@Dave @Mukesh yaa got it... so simple... i was needlessly making it complex.... thankyou guys .....
On Tue, Oct 5, 2010 at 2:09 AM, Dave <dave_and_da...@juno.com> wrote: > @Coolfrog$: It sounds like you think there are n items in each list, > but the problem statement says that the total number of items in the > lists is n. In your example, n = 12. > > Dave > > On Oct 4, 2:16 pm, "coolfrog$" <dixit.coolfrog.div...@gmail.com> > wrote: > > @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. 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