Kth smallest positive integer Y such that its sum with X is same as its bitwise OR with X
Last Updated :
14 Sep, 2021
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:
(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++
#include <bits/stdc++.h>
using namespace std;
long long KthSolution( long long X,
long long K)
{
long long ans = 0;
for ( int i = 0; i < 64; i++) {
if (!(X & (1LL << i))) {
if (K & 1) {
ans |= (1LL << i);
}
K >>= 1;
if (!K) {
break ;
}
}
}
return ans;
}
int main()
{
long long X = 10, K = 5;
cout << KthSolution(X, K);
return 0;
}
|
Java
import java.io.*;
class GFG
{
static long KthSolution( long X, long K)
{
long ans = 0 ;
for ( int i = 0 ; i < 64 ; i++) {
if ((X & ( 1 << i)) == 0 ) {
if ((K & 1 ) > 0 ) {
ans |= ( 1 << i);
}
K >>= 1 ;
if (K == 0 ) {
break ;
}
}
}
return ans;
}
public static void main(String[] args)
{
long X = 10 ;
long K = 5 ;
System.out.println(KthSolution(X, K));
}
}
|
Python3
def KthSolution(X, K):
ans = 0
for i in range ( 64 ):
if not (X & ( 1 << i)):
if (K & 1 ):
ans | = ( 1 << i)
K >> = 1
if not K:
break
return ans
X = 10
K = 5
print (KthSolution(X, K))
|
C#
using System;
public class GFG
{
static long KthSolution( long X, long K)
{
int ans = 0;
for ( int i = 0; i < 64; i++) {
if ((X & (1 << i)) == 0) {
if ((K & 1) > 0) {
ans |= (1 << i);
}
K >>= 1;
if (K == 0) {
break ;
}
}
}
return ans;
}
public static void Main(String[] args)
{
long X = 10;
long K = 5;
Console.WriteLine(KthSolution(X, K));
}
}
|
Javascript
<script>
function KthSolution(X,K)
{
let ans = 0;
for (let i = 0; i < 64; i++) {
if (!(X & (1 << i))) {
if (K & 1) {
ans |= (1 << i);
}
K >>= 1;
if (!K) {
break ;
}
}
}
return ans;
}
let X = 10, K = 5;
document.write(KthSolution(X, K));
</script>
|
Time Complexity: O(log(X))
Auxiliary Space: O(1)
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