Given a number N, the task is to find four distinct numbers A, B, C, and D such that all are distinct and non-negative satisfies the condition: (((A|B)& C)^ D) = N, where ‘ | ‘ is Bitwise ‘or’, ‘&’ is Bitwise ‘and’, ‘^’ is Bitwise ‘xor’.
Examples:
Input: N = 5
Output: A = 15, B = 0, C = 16, D = 5Input: N = 16
Output: A = 0, B = 4, C = 3, D = 16A = 0 , B = 63 , C = 64 , D = 16
Naive Approach: One approach to solving the given problem is to use brute force to generate all possible combinations of four distinct non-negative integers A, B, C, and D, and check whether they satisfy the given condition (((A|B)& C)^ D) = N or not. However, this approach will not be efficient for larger values of N, and hence we need a better algorithmic approach.
Efficient Approach: This can be solved with the following idea:
- We can do this by bit manipulation concept. Let’s assume C is a number of power of two. We can make (A|B) & C equal to 0. For that A|B must be equal to (C-1). n & (n-1) = 0, when n is the power of two.
- For that, we can make A or B equal to C – 1 and another equal to 0. By doing this we get (A|B) & C = 0. Now we know a number of two numbers is equal to 0 when the number is the same, as A^A = 0; 0^A = A. So D must be equal to N.
- If N == 0 then a special case arises, then we have to find A as a power of two and B = A – 1. By that, we can find an integer X-1 where X is the power of two. for that, we can make C = X and D = 0. And we can make the answer.
Below is the implementation of the above approach :
C++
// C++ Implementation of the above approach#include <bits/stdc++.h>using namespace std;int bitcount(int n) { return (int)log2(n) + 1; }void FindSquad(int N){ // Initialized the squad // variable a, b, c, d int a, b, c, d; // Find the bit of the given number int power = bitcount(N) + 1; c = (1 << power); // Taking a = 0 a = 0; // B = C-1 by that we made // (A|B) & C == 0 b = c - 1; // Now make the D = N d = N; // For N == 0 taking the special // case which is already initialized // at explanation if (N == 0) { a = 4; b = 3; c = 8; d = 0; } // Giving the output cout << "A = " << a << " " << ", B = " << b << " " << ", C = " << c << " " << ", D = " << d << endl; return;}// Drivers codesigned main(){ int n = 5; // Calling the function findsquad FindSquad(n); return 0;} |
Java
// Java Implementation of the above approachimport java.util.*;class Main { static int bitcount(int n) { return (int)(Math.log(n) / Math.log(2)) + 1; } static void FindSquad(int N) { // Initialized the squad // variable a, b, c, d int a, b, c, d; // Find the bit of the given number int power = bitcount(N) + 1; c = (1 << power); // Taking a = 0 a = 0; // B = C-1 by that we made // (A|B) & C == 0 b = c - 1; // Now make the D = N d = N; // For N == 0 taking the special // case which is already initialized // at explanation if (N == 0) { a = 4; b = 3; c = 8; d = 0; } // Giving the output System.out.println("A = " + a + ", B = " + b + ", C = " + c + ", D = " + d); } // Drivers code public static void main(String[] args) { int n = 5; // Calling the function findsquad FindSquad(n); }}// This code is contributed by Prajwal Kandekar |
Python3
# Python Implementation of the above approachimport mathdef bitcount(n): return (int)(math.log2(n)) + 1def FindSquad(N): # Initialized the squad # variable a, b, c, d a, b, c, d = 0, 0, 0, 0 # Find the bit of the given number power = bitcount(N) + 1 c = (1 << power) # Taking a = 0 a = 0 # B = C-1 by that we made # (A|B) & C == 0 b = c - 1 # Now make the D = N d = N # For N == 0 taking the special # case which is already initialized # at explanation if (N == 0): a = 4 b = 3 c = 8 d = 0 # Giving the output print("A = ", a, ", B = ", b, ", C = ", c, ", D = ", d)# Driver codeif __name__ == "__main__": n = 5 # Calling the function findsquad FindSquad(n)# This code is contributed by Tapesh(tapeshdua420) |
C#
using System;public class Program{ static int bitcount(int n) { return (int)Math.Log(n, 2) + 1; } static void FindSquad(int N) { // Initialized the squad // variable a, b, c, d int a, b, c, d; // Find the bit of the given number int power = bitcount(N) + 1; c = (1 << power); // Taking a = 0 a = 0; // B = C-1 by that we made // (A|B) & C == 0 b = c - 1; // Now make the D = N d = N; // For N == 0 taking the special // case which is already initialized // at explanation if (N == 0) { a = 4; b = 3; c = 8; d = 0; } // Giving the output Console.WriteLine("A = {0}, B = {1}, C = {2}, D = {3}", a, b, c, d); } public static void Main(string[] args) { int n = 5; // Calling the function FindSquad FindSquad(n); }} |
Javascript
// Javascript Implementation of the above approachfunction bitCount(n) { return Math.floor(Math.log2(n)) + 1;}function findSquad(N) { let a, b, c, d; // Find the bit of the given number const power = bitCount(N) + 1; c = 1 << power; // Taking a = 0 a = 0; // B = C-1 by that we made (A|B) & C == 0 b = c - 1; // Now make the D = N d = N; // For N == 0 taking the special case if (N === 0) { a = 4; b = 3; c = 8; d = 0; } // Output the values console.log(`A = ${a}, B = ${b}, C = ${c}, D = ${d}`);}// Driver codeconst n = 5;findSquad(n); |
Output
A = 0, B = 15, C = 16, D = 5
Time Complexity: O(1)
Auxiliary Space: O(1)
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