Given an integer N, the task is to find bitwise and (&) of all even numbers from 1 to N.
Examples:
Input: 2
Output: 2Input :10
Output : 0
Explanation: Bitwise and of 2, 4, 6, 8 and 10 are 0.
Naive approach: Initialize result as 2. Iterate the loop from 4 to n (for all even numbers) and update the result by finding bitwise and (&).
Below is the implementation of the approach:
C++
// C++ implementation of the above approach#include <iostream>using namespace std;// Function to return the bitwise &// of all the even numbers upto Nint bitwiseAndTillN(int n){ // Initialize result as 2 int result = 2; for (int i = 4; i <= n; i = i + 2) { result = result & i; } return result;}// Driver codeint main(){ int n = 2; cout << bitwiseAndTillN(n); return 0;} |
Java
// Java implementation of the above approach class GFG { // Function to return the bitwise & // of all the even numbers upto N static int bitwiseAndTillN(int n) { // Initialize result as 2 int result = 2; for (int i = 4; i <= n; i = i + 2) { result = result & i; } return result; } // Driver code public static void main (String[] args) { int n = 2; System.out.println(bitwiseAndTillN(n)); } }// This code is contributed by AnkitRai01 |
Python3
# Python3 implementation of the above approach # Function to return the bitwise & # of all the even numbers upto N def bitwiseAndTillN(n) : # Initialize result as 2 result = 2; for i in range(4, n + 1, 2) : result = result & i; return result; # Driver code if __name__ == "__main__" : n = 2; print(bitwiseAndTillN(n)); # This code is contributed by AnkitRai01 |
C#
// C# implementation of the above approach using System;class GFG { // Function to return the bitwise & // of all the even numbers upto N static int bitwiseAndTillN(int n) { // Initialize result as 2 int result = 2; for (int i = 4; i <= n; i = i + 2) { result = result & i; } return result; } // Driver code public static void Main() { int n = 2; Console.WriteLine(bitwiseAndTillN(n)); } }// This code is contributed by AnkitRai01 |
Javascript
<script>// Javascript implementation of the above approach// Function to return the bitwise &// of all the even numbers upto Nfunction bitwiseAndTillN(n) { // Initialize result as 2 let result = 2; for (let i = 4; i <= n; i = i + 2) { result = result & i; } return result;}// Driver codelet n = 2;document.write(bitwiseAndTillN(n));</script> |
Output:
2
Time Complexity: O(n), where n is the given integer.
Auxiliary Space: O(1), no extra space is required, so it is a constant.
Efficient approach: Efficient approach is to return 2 for N less than 4 and return 0 for all N>=4 because bitwise and of 2 and 4 is 0 and bitwise and of 0 with any number is 0.
Below is the implementation of the approach:
C++
// C++ implementation of the above approach#include <iostream>using namespace std;// Function to return the bitwise &// of all the numbers upto Nint bitwiseAndTillN(int n){ if (n < 4) return 2; else return 0;}int main(){ int n = 2; cout << bitwiseAndTillN(n); return 0;} |
Java
// Java implementation of the above approachclass GFG{ // Function to return the bitwise & // of all the numbers upto N static int bitwiseAndTillN(int n) { if (n < 4) return 2; else return 0; } // Driver code public static void main (String[] args) { int n = 2; System.out.println(bitwiseAndTillN(n)); }}// This code is contributed by AnkitRai01 |
Python3
# Python3 implementation of the above approach# Function to return the bitwise &# of all the numbers upto Ndef bitwiseAndTillN( n): if (n < 4): return 2 else: return 0# Driver coden = 2print(bitwiseAndTillN(n))# This code is contributed by ANKITKUMAR34 |
C#
// C# implementation of the above approachusing System;class GFG{ // Function to return the bitwise & // of all the numbers upto N static int bitwiseAndTillN(int n) { if (n < 4) return 2; else return 0; } // Driver code public static void Main() { int n = 2; Console.WriteLine(bitwiseAndTillN(n)); }}// This code is contributed by AnkitRai01 |
Javascript
<script>// JavaScript implementation of the above approach// Function to return the bitwise &// of all the numbers upto Nfunction bitwiseAndTillN(n){ if (n < 4) return 2; else return 0;}// driver code let n = 2; document.write (bitwiseAndTillN(n)); // this code is contributed by shivanisinghss2110 </script> |
Output:
2
Time complexity: O(1)
Auxiliary Space: O(1)
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