[algogeeks] Re: find duplicate and missing element
what do u think about this ...its O(n) program..
#include
#include
# include "malloc.h"
# include
bool bRemoveDuplicates(int array[], int iSize){
if(iSize <1)
return false;
if(array == NULL)
return false;
if(iSize == 1)
return true;
int left_comp = 0;
int right_comp = 1;
bool start_move = false;
int flag=0;
int hole =0;
//{1,1,1,1,1,1,1,1,2,2,4,5,7,8,8,9,9,10,11,11,12,13,14};
for(; right_comp >i;
}
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Re: [algogeeks] Re: Simple reVerse
if input is 12345 then the output should be 54321 -- You received this message because you are subscribed to the Google Groups "Algorithm Geeks" group. To post to this group, send email to algoge...@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
Re: [algogeeks] Re: Simple reVerse
@Dave but i tried the code its not working Its giving 1 for all the input could u explain the code please -- You received this message because you are subscribed to the Google Groups "Algorithm Geeks" group. To post to this group, send email to algoge...@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
[algogeeks] Re: Simple reVerse
I presume that you mean reversing the order of the bits, so that the
low-order bit goes to the high-order position and vice-versa. Assuming
32 bit integers, this does the trick:
int r;
r = ( ( x && 0x ) >> 16 ) || ( ( x && 0x ) <<
16 );
r = ( ( r && 0xff00ff00 ) >> 8 ) || ( ( r && 0x00ff00ff ) <<
8 );
r = ( ( r && 0xf0f0f0f0 ) >> 4 ) || ( ( r && 0x0f0f0f0f ) <<
4 );
r = ( ( r && 0x ) >> 2 ) || ( ( r && 0x ) <<
2 );
r = ( ( r && 0x ) >> 1 ) || ( ( r && 0x ) <<
1 );
return r;
Dave
On Sep 3, 1:00 pm, Albert wrote:
> int reverse(int x)
> {
>
> ...
> ...
>
>
> }
>
> complete the above code such that it returns the reverse of 'x'
>
> condition here is u should not use loops and global as well as static
> variable..
>
> try to give all the possible solutions for this ...
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[algogeeks] Re: Amazon interview Question (Array yet again!)
nice explanation gene :)
On Sep 2, 8:35 am, Gene wrote:
> Okay. First, you can make the DP more efficient than the one I gave
> earlier. You don't need to scan the whole previous column when
> calculating costs of decrementing. Rather there are only two
> possibilities.
>
> I will add that rahul patil's solution looks correct, but it's
> exponential time. DP is better for this problem.
>
> Remember C(n, m) is the cost of making a[1 .. n] into a non-decreasing
> sequence with the last element being no more than m. And we always
> draw m from the set V of values in a.
>
> So here is the new DP:
>
> C(1, m) = max(a[1] - m, 0) // first row only decrement is possible
> C(n, m) = min (
> a[n] + C(n - 1, m), // delete
> (a[n] <= m) ? C(n - 1, a[n]) : C(n - 1, m) + a[n] - m // decrement
> )
>
> In case you don't follow, the "//delete" line is saying that we
> already know the cost of making everything up to element n-1 non-
> decreasing and no more than m. This, by recursive assumption, is just
> C(n-1,m). The additional cost of deleting a[n] is a[n].
>
> The "//decrement" line is similar, but there are 2 cases. If a[n] is
> more than m, we must decrement it. The cost here consists of making
> everything up to n-1 non-decreasing and no more than m, i.e. C(n-1,m).
> Plus we have the cost of chopping a[n] down to m, which is a[n]-m.
>
> In the other case, a[n] is m or less. So there's no need to
> decrement, but we must get the elements a[1..n-1] to be no more than
> a[n]. Again by recursive assumption this cost is C(n-1,a[n]).
>
> Here is an example. Suppose we have a = [5, 1, 1, 1, 3, 1]. The
> least cost here is obtained by decrementing the 5 to 1 (cost 4) and
> deleting the final 1 (cost 1) for a total cost of 5.
