I found a good question to try for bit manipulation.Try it... :)
Given an integer x, find out the smallest integer which has same
number of set bits as x and is greater than x.
For example if the input integer is 12 (1100) then your function
should return 17(10001). If the input integer is
Solutn:
1101000
Start from rightmost bit-leftmost bit
Find the starting and ending 1’s positions, here 3 and 7
If any 0 bw them… while traversing.. (bitwise r-l).. make it 1 and other
adjacent right ” 1” as “0”. And this is your new no. here 111
If you find no “0” in bw ( eg for
int count1=count2=0;
given number=n;
int m=n; //save in temp variable m
while(m!=0) //count number of bits in given number
{
m=m (m-1);
count1++;
}
m=n; //save in temp variable m
while(count1!=count2) //check no. of bits same or not
{
m=m+1;
Q1)An array A[1 n] contains all the integers from 0 to n except for one
number which is missing . In this problem, we cannot access an entire
integer in A with a single operation . The elements of A are represented in
binary, and the only operation we can use to access them is “fetch the jth
