Minimum number of elements to be removed to make XOR maximum

Given a number N where, 1\leq N\leq 10^{18}. The task is to find the minimum number of elements to be deleted in between 1 to N such that the XOR obtained from the remaining elements is maximum.

Examples:

Input: N = 5
Output: 2

Input: 1000000000000000
Output: 1


Approach: Considering the following cases:

Case 1: When n=1 or n=2, then answer is 0. No need to remove any element.
Case 2: Now, we have to find a number which is power of 2 and greater than or equal to n.
Let’s call this number be a.
So, if n=a or n=a-1 then we will just remove a-1. Hence the answer is 1.
else if n=a-2, then answer is 0. No need to remove any element.
Case 3: Otherwise, if n is even, then answer is 1.
else if n is odd, then answer is 2.

Below is the implementation of above approach:

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ implementation to find minimum number of
// elements to remove to get maximum XOR value
#include <bits/stdc++.h>
using namespace std;
  
unsigned int nextPowerOf2(unsigned int n)
{
    unsigned count = 0;
  
    // First n in the below condition
    // is for the case where n is 0
    if (n && !(n & (n - 1)))
        return n;
  
    while (n != 0) {
        n >>= 1;
        count += 1;
    }
  
    return 1 << count;
}
  
// Function to find minimum number of
// elements to be removed.
int removeElement(unsigned int n)
{
  
    if (n == 1 || n == 2)
        return 0;
  
    unsigned int a = nextPowerOf2(n);
  
    if (n == a || n == a - 1)
        return 1;
  
    else if (n == a - 2)
        return 0;
  
    else if (n % 2 == 0)
        return 1;
  
    else
        return 2;
}
  
// Driver code
int main()
{
    unsigned int n = 5;
  
    // print minimum number of elements
    // to be removed
    cout << removeElement(n);
  
    return 0;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

//Java implementation to find minimum number of
//elements to remove to get maximum XOR value
public class GFG {
  
    static int nextPowerOf2(int n)
    {
     int count = 0;
  
     // First n in the below condition
     // is for the case where n is 0
     if (n!=0 && (n& (n - 1))==0)
         return n;
  
     while (n != 0) {
         n >>= 1;
         count += 1;
     }
  
     return 1 << count;
    }
  
    //Function to find minimum number of
    //elements to be removed.
    static int removeElement(int n)
    {
  
     if (n == 1 || n == 2)
         return 0;
  
     int a = nextPowerOf2(n);
  
     if (n == a || n == a - 1)
         return 1;
  
     else if (n == a - 2)
         return 0;
  
     else if (n % 2 == 0)
         return 1;
  
     else
         return 2;
    }
  
    //Driver code
    public static void main(String[] args) {
          
         int n = 5;
  
         // print minimum number of elements
         // to be removed
         System.out.println(removeElement(n));
    }
}

chevron_right


Python 3

filter_none

edit
close

play_arrow

link
brightness_4
code

# Python 3 to find minimum number 
# of elements to remove to get 
# maximum XOR value
  
def nextPowerOf2(n) :
    count = 0
  
    # First n in the below condition 
    # is for the case where n is 0 
    if (n and not(n and (n - 1))) :
        return n
  
    while n != 0 :
        n >>= 1
        count += 1
  
    return 1 << count
  
# Function to find minimum number 
# of elements to be removed. 
def removeElement(n) :
  
    if n == 1 or n == 2 :
        return 0
  
    a = nextPowerOf2(n)
      
    if n == a or n == a - 1 :
        return 1
  
    elif n == a - 2 :
        return 0
  
    elif n % 2 == 0 :
        return 1
  
    else :
        return 2
      
# Driver Code
if __name__ == "__main__" :
  
    n = 5
  
    # print minimum number of 
    # elements to be removed 
    print(removeElement(n))
  
# This code is contributed 
# by ANKITRAI1

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

//C# implementation to find minimum number of
//elements to remove to get maximum XOR value
  
using System;
public class GFG {
   
    static int nextPowerOf2(int n)
    {
     int count = 0;
   
     // First n in the below condition
     // is for the case where n is 0
     if (n!=0 && (n& (n - 1))==0)
         return n;
   
     while (n != 0) {
         n >>= 1;
         count += 1;
     }
   
     return 1 << count;
    }
   
    //Function to find minimum number of
    //elements to be removed.
    static int removeElement(int n)
    {
   
     if (n == 1 || n == 2)
         return 0;
   
     int a = nextPowerOf2(n);
   
     if (n == a || n == a - 1)
         return 1;
   
     else if (n == a - 2)
         return 0;
   
     else if (n % 2 == 0)
         return 1;
   
     else
         return 2;
    }
   
    //Driver code
    public static void Main() {
           
         int n = 5;
   
         // print minimum number of elements
         // to be removed
         Console.Write(removeElement(n));
    }
}

chevron_right


PHP

filter_none

edit
close

play_arrow

link
brightness_4
code

<?php
// PHP implementation to find 
// minimum number of elements 
// to remove to get maximum 
// XOR value
  
function nextPowerOf2($n)
{
    $count = 0;
  
    // First n in the below condition
    // is for the case where n is 0
    if ($n && !($n & ($n - 1)))
        return $n;
  
    while ($n != 0) 
    {
        $n >>= 1;
        $count += 1;
    }
  
    return 1 << $count;
}
  
// Function to find minimum number 
// of elements to be removed.
function removeElement($n)
{
  
    if ($n == 1 || $n == 2)
        return 0;
  
    $a = nextPowerOf2($n);
  
    if ($n == $a || $n == $a - 1)
        return 1;
  
    else if ($n == $a - 2)
        return 0;
  
    else if ($n % 2 == 0)
        return 1;
  
    else
        return 2;
}
  
// Driver code
$n = 5;
  
// print minimum number of 
// elements to be removed
echo removeElement($n);
  
// This code is contributed by mits
?>

chevron_right


Output:

2

Time complexity: O(logn)



My Personal Notes arrow_drop_up

Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.



Improved By : AnkitRai01, Ita_c, Mithun Kumar