>
> So let's try the algorithm. (You must view this with a fixed font.)
>
> Table of C(n, m) values:
> m = 1 3 5
> n = 1 : 4 2 0
> n = 2 : 4 3* 1*
> n = 3 : 4 4 2*
> n = 4 : 4 4 3*
> n = 5 : 6m 4 4
> n = 6 : 6 5* 5*
>
> Here * means C resulted from decrementing and "m" means that a
> decrement was based on the value of m rather than a[n].
>
> We take the answer from C(6,5) = 5.
>
> Implementing this is a little tricky because m values are drawn from
> V. You could use a hash table for the m-axis. But it's more
> efficient to store V in an array and convert all the values of m in
> the DP into indices of V. Because all the indices lie in [ 1 .. |
> V| ], we can use simple arrays rather than hash tables to represent
> the rows of the table C.
>
> We only need 2 rows at a time, so O(|V|) storage does the job.
>
> For C, we also need to convert all the indices to origin 0.
>
> So here's the final O(n^2) code. I think this is a correct
> implementation. If anyone has an example that breaks it, I'd like to
> see.
>
> #include
>
> #define NMAX 1
>
> int cost(int *a, int N)
> {
> int i, j, max, Nv;
> int V[NMAX], A[NMAX], B[NMAX];
> int *Cm = A, *Cn = B; // (n-1)'th and n'th rows of C
>
> // Compute set V with no duplicates.
> // Remember where max element is.
> Nv = max = 0;
> for (i = 0; i < N; i++) {
> for (j = 0; j < Nv; j++)
> if (a[i] == V[j])
> break;
> if (j == Nv) {
> V[Nv++] = a[i];
> if (V[j] > V[max])
> max = j;
> }
> a[i] = j; // Convert a to indices.
> }
> // Fill in first row of C.
> for (i = 0; i < Nv; i++)
> Cm[i] = (V[a[0]] >= V[i]) ? V[a[0]] - V[i] : 0;
>
> // Fill in the rest of the rows of C.
> for (i = 1; i < N; i++) {
> for (j = 0; j < Nv; j++) {
> int del = Cm[j] + V[a[i]];
> int dec = (V[a[i]] <= V[j]) ? Cm[a[i]] : Cm[j] + V[a[i]] - V[j];
> Cn[j] = (del < dec) ? del : dec;
> }
> // Swap row buffers so current becomes previous.
> int *tmp = Cn; Cn = Cm; Cm = tmp;
> }
> return Cm[max];
>
> }
>
> int main(void)
> {
> static int a[] = { 5, 1, 1, 1, 3, 1 };
> printf("Min cost = %d\n", cost(a, sizeof a / sizeof a[0]));
> return 0;
>
> }
>
> On Aug 30, 9:19 pm, vikash jain wrote:
>
>
>
> > @gene ..if u just give an example herethat will make things more
> > clear...
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[algogeeks] Re: Amazon interview Question (Array yet again!)
@gene: nice solution..
but it's not working for a[]={20,22,13,11};
ur code will give soln : 24
but ans should be: 22 {11,11,11,11}
pls correct me if i m wrong
On Aug 28, 8:26 am, jagadish wrote:
> @Gene: Thanks alot! :-) your solution works like charm!
>
> On Aug 28, 7:09 am, Gene wrote:
>
>
>
> > This is a nice problem. It looks like a trivial matroid, so a greedy
> > algorithm will work fine. The obvious greedy algorithm is to work
> > left-to-right and incorporate elements into the sorted order one-by-
> > one. In each case, you have 2 choices. The first is to decrement
> > elements to the left by the amount needed to restore non-decreasing
> > order. The second is to delete the new element. The cost of each is
> > easy to calculate. Pick the choice with least cost and continue.
> > This algorithm is O(n^2). There may be a faster way to do it, but I
> > can't see one.
