Kth smallest positive integer Y such that its sum with X is same as its bitwise OR with X
Given two positive integers X and K, the task is to find the Kth smallest positive integer (Y) such that the sum of X and Y is equal to Bitwise OR of X and Y, i.e. (X + Y = X | Y)
Examples:
Input: X = 5, K = 1
Output: 2
Explanation:
Consider the smallest number as 2. The sum of 2 and 5 is 2 + 5 = 7 and the Bitwise OR of 2 and 5 is 7.Input: X = 5, K = 5
Output: 18
Approach: The given problem can be solved by below observation:
- Using the properties of bitwise operator on two numbers X and Y:
(X + Y) = (X & Y) + (X | Y)
- Therefore, for the value of X + Y to be same as X | Y, the value of X & Y must be 0.
- Also, we know that for X & Y to be 0, the position of set bits in X must be unset in Y, and the positions of unset bits of X can be either 0 or 1 in Y.
- Hence, to find Kth smallest positive number meeting above conditions, we can use combinations for the bit positions of Y that are unset in X.
To implement the above, iterate the binary representation of X from right to left, and at every iteration, consider the following cases:
- If the current bit of X is 1, the corresponding bit in Y will be 0, to get (X & Y) as 0.
- If the current bit of X is 0 and rightmost bit of K is 1, the corresponding bit of Y will be 1. This means that two combinations at current bit in Y has been used – first 0 and then 1. Then divide K by 2
- If the current bit of X is 0 and rightmost bit of K is 0, the corresponding bit of Y will be 0. This means that only one combination at current bit in Y has been used – 0. Then divide K by 2
- If K becomes 0, break the loop, stating that the Kth number has been found.
Finally, print the decimal conversion of Y as the required answer.
Below is the implementation of the above approach:
C++
// C++ implementation for the above approach#include <bits/stdc++.h>using namespace std;// Function to calculate K-th smallest // solution(Y) of equation X+Y = X|Ylong long KthSolution(long long X, long long K){ // Initialize the variable // to store the answer long long ans = 0; for (int i = 0; i < 64; i++) { // The i-th bit of X is off if (!(X & (1LL << i))) { // The i-bit of K is on if (K & 1) { ans |= (1LL << i); } // Divide K by 2 K >>= 1; // If K becomes 0 then break if (!K) { break; } } } return ans;}// Driver Codeint main(){ long long X = 10, K = 5; cout << KthSolution(X, K); return 0;} |
Java
/*package whatever //do not write package name here */import java.io.*;class GFG { // Function to calculate K-th smallest // solution(Y) of equation X+Y = X|Y static long KthSolution(long X, long K) { // Initialize the variable // to store the answer long ans = 0; for (int i = 0; i < 64; i++) { // The i-th bit of X is off if ((X & (1 << i)) == 0) { // The i-bit of K is on if ((K & 1) > 0) { ans |= (1 << i); } // Divide K by 2 K >>= 1; // If K becomes 0 then break if (K == 0) { break; } } } return ans; } // Driver Code public static void main(String[] args) { long X = 10; long K = 5; System.out.println(KthSolution(X, K)); }}// This code is contributed by maddler. |
Python3
# Python Program to implement# the above approach# Function to calculate K-th smallest# solution(Y) of equation X+Y = X|Ydef KthSolution(X, K): # Initialize the variable # to store the answer ans = 0 for i in range(64): # The i-th bit of X is off if not (X & (1 << i)): # The i-bit of K is on if (K & 1): ans |= (1 << i) # Divide K by 2 K >>= 1 # If K becomes 0 then break if not K: break return ans# Driver CodeX = 10K = 5print(KthSolution(X, K))# This code is contributed by Saurabh Jaiswal |
C#
/*package whatever //do not write package name here */using System;public class GFG { // Function to calculate K-th smallest // solution(Y) of equation X+Y = X|Y static long KthSolution(long X, long K) { // Initialize the variable // to store the answer int ans = 0; for (int i = 0; i < 64; i++) { // The i-th bit of X is off if ((X & (1 << i)) == 0) { // The i-bit of K is on if ((K & 1) > 0) { ans |= (1 << i); } // Divide K by 2 K >>= 1; // If K becomes 0 then break if (K == 0) { break; } } } return ans; } // Driver Code public static void Main(String[] args) { long X = 10; long K = 5; Console.WriteLine(KthSolution(X, K)); }}// This code is contributed by Princi Singh |
Javascript
<script> // JavaScript Program to implement // the above approach// Function to calculate K-th smallest // solution(Y) of equation X+Y = X|Yfunction KthSolution(X,K){ // Initialize the variable // to store the answer let ans = 0; for (let i = 0; i < 64; i++) { // The i-th bit of X is off if (!(X & (1 << i))) { // The i-bit of K is on if (K & 1) { ans |= (1 << i); } // Divide K by 2 K >>= 1; // If K becomes 0 then break if (!K) { break; } } } return ans;}// Driver Code let X = 10, K = 5; document.write(KthSolution(X, K)); // This code is contributed by Potta Lokesh </script> |
Output:
17
Time Complexity: O(log(X))
Auxiliary Space: O(1)
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