Find a pair of numbers with set bit count as at most that of N and whose Bitwise XOR is N Last Updated : 12 Oct, 2022 Comments Improve Suggest changes Like Article Like Report Given a positive integer N, the task is to find the pair of integers (X, Y) such that the Bitwise XOR of X and Y is N and X * Y is maximum where the count of bits in X and Y is less than or equal to N. Examples: Input: N = 10Output: 13 7Explanation: The case where X = 13 and Y = 7 is the most optimal choice as 13 xor 7 = 10 and 13 * 7 = 91 which is maximum possible. Input: N = 45Output: 50 31 Approach: The given problem can be solved using the following observations: If the ith bit of N is 0, then the ith bit of both X and Y must be either 0 or 1. In order to maximize the product, set such bits as 1.If the ith bit of N is 1, then one of the ith bits of X or Y must be 1 and the other must be 0. Since N must have a constant number of set bits, therefore the sum of X and Y must be constant.If the sum of two numbers is constant, their product is maximum when the difference between the two numbers is minimized. According to the above observations, initialize two integers X and Y as 0. In order to minimize the difference between X and Y, X must be equal to the MSBN and Y must be equal to N - MSBN where MSB represents the Most Significant Bit. Also, for all the 0 bits in N, set the respective bits in both X and Y as 1. Below is the implementation of the above approach: C++ // C++ program for the above approach #include <bits/stdc++.h> using namespace std; // Function to find the pair (X, Y) such // that X xor Y = N and the count of set // bits in X and Y is less than count of // set bit in N void maximizeProduct(int N) { // Stores MSB (Most Significant Bit) int MSB = (int)log2(N); // Stores the value of X int X = 1 << MSB; // Stores the value of Y int Y = N - (1 << MSB); // Traversing over all bits of N for (int i = 0; i < MSB; i++) { // If ith bit of N is 0 if (!(N & (1 << i))) { // Set ith bit of X to 1 X += 1 << i; // Set ith bit of Y to 1 Y += 1 << i; } } // Print Answer cout << X << " " << Y; } // Driver Code int main() { int N = 45; maximizeProduct(N); return 0; } Java // Java program for the above approach import java.io.*; class GFG { // Function to find the pair (X, Y) such // that X xor Y = N and the count of set // bits in X and Y is less than count of // set bit in N static void maximizeProduct(int N) { // Stores MSB (Most Significant Bit) int MSB = (int)(Math.log(N) / Math.log(2)); // Stores the value of X int X = 1 << MSB; // Stores the value of Y int Y = N - (1 << MSB); // Traversing over all bits of N for (int i = 0; i < MSB; i++) { // If ith bit of N is 0 if ((N & (1 << i))==0) { // Set ith bit of X to 1 X += 1 << i; // Set ith bit of Y to 1 Y += 1 << i; } } // Print Answer System.out.println(X+" "+Y); } // Driver Code public static void main(String[] args) { int N = 45; maximizeProduct(N); } } // This code is contributed by dwivediyash Python3 # python 3 program for the above approach import math # Function to find the pair (X, Y) such # that X xor Y = N and the count of set # bits in X and Y is less than count of # set bit in N def maximizeProduct(N): # Stores MSB (Most Significant Bit) MSB = (int)(math.log2(N)) # Stores the value of X X = 1 << MSB # / Stores the value of Y Y = N - (1 << MSB) # Traversing over all bits of N for i in range(MSB): # If ith bit of N is 0 if (not (N & (1 << i))): # / Set ith bit of X to 1 X += 1 << i # Set ith bit of Y to 1 Y += 1 << i # Print Answer print(X, Y) # Driver Code if __name__ == "__main__": N = 45 maximizeProduct(N) # This code is contributed by ukasp. C# // C# program for the above approach using System; class GFG { // Function to find the pair (X, Y) such // that X xor Y = N and the count of set // bits in X and Y is less than count of // set bit in N static void maximizeProduct(int N) { // Stores MSB (Most Significant Bit) int MSB = (int)(Math.Log(N) / Math.Log(2)); // Stores the value of X int X = 1 << MSB; // Stores the value of Y int Y = N - (1 << MSB); // Traversing over all bits of N for (int i = 0; i < MSB; i++) { // If ith bit of N is 0 if ((N & (1 << i))==0) { // Set ith bit of X to 1 X += 1 << i; // Set ith bit of Y to 1 Y += 1 << i; } } // Print Answer Console.Write(X+" "+Y); } // Driver Code public static void Main() { int N = 45; maximizeProduct(N); } } // This code is contributed by code_hunt. JavaScript <script> // JavaScript Program to implement // the above approach // Function to find the pair (X, Y) such // that X xor Y = N and the count of set // bits in X and Y is less than count of // set bit in N function maximizeProduct(N) { // Stores MSB (Most Significant Bit) let MSB = Math.log2(N); // Stores the value of X let X = 1 << MSB; // Stores the value of Y let Y = N - (1 << MSB); // Traversing over all bits of N for (let i = 0; i < MSB; i++) { // If ith bit of N is 0 if (!