@yogesh
your code is simple one and excellent to understand;
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@DK. Yes. But you must admit that it is a bit-twiddler's delight!
Dave
On Jul 10, 3:58 pm, DK wrote:
> @Dave: Thanks for the link.
> Just a point of discussion - this kind of code would probably never pass
> code-review (or would be heavily documented with references and warnings
> that say HAND
@Dave: Thanks for the link.
Just a point of discussion - this kind of code would probably never pass
code-review (or would be heavily documented with references and warnings
that say HANDS OFF ;) )
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DK
http://twitter.com/divyekapoor
http://www.divye.in
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@Piyush: The referenced web page includes a powerpoint presentation
that explains it. Did you read the presentation before asking? If not,
why not?
Dave
On Jul 9, 5:04 pm, Piyush Sinha wrote:
> @Dave..canu u explain ur algo to approach this formula??
>
> On 7/10/11, Dave wrote:
>
>
>
>
>
> > @P
@Dave..canu u explain ur algo to approach this formula??
On 7/10/11, Dave wrote:
> @Piyush: See http://groups.google.com/group/algogeeks/msg/2b64c4f96fa3598e
> for a one-line algorithm.
>
> Dave
>
> On Jul 9, 4:23 am, Piyush Sinha wrote:
>> I found a good question to try for bit manipulation
@Piyush: See http://groups.google.com/group/algogeeks/msg/2b64c4f96fa3598e
for a one-line algorithm.
Dave
On Jul 9, 4:23 am, Piyush Sinha 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
No. It can be applied to any positive number N.
The solution above splits the numbers by the least significant bit,
choose the group with missing number, and then drop the last bit and
continue.
In this way the set [0,N] is partitioned (almost) evenly into 2 groups:
{2k | k=0,1,...,N/2} and
Yes, Following should do for next smallest:
1. Find rightmost 01 pair
2. swap these two bits (make it 10)
e.g.
N=3(011), Next smallest: 5(101)
N=10(1010), Next smallest: 12(1100)
N=14(01110), Next Smallest: 22(10110)
On Jan 18, 12:48 am, juver++ wrote:
> @abobe
> my solution is wrong when number
Yes, I guess following correction in step 1 for Next Smallest should
fix it:
1. Find the leftmost 1 [ Not rightmost].
On Jan 18, 12:48 am, juver++ wrote:
> @abobe
> my solution is wrong when number is even (but it can be avoided with some
> corrections into an implementation),
> btw, you have a m
Yes, I guess following correction in step 1 for Next Smallest should
fix it:
1. Find the leftmost 1 [ Not rightmost].
On Jan 18, 12:48 am, juver++ wrote:
> @abobe
> my solution is wrong when number is even (but it can be avoided with some
> corrections into an implementation),
> btw, you have a m
@abobe
my solution is wrong when number is even (but it can be avoided with some
corrections into an implementation),
btw, you have a mistake: N=3(011), Next smallest: 6(110) ,* Should be 101
(5)!*
In other cases my version is correct.
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Q1 is very well answered by Sunny but looks like Q2 is still open.
IMHO, juver++ soln on Q2 doesn't look correct to me (OR I didn't
understand the problem).
if given no is N=2 (010), then next smallest no N+1 = 3 (011) is not
the right answer as it has TWO 1's.
I guess, for N=2 (010), next smallest
@Sunny Good!
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Q1: initially compute xor of all the values from 0 to n in a variable Temp
so temp = 0^1^2^n
let result is used to store the missing number
for each ith bit of missing number where i = 0-31 we can find it as
following
ith bit of result = (xor of all ith bits of values of array) xored with (
@juver i am really sorry ,i forget to mention.ya this soln will work
only if n is even power of 2.
Regards
Priyaranjan
http://code-forum.blogspot.com
On Jan 17, 12:43 pm, juver++ wrote:
> @awesomeandroid
> Your solution for Q1 is wrong. It can be applied only for such numbers N =
> 2^k, so numb
@above, no :) it is solvable in linear time.
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I think Q1 is NP hard problem since the number of bits grows exponentially
as the array size increases.
On Mon, Jan 17, 2011 at 1:13 PM, juver++ wrote:
> @awesomeandroid
> Your solution for Q1 is wrong. It can be applied only for such numbers N =
> 2^k, so number should power of 2.
>
> --
> You
@awesomeandroid
Your solution for Q1 is wrong. It can be applied only for such numbers N =
2^k, so number should power of 2.
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for Q1. you can do it in o(n) time
Call your missing number M.
You can split your array into two parts depending on whether the least
significant bit of A[i] is a 1 or a 0. The smaller of the two parts
(call it P_1) is at most (n-1)/2 elements in size, and it tells you
whether M's least significan
Q2.next higher number having equal number of 1's
http://code-forum.blogspot.com/2011/01/next-higher-number-having-equal-number.html
Regards
Priyaranjan
On Jan 16, 9:09 pm, Decipher wrote:
> Q1)An array A[1 n] contains all the integers from 0 to n except for one
> number which is missing . In t
Q1: one of the solution implies DP: count[i][t] - how many bits with value t
(t is either 0 or 1) are in all numbers from 0 to n at position i.
Then for each position count actual number of bits 0 and 1, and determine
missing bit.
Set up missing bit in the final result.
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Q2:
next smallest:
if N is even (so, its least significant bit is 0) simply add one
bit, Next_Smallest = N + 1
else {
//locate rightmost 0 in the binary representation of N at index X,
//at index X-1 there is 1, so you need to swap it with 0 at X.
K = N+1, after this rightmost consecu
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