We can do this application using bit-manipulation using two method:-

  1. Left Shift operator.
Note:- In the Left Shift method , we take example n=5 and k=3
n= 00000.....0101 (5)
1= 00000.....0100 (using left shift we shift the value 1)
o/p= 00000.....0100 ( output is any +ve integer that means the at k-th position its set , 1 & 1= 1).

2. Right Shift operator.

Note:- In the Right shift method , we take the same example n=5 and k=3
n= 0000....0101 (5)
n>>(3-1) (right shift of n>>2)
After the right shift operator we have
n= 0000....0001
1= 0000....0001
n&1= 0000....0001 ( we have to get the value equivalent to 1 for the kth bit is set or not judgement.)

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