(N & (1 << i))) { // Set ith bit of X to 1 X += 1 << i; // Set ith bit of Y to 1 Y += 1 << i; } } // Print Answer document.write(X + " " + Y); } // Driver Code let N = 45; maximizeProduct(N); // This code is contributed by Potta Lokesh </script> Output: 50 31 Time Complexity: O(log N)Auxiliary Space: O(1) Comment More infoAdvertise with us Next Article Find a pair of numbers with set bit count as at most that of N and whose Bitwise XOR is N vedant280 Follow Improve Article Tags : Bit Magic Greedy Mathematical DSA setBitCount binary-representation Bitwise-XOR +3 More Practice Tags : Bit MagicGreedyMathematical Similar Reads Count integers less than A so that Bitwise AND of B with XOR of A and the number is 0 Given two positive integers A and B, the task is to find the number of integers X such that (A⊕X ) & B = 0 and 0 ≤ X < A where ⊕ is bitwise XOR and & is bitwise AND operator. Examples: Input: A = 3, B = 5Output: 1
Explanation: Here A = 3, B = 5. So we need to find the number of X, which s 5 min read Count Number of Pairs where Bitwise AND and Bitwise XOR is Equal Given an integer array arr of size N, the task is to count the number of pairs whose BITWISE AND and BITWISE XOR are equal. Example: Input: N = 3, arr[] = {0,0,1}Output: 1Explanation: There is only one pair arr[0] and arr[1] as 0&0=0 and 0^0=0 Input: N = 4, arr[] = {1, 2, 4, 8}Output: 0Explanati 4 min read Total pairs in an array such that the bitwise AND, bitwise OR and bitwise XOR of LSB is 1 Given an array arr[] of size N. The task is to find the number of pairs (arr[i], arr[j]) as cntAND, cntOR, and cntXOR such that: cntAND: Count of pairs where bitwise AND of least significant bits is 1.cntOR: Count of pairs where bitwise OR of least significant bits is 1.cntXOR: Count of pairs where 7 min read Count of pairs having bit size at most X and Bitwise OR equal to X Given a number X, calculate number of possible pairs (a, b) such that bitwise or of a and b is equal to X and number of bits in both a and b is less than equal to number of bits in X. Examples: Input: X = 6 Output: 9 Explanation: The possible pairs of (a, b) are (4, 6), (6, 4), (6, 6), (6, 2), (4, 2 4 min read Count of pairs in Array such that bitwise AND of XOR of pair and X is 0 Given an array arr[] consisting of N positive integers and a positive integer X, the task is to find the number of pairs (i, j) such that i < j and (arr[i]^arr[j] )&X is 0. Examples: Input: arr[] = {1, 3, 4, 2}, X = 2 Output: 2Explanation:Following are the possible pairs from the given array: 11 min read Count of pairs with bitwise XOR value greater than its bitwise AND value | Set 2 Given an array arr that contains N positive Integers. Find the count of all possible pairs whose bit wise XOR value is greater than bit wise AND value Examples: Input : arr[]={ 12, 4 , 15}Output: 2Explanation: 12 ^ 4 = 8 , 12 & 4 = 4 . so 12 ^ 4 > 12 & 4 4 ^ 15 = 11, 4 & 15 = 4. so 4 6 min read Count of all possible values of X whose Bitwise XOR with N is greater than N Given an integer N count the number of values of X such that X⊕N > N, where ⊕ denotes bitwise XOR operation Examples: Input: N = 10Output: 5 Explanation: The five possible value satisfying the above condition are:1⊕10 = 11, 4⊕10 = 14, 5⊕10 = 15, 6⊕10 = 12, 7⊕10 = 13 Input: N = 8Output: 7Explanati 6 min read Maximize count of pairs whose Bitwise AND exceeds Bitwise XOR by replacing such pairs with their Bitwise AND Given an array arr[] consisting of N positive integers, replace pairs of array elements whose Bitwise AND exceeds Bitwise XOR values by their Bitwise AND value. Finally, count the maximum number of such pairs that can be generated from the array. Examples: Input: arr[] = {12, 9, 15, 7}Output: 2Expla 9 min read Count of pairs with bitwise XOR value greater than its bitwise AND value Given an array arr that contains N positive Integers. Find the count of all possible pairs whose bitwise XOR value is greater than bitwise AND value Examples: Input : arr[]={ 12, 4, 15}Output: 2Explanation: 12 ^ 4 = 8, 12 & 4 = 4. so 12 ^ 4 > 12 & 4 4 ^ 15 = 11, 4 & 15 = 4. so 4 ^ 15 4 min read Count pairs with bitwise XOR exceeding bitwise AND from a given array Given an array, arr[] of size N, the task is to count the number of pairs from the given array such that the bitwise AND(&) of each pair is less than its bitwise XOR(^). Examples: Input: arr[] = {1, 2, 3, 4, 5} Output: 8Explanation: Pairs that satisfy the given conditions are: (1 & 2) < ( 10 min read Like