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; //increment number everytime
result=m; //if exits from loop then this contain
result
count2=0;
while(m!=0)
{
m=m & (m-1);
count2++;
}
}
On Sat, Jul 9, 2011 at 5:38 PM, Aman Goyal <[email protected]> wrote:
> 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 1110000
>
>
> If you find no “0” in bw ( eg for 111100)
>
> Then ,a bit will be increased.
>
> Make first bit from left 1 and set the lower order (total_set_bits-1)
> bits.. this will be the new no.
>
> Here : 1000111.
>
>
> On Sat, Jul 9, 2011 at 2:53 PM, Piyush Sinha <[email protected]>wrote:
>
>> 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 3(11) then your
>> function should return 5(101)
>>
>> --
>> *Piyush Sinha*
>> *IIIT, Allahabad*
>> *+91-8792136657*
>> *+91-7483122727*
>> *https://www.facebook.com/profile.php?id=100000655377926 *
>>
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>>
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