Given a positive integer N, the task is to print all the distinct even and odd values of prefix Bitwise XORs of first N natural numbers.
Examples:
Input: N = 6
Output:
Even: 0 4
Odd: 1 3 7
Explanation:
The prefix Bitwise XOR of the first 6 natural number si {1, 3, 0, 4, 1, 7}.
Even prefix Bitwise XORs are 0, 4.
Odd prefix Bitwise XORs are 1, 3, 7.Input: N = 9
Output:
Even: 0 4 8
Odd: 1 3 7
Approach: The given problem can be solved based on the below observations:
- If the value of N modulo 4 is 0, then the value of Bitwise XOR of the first N natural numbers is N.
- If the value of N modulo 4 is 1, then the value of Bitwise XOR of the first N natural numbers is 1.
- If the value of N modulo 4 is 2, then the value of Bitwise XOR of the first N natural numbers is (N + 1).
- If the value of N modulo 4 is 3, then the value of Bitwise XOR of the first N natural numbers is 0.
Therefore, from the above principle, it can be said that Bitwise XOR as even numbers will always come as 0 or multiples of 4, and Bitwise XOR as odd numbers will always come as 1 or 1 less than multiples of 4.
Below is the implementation of the above approach.
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Print all distinct even & odd// prefix Bitwise XORs from 1 to Nvoid evenOddBitwiseXOR(int N){ cout << "Even: " << 0 << " "; // Print the even number for (int i = 4; i <= N; i = i + 4) { cout << i << " "; } cout << "\n"; cout << "Odd: " << 1 << " "; // Print the odd number for (int i = 4; i <= N; i = i + 4) { cout << i - 1 << " "; } if (N % 4 == 2) cout << N + 1; else if (N % 4 == 3) cout << N;}// Driver Codeint main(){ int N = 6; evenOddBitwiseXOR(N); return 0;} |
Java
// Java approach for the above approachclass GFG{// Print all distinct even & odd// prefix Bitwise XORs from 1 to Nstatic void evenOddBitwiseXOR(int N){ System.out.print("Even: " + 0 + " "); // Print the even number for(int i = 4; i <= N; i = i + 4) { System.out.print(i + " "); } System.out.print("\n"); System.out.print("Odd: " + 1 + " "); // Print the odd number for(int i = 4; i <= N; i = i + 4) { System.out.print(i - 1 + " "); } if (N % 4 == 2) System.out.print(N + 1); else if (N % 4 == 3) System.out.print(N);}// Driver Codepublic static void main(String[] args){ int N = 6; evenOddBitwiseXOR(N);}}// This code is contributed by abhinavjain194 |
Python3
# Python3 program for the above approach# Print all distinct even & odd# prefix Bitwise XORs from 1 to Ndef evenOddBitwiseXOR(N): print("Even: ", 0, end = " ") # Print the even number for i in range(4, N + 1, 4): print(i, end = " ") print() print("Odd: ", 1, end = " ") # Print the odd number for i in range(4, N + 1, 4): print(i - 1, end = " ") if (N % 4 == 2): print(N + 1) elif (N % 4 == 3): print(N)# Driver CodeN = 6evenOddBitwiseXOR(N)# This code is contributed by sanjoy_62 |
C#
// C# program for the above approachusing System;class GFG{ // Print all distinct even & odd// prefix Bitwise XORs from 1 to Nstatic void evenOddBitwiseXOR(int N){ Console.Write("Even: " + 0 + " "); // Print the even number for(int i = 4; i <= N; i = i + 4) { Console.Write(i + " "); } Console.Write("\n"); Console.Write("Odd: " + 1 + " "); // Print the odd number for(int i = 4; i <= N; i = i + 4) { Console.Write(i - 1 + " "); } if (N % 4 == 2) Console.Write(N + 1); else if (N % 4 == 3) Console.Write(N);}// Driver codepublic static void Main(){ int N = 6; evenOddBitwiseXOR(N);}}// This code is contributed by splevel62 |
Javascript
<script>// Javascript implementation of the above approach// Print all distinct even & odd// prefix Bitwise XORs from 1 to Nfunction evenOddBitwiseXOR(N){ document.write("Even: " + 0 + " "); // Print the even number for(let i = 4; i <= N; i = i + 4) { document.write(i + " "); } document.write("<br/>"); document.write("Odd: " + 1 + " "); // Print the odd number for(let i = 4; i <= N; i = i + 4) { document.write(i - 1 + " "); } if (N % 4 == 2) document.write(N + 1); else if (N % 4 == 3) document.write(N);} // Driver Code let N = 6; evenOddBitwiseXOR(N); </script> |
Output:
Even: 0 4 Odd: 1 3 7
Time Complexity: O(N)
Auxiliary Space: O(1)
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