of N
static int findN(int k)
{

    // variable to store
    // the result
    int ans;

    // handling case for '0'
    if (k == 0)
        ans = 3;

    // handling case for '1'
    if (k == 1)
        ans = 1;

    // when number is completely
    // divided by 4 then minimum
    // 'x' will be 'k'
    else if (k % 4 == 0)
        ans = k;

    // when number divided by 4
    // gives 3 as remainder then
    // minimum 'x' will be 'k-1'
    else if (k % 4 == 3)
        ans = k - 1;

    // else it is not possible to
    // get k for any value of x
    else
        ans = -1;

    return ans;
}

// Driver code
public static void main (String[] args)
{
    // let the given number be 7
    int k = 7;

    int res = findN(k);
    if (res == -1)
        System.out.println("Not possible");
    else
        System.out.println(res);
}
}

// This code is contributed
// by inder_verma

Python3

# Python3 implementation of
# above approach

# Function to find the value of N
def findN(k) :

    # handling case for '0'
    if (k == 0) :
        ans = 3

    # handling case for '1'
    if (k == 1) :
        ans = 1

    # when number is completely
    # divided by 4 then minimum
    # 'x' will be 'k'
    elif (k % 4 == 0) :
        ans = k

    # when number divided by 4
    # gives 3 as remainder then
    # minimum 'x' will be 'k-1'
    elif (k % 4 == 3) :
        ans = k - 1

    # else it is not possible to
    # get k for any value of x
    else:
        ans = -1

    return ans

# Driver code

# let the given number be 7
k = 7

res = findN(k)
if (res == -1):
    print("Not possible")
else:
    print(res)

# This code is contributed
# by Smitha

C#

// C# implementation of
// above approach
using System;

class GFG
{

// Function to find the
// value of N
static int findN(int k)
{

    // variable to store
    // the result
    int ans;

    // handling case for '0'
    if (k == 0)
        ans = 3;

    // handling case for '1'
    if (k == 1)
        ans = 1;

    // when number is completely
    // divided by 4 then minimum
    // 'x' will be 'k'
    else if (k % 4 == 0)
        ans = k;

    // when number divided by 4
    // gives 3 as remainder then
    // minimum 'x' will be 'k-1'
    else if (k % 4 == 3)
        ans = k - 1;

    // else it is not possible to
    // get k for any value of x
    else
        ans = -1;

    return ans;
}

// Driver code
public static void Main ()
{
    // let the given number be 7
    int k = 7;

    int res = findN(k);
    if (res == -1)
        Console.WriteLine("Not possible");
    else
        Console.WriteLine(res);
}
}

// This code is contributed
// by inder_verma

PHP

<?php
// PHP implementation of above approach

// Function to find the value of N
function findN($k)
{

    // variable to store the result
    $ans;

    // handling case for '0'
    if ($k == 0)
        $ans = 3;

    // handling case for '1'
    if ($k == 1)
        $ans = 1;

    // when number is completely divided
    // by 4 then minimum 'x' will be 'k'
    else if ($k % 4 == 0)
        $ans = $k;

    // when number divided by 4 gives 3 as
    // remainder then minimum 'x' will be 'k-1'
    else if ($k % 4 == 3)
        $ans = $k - 1;

    // else it is not possible to
    // get k for any value of x
    else
        $ans = -1;

    return $ans;
}

// Driver code

// let the given number be 7
$k = 7;

$res = findN($k);
if ($res == -1)
    echo "Not possible";
else
    echo $res;

// This code is contributed by Mahadev
?>

Javascript

<script>
// javascript implementation of
// above approach


// Function to find the
// value of N
function findN(k)
{

    // variable to store
    // the result
    var ans;

    // handling case for '0'
    if (k == 0)
        ans = 3;

    // handling case for '1'
    if (k == 1)
        ans = 1;

    // when number is completely
    // divided by 4 then minimum
    // 'x' will be 'k'
    else if (k % 4 == 0)
        ans = k;

    // when number divided by 4
    // gives 3 as remainder then
    // minimum 'x' will be 'k-1'
    else if (k % 4 == 3)
        ans = k - 1;

    // else it is not possible to
    // get k for any value of x
    else
        ans = -1;

    return ans;
}

// Driver code
// let the given number be 7
var k = 7;

var res = findN(k);
if (res == -1)
    document.write("Not possible");
else
    document.write(res);


// This code contributed by shikhasingrajput
</script>

Output:

6

Time Complexity: O(1)

Auxiliary Space: O(1)

How does this work?
When we do XOR of numbers, we get 0 as XOR value just before a multiple of 4. This keeps repeating before every multiple of 4. 

Number Binary-Repr  XOR-from-1-to-n
1         1           [0001]
2        10           [0011]
3        11           [0000]  <----- We get a 0
4       100           [0100]  <----- Equals to n
5       101           [0001]
6       110           [0111]
7       111           [0000]  <----- We get 0
8      1000           [1000]  <----- Equals to n
9      1001           [0001]
10     1010           [1011]
11     1011           [0000] <------ We get 0
12     1100           [1100] <------ Equals to n