>
> > #include
>
> > int make_nondecreasing(int *a, int n)
> > {
> > int i, j, dec, dec_cost, total_cost;
>
> > total_cost = 0;
> > for (i = 0; i < n - 1; i++) {
>
> > // If we find a decrease...
> > if (a[i] > a[i + 1]) {
>
> > // Find cost of decrementing all to the left.
> > dec_cost = dec = a[i] - a[i + 1];
> > for (j = i - 1; j >= 0; j--) {
>
> > // Find decrement that would be needed.
> > dec += a[j] - a[j + 1];
>
> > // If no decement, we're done.
> > if (dec <= 0)
> > break;
>
> > // Count cost of decrement.
> > dec_cost += dec;
> > }
>
> > // Compare decrement cost with deletion cost.
> > if (dec_cost < a[i + 1]) {
>
> > // Decrement is cheaper. Do it.
> > for (j = i; j >= 0; j--) {
> > if (a[j] > a[i + 1])
> > a[j] = a[i + 1];
> > }
> > total_cost += dec_cost;
> > }
> > else {
>
> > // Deletion is cheaper. Do it.
> > total_cost += a[i + 1];
> > for (j = i + 1; j < n - 1; j++)
> > a[j] = a[j + 1];
> > --n;
> > }
> > }
> > }
> > return total_cost;
>
> > }
>
> > int main(void)
> > {
> > int a[] = { 14, 15, 16, 13, 11, 18 };
> > //int a[] = { 4, 3, 5, 6};
> > //int a[] = { 10, 3, 11, 12 };
> > int cost = make_nondecreasing(a, sizeof a / sizeof a[0]);
> > printf("cost=%d\n", cost);
> > return 0;
>
> > }
>
> > On Aug 27, 12:15 pm, jagadish wrote:
>
> > > You are given an array of positive integers. Convert it to a sorted
> > > array with minimum cost. Only valid operation are
> > > 1) Decrement -> cost = 1
> > > 2) Delete an element completely from the array -> cost = value of
> > > element
>
> > > For example:
> > > 4,3,5,6, -> cost 1
> > > 10,3,11,12 -> cost 3
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[algogeeks] Simple reVerse
int reverse(int x)
{
...
...
}
complete the above code such that it returns the reverse of 'x'
condition here is u should not use loops and global as well as static
variable..
try to give all the possible solutions for this ...
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[algogeeks] Re: Take 5 digit number and find 2 power that number.............
But how the number(in decimal form) will be displayedif ques demands so. On Sep 2, 1:49 pm, saurabh singh wrote: > Suppose the number of shifts be x. > Also let the integer be represented by 16 bits on that machine. > Now take int n= (int)(x/16 + 0.5), to take the upper cap on result :) . > SO having 2^x will be same as doing 2<<(x-1) so essentially if we represent > the resultant number in a linked list of nodes, where each node can store an > integer, then the bit position at 2+x-1=x+1 will be set as 1 and this > (x+1)th nit will fall in (x+1)/16th node in linked list also in that node > (x+1)%16 will give the position of bit to be set. > > On Thu, Sep 2, 2010 at 1:32 PM, ashish agarwal > > > > > > wrote: > > I think it will be 1< > > On Wed, Sep 1, 2010 at 10:53 PM, Yan Wang wrote: > > >> Maybe you misunderstand the question. > >> The question is how to compute 2^X where 0 <= X <= 9? > >> How? > > >> On Wed, Sep 1, 2010 at 10:48 PM, Ruturaj wrote: > >> > a 5 digit number is of order 10^5 which is << 10^16 of which int in C > >> > is of size. > >> > Just multiply both numbers. > > >> > On Sep 2, 10:39 am, prasad rao wrote: > >> >> Program that takes a 5 digit number and calculates 2 power that number > >> and > >> >> prints it and should not use the Big-integer and Exponential > >> Function's. > > >> > -- > >> > You received this message because you are subscribed to the Google > >> Groups "Algorithm Geeks" group. > >> > To post to this group, send email to algoge...