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Minimum number of elements to be removed to make XOR maximum

  • Difficulty Level : Medium
  • Last Updated : 29 Apr, 2021

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 the above approach: 
 



C++




// 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;
}

Java




//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));
    }
}

Python 3




# 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

C#




//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));
    }
}

PHP




<?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
?>

Javascript




<script>
 
// Javascript implementation to
// find minimum number of
// elements to remove to get
// maximum XOR value
 
function nextPowerOf2(n)
{
    let 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;
 
    let 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
    let n = 5;
 
    // print minimum number of elements
    // to be removed
    document.write(removeElement(n));
 
</script>
Output: 
2

 

Time complexity: O(logn)
 

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