Given a number N, the task is to find the remainder when N is divided by 4 using Bitwise AND operator.
Examples:
Input: N = 98 Output: 2 Explanation: 98 % 4 = 2. Hence the output is 2. Input: 200 Output: 0 Explanation: 200 % 4 = 0. Hence output is 0.
Naive approach:
For solving the above-mentioned problem we can use a naive method by using the Modulo (%) operator to find the remainder. But, the Modulo operator is computationally expensive and the method is inefficient.
Efficient Approach:
If we carefully observe the binary representation of N and its remainder with 4, we observe that remainder is simply the rightmost two bits in N. To get the rightmost two bits in number N, we perform bitwise AND (&) with 3 because 3 in binary is 0011. To understand the approach better let us have a look at the image below:
Below is the implementation of the above approach:
C
// C implementation to find N // modulo 4 using Bitwise AND operator #include <stdio.h> // Function to find the remainder int findRemainder(int n) { // Bitwise AND with 3 int x = n & 3; // return x return x; } // Driver code int main() { int N = 43; int ans = findRemainder(N); printf("%d", ans); return 0; } |
C++
// C++ implementation to find N // modulo 4 using Bitwise AND operator #include <bits/stdc++.h> using namespace std; // Function to find the remainder int findRemainder(int n) { // Bitwise AND with 3 int x = n & 3; // Return x return x; } // Driver code int main() { int N = 43; int ans = findRemainder(N); cout << ans << endl; return 0; } |
Java
// Java implementation to find N // modulo 4 using Bitwise AND operator class Main { // Driver code public static void main(String[] args) { int N = 43; int ans = findRemainder(N); System.out.println(ans); } // Function to find the remainder public static int findRemainder(int n) { // Bitwise AND with 3 int x = n & 3; // return x return x; } } |
Python 3
# Python 3 implementation to find N # modulo 4 using Bitwise AND operator # Function to find the remainder def findRemainder(n): # Bitwise AND with 3 x = n & 3 # Return x return x # Driver code if __name__ == '__main__': N = 43 ans = findRemainder(N) print(ans) # This code is contributed by Surendra_Gangwar |
C#
// C# implementation to find N // modulo 4 using Bitwise AND operator using System; class GFG { // Driver code public static void Main() { int N = 43; int ans = findRemainder(N); Console.Write(ans); } // Function to find the remainder public static int findRemainder(int n) { // Bitwise AND with 3 int x = n & 3; // return x return x; } } # This code is contributed by chitranayal |
3
Time Complexity: O(1)
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