@googlegroups.com. > >> > To unsubscribe from this group, send email to > >> algogeeks+unsubscr...@googlegroups.com >> .com> > >> . > >> > For more options, visit this group at > >>http://groups.google.com/group/algogeeks?hl=en. > > >> -- > >> You received this message because you are subscribed to the Google Groups > >> "Algorithm Geeks" group. > >> To post to this group, send email to algoge...@googlegroups.com. > >> To unsubscribe from this group, send email to > >> algogeeks+unsubscr...@googlegroups.com >> .com> > >> . > >> For more options, visit this group at > >>http://groups.google.com/group/algogeeks?hl=en. > > > -- > > You received this message because you are subscribed to the Google Groups > > "Algorithm Geeks" group. > > To post to this group, send email to algoge...@googlegroups.com. > > To unsubscribe from this group, send email to > > algogeeks+unsubscr...@googlegroups.com > .com> > > . > > For more options, visit this group at > >http://groups.google.com/group/algogeeks?hl=en. > > -- > Thanks & Regards, > Saurabh -- You received this message because you are subscribed to the Google Groups "Algorithm Geeks" group. To post to this group, send email to algoge...@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
Re: [algogeeks] Re: Amazon intern Question
Trie-traverse search will be best.. On Thu, Sep 2, 2010 at 8:10 PM, Manjunath Manohar wrote: > trie will be the best choice for this.. > > -- > You received this message because you are subscribed to the Google Groups > "Algorithm Geeks" group. > To post to this group, send email to algoge...@googlegroups.com. > To unsubscribe from this group, send email to > algogeeks+unsubscr...@googlegroups.com > . > For more options, visit this group at > http://groups.google.com/group/algogeeks?hl=en. > -- Dhruvesh Kumar Mobile No.: +919911314113 M.C.A., Department of Computer Science, Delhi University, INDIA -- You received this message because you are subscribed to the Google Groups "Algorithm Geeks" group. To post to this group, send email to algoge...@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
Re: [algogeeks] Re: find duplicate and missing element
@ankit
Yup I got your point. I didn't see the algo given by dhritiman previously. I
think that is better than my solution , where it fits in all cases.
Cheers
Kartheek
On Fri, Sep 3, 2010 at 1:32 PM, Ankit Sinha wrote:
> @kartheek, thanks for ur input!! Certainly, ur soln is fine but only
> will cater when array is 1...n but what if it ranges for 0...n-1. The
> algo given by dhritiman fits in all the scenario. but ofcourse for
> given question ( 1 to 100) your mathematical approach is good. :)...
>
> Cheers,
> Ankit Sinha!!!
>
> On Fri, Sep 3, 2010 at 8:14 AM, kartheek muthyala
> wrote:y
> >
> > Yeah
> > The solution given by Ankit is gr8. But inorder to not destroy the array
> we
> > need to go by the geek method where
> >
> > Suppose x is the duplicated element and y is the missing element in the
> > array.
> > Multiply all the elements in the array to prod and sum all the elements
> in
> > the array to sum.
> > Divide prod with 100! that is gonna give value of x/y
> > Subtract sum from 5050 that is gonna give 5050 + x - y
> > Solve the two equations to get x and y.
> > It is gonna take O(N) ideally.
> > Cheers
> > Kartheek
> >
> > On Fri, Sep 3, 2010 at 11:09 AM, Ankit Sinha
> wrote:
> >>
> >> @Dhritiman, It's good algo man!!!The only thing is we are destroying
> >> the array but also that's mandatory as only o(n) complexity we are
> >> interested in.
