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
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 1000000007using namespace std;// Method to find smallest number Y for// a given value of X such that X AND Y is zeroint 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 Codeint 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 zeroimport 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 zerodef 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 Codeif __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 zerousing 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 zerofunction 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 codelet 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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