Construct a sorted Array such that setbit in bitwise XOR of any pair is even
Last Updated :
07 Nov, 2022
Given an integer N(1 ≤ N ≤ 500), Construct an integer arr[] of length N, such that it should follow all the given conditions below:
- Each element of arr[] should be distinct.
- Binary representation of (arr[i]^arr[j]) should contain even number of set bits for all ( 1≤i≤N ) and ( i≠j ).
- All the elements in arr[] should be in sorted order.
- Ai > 0 for all (1 ≤ i ≤ N).
Examples:
Input: N = 2
Output: {10, 15}
Explanation: It can be verified that all elements of arr[]
in each output are distinct and in sorted order and
all the possible pairs of arr[] fulfill the condition mentioned in problem statement..
Input: N = 3
Output: {3, 5, 12}
Approach: To solve the problem follow the below observation:
It can be observe from outputs that arr[] contains only elements having even parity of set bits in their binary representation. If we try to get some such type of elements under range 1 to 10 using a brute-force code we will get a mathematical series as { 3, 5, 6, 9}.
This series is called "Evil Number" series and related to Number theory. This series follows all the given conditions of the problem. Therefore, This problem can be solve by printing first N terms of Evil Number series(Excluding 0, As Arr[i] should be greater than zero).
Follow the steps to solve the problem:
- Print the first N terms of Evil number Series or First N integers greater than zero having even parity of set bits.
Below is the implementation for the above approach:
C++
// c++ implementation
#include <bits/stdc++.h>
using namespace std;
string toBinary(int n)
{
string r;
while(n!=0) {r=(n%2==0 ?"0":"1")+r; n/=2;}
return r;
}
// Function which takes binary
// representation of a number
// as String argument and returns
// total number of set bits
int count1(string str)
{
// Counter variable to store
// number of set bits in
// binary representation
int counter = 0;
// Loop for traversing on
// Binary String
for (int i = 0; i < str.size(); i++)
{
// Condition when character
// '1' found in string
if (str[i] == '1')
{
// Incrementing counter
counter++;
}
}
// Returning count of set bits
return counter;
}
int main() {
// Input value of N
int N = 10;
// Counter variable
int counter = 1;
// Loop for finding first N terms
// of Evil Numbers Series
for (int i = 1; i <= N; i++) {
// While Loop which executes
// till counter is not a member
// of Evil Number series
while ((count1(toBinary(counter)))
% 2
!= 0) {
// Incrementing counter
counter++;
}
// Printing current value
// of counter
cout<<counter<<" ";
// Incrementing counter
counter++;
} //for loop end
return 0;
}
// this code is contributed by ksam24000
// Java code to implement the approach
// Brute force solution to find first N
// terms such that all the terms have even
// parity of set bits in their binary
// representation
import java.io.*;
import java.lang.*;
import java.util.*;
class GFG {
// Driver code
public static void main(String[] args)
throws java.lang.Exception
{
// Input value of N
int N = 10;
// Counter variable
int counter = 1;
// Loop for finding first N terms
// of Evil Numbers Series
for (int i = 1; i <= N; i++) {
// While Loop which executes
// till counter is not a member
// of Evil Number series
while ((count1(Integer.toBinaryString(counter)))
% 2
!= 0) {
// Incrementing counter
counter++;
}
// Printing current value
// of counter
System.out.print(counter + " ");
// Incrementing counter
counter++;
}
}
// Function which takes binary
// representation of a number
// as String argument and returns
// total number of set bits
static int count1(String str)
{
// Counter variable to store
// number of set bits in
// binary representation
int counter = 0;
// Loop for traversing on
// Binary String
for (int i = 0; i < str.length(); i++) {
// Condition when character
// '1' found in string
if (str.charAt(i) == '1')
// Incrementing counter
counter++;
}
// Returning count of set bits
return counter;
}
}
class GFG :
# Driver code
@staticmethod
def main( args) :
# Input value of N
N = 10
# Counter variable
counter = 1
# Loop for finding first N terms
# of Evil Numbers Series
i = 1
while (i <= N) :
# While Loop which executes
# till counter is not a member
# of Evil Number series
while ((GFG.count1(str(bin(counter)))) % 2 != 0) :
# Incrementing counter
counter += 1
# Printing current value
# of counter
print(str(counter) + " ", end ="")
# Incrementing counter
counter += 1
i += 1
# Function which takes binary
# representation of a number
# as String argument and returns
# total number of set bits
@staticmethod
def count1( str) :
# Counter variable to store
# number of set bits in
# binary representation
counter = 0
# Loop for traversing on
# Binary String
i = 0
while (i < len(str)) :
# Condition when character
# '1' found in string
if (str[i] == '1') :
# Incrementing counter
counter += 1
i += 1
# Returning count of set bits
return counter
if __name__=="__main__":
GFG.main([])
# This code is contributed by aadityaburujwale.
// Include namespace system
using System;
public class GFG
{
// Driver code
public static void Main(String[] args)
{
// Input value of N
var N = 10;
// Counter variable
var counter = 1;
// Loop for finding first N terms
// of Evil Numbers Series
for (int i = 1; i <= N; i++)
{
// While Loop which executes
// till counter is not a member
// of Evil Number series
while ((GFG.count1(Convert.ToString(counter, 2))) % 2 != 0)
{
// Incrementing counter
counter++;
}
// Printing current value
// of counter
Console.Write(counter.ToString() + " ");
// Incrementing counter
counter++;
}
}
// Function which takes binary
// representation of a number
// as String argument and returns
// total number of set bits
public static int count1(String str)
{
// Counter variable to store
// number of set bits in
// binary representation
var counter = 0;
// Loop for traversing on
// Binary String
for (int i = 0; i < str.Length; i++)
{
// Condition when character
// '1' found in string
if (str[i] == '1')
{
// Incrementing counter
counter++;
}
}
// Returning count of set bits
return counter;
}
}
// This code is contributed by aadityaburujwale.
// js implementation
function toBinary(n)
{
let r = "";
while(n != 0) {r = (n % 2 == 0 ? "0":"1") + r; n = Math.floor(n/2);}
return r;
}
// Function which takes binary
// representation of a number
// as String argument and returns
// total number of set bits
function count1(str)
{
// Counter variable to store
// number of set bits in
// binary representation
let counter = 0;
// Loop for traversing on
// Binary String
for (let i = 0; i < str.length; i++)
{
// Condition when character
// '1' found in string
if (str[i] == '1')
{
// Incrementing counter
counter++;
}
}
// Returning count of set bits
return counter;
}
// driver code
// Input value of N
let N = 10;
// Counter variable
let counter = 1;
// Loop for finding first N terms
// of Evil Numbers Series
for (let i = 1; i <= N; i++) {
// While Loop which executes
// till counter is not a member
// of Evil Number series
while ((count1(toBinary(counter)))
% 2
!= 0) {
// Incrementing counter
counter++;
}
// Printing current value
// of counter
console.log(counter);
// Incrementing counter
counter++;
}
// This code is contributed by ksam24000
Output3 5 6 9 10 12 15 17 18 20
Time Complexity: O(N2)
Auxiliary Space: O(1)
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