Related Articles

# Minimum number of elements to be removed to make XOR maximum

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

Given a number where . The task is to find the minimum number of elements to be deleted in between to 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 or , 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 Let’s call this number be .
So, if or then we will just remove . Hence the answer is 1.
else if , then answer is 0. No need to remove any element.
Case 3: Otherwise, if is , then answer is 1
else if is , 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 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 codeint 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 valuepublic 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 Codeif __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

 >= 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$n = 5; // print minimum number of// elements to be removedecho removeElement($n); // This code is contributed by mits?>

## Javascript

 
Output:
2

Time complexity: O(logn)

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.

My Personal Notes arrow_drop_up