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Find smallest positive number Y such that Bitwise AND of X and Y is Zero

  • Last Updated : 09 Aug, 2021

Given an integer X. The task is to find the smallest positive number Y(> 0) such that X AND Y is zero.
Examples: 
 

Input : X = 3 
Output :
4 is the smallest positive number whose bitwise AND with 3 is zero 
Input : X = 10 
Output : 1  

Approach : 
There are 2 cases : 
 

  • If the binary representation of X contains all 1s, in that case, all the bits of Y should be 0 to make the result of AND operation is zero. Then X+1 is our answer which is the first positive integer. 
     
  • If the binary representation of X doesn’t contain all 1s, in that case, find the first position in X at which bit is 0. Then our answer will be power(2, position) 
     

Below is the implementation of the above approach : 

C++




// C++ program to find smallest number Y for
// a given value of X such that X AND Y is zero
#include <bits/stdc++.h>
#define mod 1000000007
using namespace std;
 
// Method to find smallest number Y for
// a given value of X such that X AND Y is zero
int findSmallestNonZeroY(int A_num)
{
 
    // Convert the number into its binary form
    string A_binary = bitset<8>(A_num).to_string();
    int B = 1;
    int length = A_binary.size();
    int no_ones = __builtin_popcount(A_num);
 
    // Case 1 : If all bits are ones,
    // then return the next number
    if (length == no_ones )
        return A_num + 1;
 
    // Case 2 : find the first 0-bit
    // index and return the Y
    for (int i=0;i<length;i++)
    {
            char ch = A_binary[length - i - 1];
 
            if (ch == '0')
            {
                B = pow(2.0, i);
                break;
            }
        }
    return B;
}
 
// Driver Code
int main()
{
    int X = findSmallestNonZeroY(10);
    cout << X;
}
 
// This code is contributed by mohit kumar 29

Java




// Java program to find smallest number Y for
// a given value of X such that X AND Y is zero
import java.lang.*;
 
public class Main {
     
    // Method to find smallest number Y for
    // a given value of X such that X AND Y is zero
    static long findSmallestNonZeroY(long A_num)
    {
        // Convert the number into its binary form
        String A_binary = Long.toBinaryString(A_num);
        long B = 1;
        int len = A_binary.length();
        int no_ones = Long.bitCount(A_num);
 
        // Case 1 : If all bits are ones,
        // then return the next number
        if (len == no_ones) {
            return A_num + 1;
        }
 
        // Case 2 : find the first 0-bit
        // index and return the Y
        for (int i = 0; i < len; i++) {
            char ch = A_binary.charAt(len - i - 1);
            if (ch == '0') {
                B = (long)Math.pow(2.0, (double)i);
                break;
            }
        }
        return B;
    }
     
    // Driver code
    public static void main(String[] args)
    {
        long X = findSmallestNonZeroY(10);
        System.out.println(X);
    }
}

Python3




# Python3 program to find smallest number Y for
# a given value of X such that X AND Y is zero
 
# Method to find smallest number Y for
# a given value of X such that X AND Y is zero
def findSmallestNonZeroY(A_num) :
     
    # Convert the number into its binary form
    A_binary = bin(A_num)
    B = 1
    length = len(A_binary);
    no_ones = (A_binary).count('1');
     
    # Case 1 : If all bits are ones,
    # then return the next number
    if length == no_ones :
        return A_num + 1;
         
    # Case 2 : find the first 0-bit
    # index and return the Y
    for i in range(length) :
            ch = A_binary[length - i - 1];
             
            if (ch == '0') :
                B = pow(2.0, i);
                break;
                 
    return B;
 
# Driver Code
if __name__ == "__main__" :
    X = findSmallestNonZeroY(10);
    print(X)
     
# This code is contributed by AnkitRai01

C#




// C# program to find smallest number Y for
// a given value of X such that X AND Y is zero
using System;
     
class GFG
{
     
    // Method to find smallest number Y for
    // a given value of X such that X AND Y is zero
    static long findSmallestNonZeroY(long A_num)
    {
        // Convert the number into its binary form
        String A_binary = Convert.ToString(A_num, 2);
        long B = 1;
        int len = A_binary.Length;
        int no_ones = bitCount(A_num);
 
        // Case 1 : If all bits are ones,
        // then return the next number
        if (len == no_ones)
        {
            return A_num + 1;
        }
 
        // Case 2 : find the first 0-bit
        // index and return the Y
        for (int i = 0; i < len; i++)
        {
            char ch = A_binary[len - i - 1];
            if (ch == '0')
            {
                B = (long)Math.Pow(2.0, (double)i);
                break;
            }
        }
        return B;
    }
     
    static int bitCount(long x)
    {
        // To store the count
        // of set bits
        int setBits = 0;
        while (x != 0)
        {
            x = x & (x - 1);
            setBits++;
        }
        return setBits;
    }
     
    // Driver code
    public static void Main(String[] args)
    {
        long X = findSmallestNonZeroY(10);
        Console.WriteLine(X);
    }
}
 
// This code is contributed by 29AjayKumar

Javascript




<script>
// Javascript program to find smallest number Y for
// a given value of X such that X AND Y is zero
 
// Method to find smallest number Y for
    // a given value of X such that X AND Y is zero
function findSmallestNonZeroY(A_num)
{
    // Convert the number into its binary form
        let A_binary = (A_num >>> 0).toString(2);
        let B = 1;
        let len = A_binary.length;
        let no_ones = bitCount(A_num);
   
        // Case 1 : If all bits are ones,
        // then return the next number
        if (len == no_ones) {
            return A_num + 1;
        }
   
        // Case 2 : find the first 0-bit
        // index and return the Y
        for (let i = 0; i < len; i++) {
            let ch = A_binary[len - i - 1];
            if (ch == '0') {
                B = Math.floor(Math.pow(2.0, i));
                break;
            }
        }
        return B;
}
function bitCount(x)
{
    // To store the count
        // of set bits
        let setBits = 0;
        while (x != 0)
        {
            x = x & (x - 1);
            setBits++;
        }
        return setBits;
}
 
// Driver code
let X = findSmallestNonZeroY(10);
document.write(X);
 
 
// This code is contributed by unknown2108
</script>
Output: 
1

 

Time Complexity: O(1)
Auxiliary Space: O(1)


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