Check whether the bit at given position is set or unset

Given two positive integers n and k. The problem is to check whether the bit at position k from the right in the binary representation of n is set (‘1’) or unset (‘0’).
Constraints: 1 <= k <= number of bits in the binary representation of n.

Examples:

Input : n = 10, k = 2
Output : Set
(10)10 = (1010)2
The 2nd bit from the right is set.

Input : n = 21, k = 4
Output : Unset

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: Following are the steps:

1. Calculate new_num = (n >> (k – 1)).
2. if (new_num & 1) == 1 then bit is “Set”, else “Unset”.

C++

 // C++ implementation to check whether the bit // at given position is set or unset #include using namespace std;    // function to check whether the bit // at given position is set or unset bool bitAtGivenPosSetOrUnset(unsigned int n,                              unsigned int k) {     int new_num = n >> (k - 1);        // if it results to '1' then bit is set,     // else it results to '0' bit is unset     return (new_num & 1); }    // Driver program to test above int main() {     unsigned int n = 10, k = 2;     if (bitAtGivenPosSetOrUnset(n, k))         cout << "Set";     else         cout << "Unset";     return 0; }

Java

 // Java program to // check the set bit // at kth position import java.io.*;    class GFG {        // function to check whether // the bit at given position // is set or unset static int bitAtGivenPosSetOrUnset                    ( int n, int k) {        // to shift the kth bit     // at 1st position     int new_num = n >> (k - 1);         // Since, last bit is now      // kth bit, so doing AND with 1     // will give result.     return (new_num & 1); }     public static void main (String[] args)     {          // K and n must be greater than 0          int n = 10, k = 2;                 if (bitAtGivenPosSetOrUnset(n, k)==1)         System.out.println("Set");     else         System.out.println("Unset");     } }    //This code is contributed by Gitanjali

Python3

 # python implementation to check # whether the bit at given # position is set or unset    import math #function to check whether the bit # at given position is set or unset def bitAtGivenPosSetOrUnset(  n,  k):      new_num = n >> (k - 1)          #if it results to '1' then bit is set,      #else it results to '0' bit is unset      return (new_num & 1)    # Driver code n = 10 k = 2 if (bitAtGivenPosSetOrUnset(n, k)):      print("Set") else:     print("Unset")    #This code is contributed by Gitanjali

C#

 // C# program to check the set bit // at kth position using System;    class GFG {            // function to check whether     // the bit at given position     // is set or unset     static int bitAtGivenPosSetOrUnset(                            int n, int k)     {                // to shift the kth bit         // at 1st position         int new_num = n >> (k - 1);                // Since, last bit is now          // kth bit, so doing AND with 1         // will give result.         return (new_num & 1);     }            // Driver code     public static void Main ()     {                    // K and n must be greater          // than 0         int n = 10, k = 2;                    if (bitAtGivenPosSetOrUnset(n, k)==1)             Console.Write("Set");         else             Console.Write("Unset");     } }    // This code is contributed by Sam007.

PHP

 > (\$k - 1);        // if it results to '1' then bit is set,     // else it results to '0' bit is unset     return (\$new_num & 1); }        // Driver Code     \$n = 10;     \$k = 2;     if (bitAtGivenPosSetOrUnset(\$n, \$k))         echo "Set";     else         echo "Unset";            // This code is contributed by Sam007 ?>

Output:

Set

Time Complexity: O(1).

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