Construct a sorted Array such that setbit in bitwise XOR of any pair is even
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
// 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; } } |
Python3
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. |
C#
// 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. |
Javascript
// 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 |
3 5 6 9 10 12 15 17 18 20
Time Complexity: O(N2)
Auxiliary Space: O(1)
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