@Dave
yes it help very much...i thank you for these...
but still one doubt
1.from first element of each sequence we have made a heap
2. now only n-1 element remains in all k seqences. ====> total (= k*(n-1))
elememts
3.*"loop is executed once for each output element = once for each
element in the input sequences" but there are k sequence...
so the loop must run for *k*(n-1) and each iteration of for loop running
cost to O(log k).
overall complexity can be
O(k(n-1)log k)..
plz do correct me Dave....
thanks..
On Mon, Oct 4, 2010 at 11:20 PM, Dave <dave_and_da...@juno.com> wrote:
> @Coolfrog$: There must have been a communication gap.
> The initial heap consists of the first element of each sequence: 2,
> 17, 6.
> Looping, we output 2, then replace it in the heap with 3 and restore
> the heap condition: 3, 17, 6.
> Output 3, replace it with 4: 4, 17, 6.
> Output 4, replace it with 5: 5, 17, 6.
> Output 5. There is nothing left in sequence 1, so the heap now has 2
> elements: 17, 6. Restore the heap condition: 6, 17. Output 6, replace
> it with 9: 9, 17.
> Etc. Etc.
> Finally, after outputting 15, list 3 is exhausted, and the heap
> consists of 1 element. At this point, the effect of the loop is that
> list 2 is output.
>
> The loop is executed once for each output element = once for each
> element in the input sequences; i.e., n times.
>
> Does that help?
>
> Dave
>
> On Oct 4, 12:09 pm, "coolfrog$" <dixit.coolfrog.div...@gmail.com>
> wrote:
> > @Dave
> > 1. if three sequence given are 2,3,4,5
> > 17,20,31,50
> > 6,9,10,15
> > while running we will get 2,3,4,5 as sequence no.1 vanishes form where to
> > choose next element. for heap . (algo.??)
> > 2.
> > Loop until the heap is empty:
> > output the top of the heap,
> > replace it with the next element from the list it came from (if
> > non-empty, of course),
> > re-establishing the heap condition.*========> O(log k).
> >
> > "**loop is executed n times**" .. i am not getting because we have
> already
> > executed loop n-1 times in arranging 3,4,5. **what about rest of
> elements...
> >
> > Correct me plz
> >
> > Regards...**
> > *
> >
> >
> >
> >
> >
> > On Mon, Oct 4, 2010 at 7:35 PM, Dave <dave_and_da...@juno.com> wrote:
> > > @saurabh:
> >
> > > For the base conversion problem, simulate long division in base B1 of
> > > dividing the number by B2. The remainder is the rightmost digit of the
> > > conversion. To get the next digit, divide the quotient from the first
> > > long division by B2 to get the remainder. Repeat for successive digits
> > > until the quotient is zero.
> >
> > > For the merge problem, assuming ascending order:
> >
> > > form a min-heap from the first items from each list.
> > > Loop until the heap is empty:
> > > output the top of the heap,
> > > replace it with the next element from the list it came from (if
> > > non-empty, of course),
> > > re-establishing the heap condition.
> >
> > > The loop is executed n times, and each involves O(k) operations.
> >
> > > Dave
> >
> > > On Oct 4, 7:16 am, saurabh agrawal <saurabh...@gmail.com> wrote:
> > > > hi mukesh...gr8 solution....can u pls help me with some other
> questions:
> > > > --Given an array of n integers find all the inversion pairs in O(n)
> > > > Inversion pair is one where a[i]>a[j], i<j
> >
> > > > --Convert a number given in base B1 to a number in base B2 without
> using
> > > any
> > > > intermediate base
> >
> > > > --You are given k sorted lists with total n inputs in all the lists
> > > devise a
> > > > algorithm to merge them into one single sorted list in O(n logk)
> >
> > > > On Mon, Oct 4, 2010 at 5:31 PM, Mukesh Gupta <
> mukeshgupta.2...@gmail.com
> > > >wrote:
> >
> > > > > The problem could be solved using xor logic. First take xor of all
> the
> > > > > elements .Doing that we get a value which is xor of the two non
> > > repeating
> > > > > elements(as xor of similar values is 0). Now xor of two non
> repeating
> > > > > numbers contains bits set at those places where the two numbers
> differ.
> > > Now
> > > > > if we take any set bit of our result and again xor all those values
> of
> > > set
> > > > > where that bit is set we get first non-repeating value. Taking xor
> of
> > > all
> > > > > those values where that bit is not set gives the another
> non-repeating
> > > > > number..
> > > > > For ex
> > > > > let a[]={2,3,4,3,2,6}
> >
> > > > > xor of all values=0010
> > > > > Now we need to get any set bit. We can extract the rightmost set
> bit of
> > > any
> > > > > number n by taking ( n & ~(n-1))
> >
> > > > > Here 2nd bit is the rightmost set bit.
> >
> > > > > Now when we take xor of all values where 2nd bit is set(this could
> be
> > > done
> > > > > as (a[i] & 0010) , we get 6
> > > > > Taking xor of all values where 2nd bit is not set yields 4.
> >
> > > > > Mukesh Gupta
> > > > > Delhi College of Engineering
> >
> > > > > On Mon, Oct 4, 2010 at 3:17 PM, malli <mallesh...@gmail.com>
> wrote:
> >
> > > > >> I have an array. All numbers in the array repeat twice except two
> > > > >> numbers which repeat only once. All the numbers are placed
> randomly.
> > > > >> Goal is to find out efficiently the two numbers that have not
> > > > >> repeated with O(1) extra memory. Expecting linear solution.
> >
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