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

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 ` `#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

## 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

 ``

Output:

`1`

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

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