> >>
> >> As Somebody wanted the code, here I am attaching below: -
> >>
> >> int a[SIZE_A] = {0,2,1,4,0};
> >> int i = 0, dup = 0, pos = 0, t =0;
> >> while (i< 5)
> >> {
> >> if (a[i] == i)
> >> i++;
> >> else if (a[a[i]] == a[i])
> >> {
> >> dup = a[i];
> >>
> >> printf ("\nduplicated element is [%d]", dup);
> >> pos = i;
> >> i++;
> >> }
> >> else
> >> {
> >> t= a[i];
> >> a[i] = a[a[i]];
> >> a[t] = t;
> >> }
> >> }
> >> printf ("\nmissing element is [%d]", pos);
> >>
> >>
> >> Cheers,
> >> Ankit Sinha
> >>
> >> On Thu, Sep 2, 2010 at 7:08 AM, Dhritiman Das
> >> wrote:
> >> > @Dinesh,
> >> > Yes, we can't apply this method, if that's not allowed.
> >> >
> >> > On Thu, Sep 2, 2010 at 10:31 AM, dinesh bansal
> >> > wrote:
> >> >>
> >> >>
> >> >> On Wed, Sep 1, 2010 at 11:08 AM, Dhritiman Das <
> dedhriti...@gmail.com>
> >> >> wrote:
> >> >>>
> >> >>> Given a array, A[1..n], do the following.
> >> >>> Start from i=1 and try placing each number in its correct position
> in
> >> >>> the
> >> >>> array.
> >> >>> If the number is already in its correct position, go ahead. if (A[i]
> >> >>> ==
> >> >>> i) , i++
> >> >>> else if the number was already restored to its correct position,
> then
> >> >>> it
> >> >>> is
> >> >>> a duplicate , so remember it and move ahead if (A[A[i]] == A[i]),
> dup
> >> >>> =
> >> >>> A[i], i++ ;
> >> >>> else
> >> >>> swap the elements at the current index i and that at A[i]'s
> correct
> >> >>> position in the array.
> >> >>> continue this until all numbers are placed in their correct position
> >> >>> in
> >> >>> the array
> >> >>> Finally, A[missing] will be dup.
> >> >>> O(n)
> >> >>
> >> >>
> >> >> @Dharitiman, good solution man. No expensive computation, no extra
> >> >> memory
> >> >> required. Only problem is it will change the input array to sorted
> >> >> order
> >> >> which may not be desired.
> >> >>
> >> >>>
> >> >>> On Wed, Sep 1, 2010 at 7:14 AM, Dave
> wrote:
> >>
> >> Suppose that x is duplicated and y is omitted. Then the sum of the
> >> numbers would be 5050 + x - y. Similarly, the sums of the squares
> of
> >> the numbers would be 338,350 + x^2 - y^2. Calculate the sum and sum
> >> of
> >> squares of the numbers and solve the resulting equations for x and
> y.
> >>
> >> Dave
> >>
> >> On Aug 31, 1:57 pm, Raj Jagvanshi wrote:
> >> > There is an array having distinct 100 elements from 1 to 100
> >> > but by mistake some no is duplicated and a no is missed.
> >> > Find the duplicate no and missing no.
> >> >
> >> > Thanks
> >> > Raj Jagvanshi
> >>
> >> --
> >> You received this message because you are subscribed to the Google
> >> Groups "Algorithm Geeks" group.
> >> To post to this group, send email to algoge...@googlegroups.com.
> >> To unsubscribe from this group, send email to
> >> algogeeks+unsubscr...@googlegroups.com
> .
> >> For more options, visit this group at
> >> http://groups.google.com/group/algogeeks?hl=en.
> >>
> >> >>>
> >> >>> --
> >> >>> You received this message because you are subscribed to the Google
> >> >>> Groups
> >> >>> "Algorithm Geeks" group.
> >> >>> To post to this group, send email to algoge...@googlegroups
Re: [algogeeks] Re: find duplicate and missing element
@kartheek, thanks for ur input!! Certainly, ur soln is fine but only
will cater when array is 1...n but what if it ranges for 0...n-1. The
algo given by dhritiman fits in all the scenario. but ofcourse for
given question ( 1 to 100) your mathematical approach is good. :)...
Cheers,
Ankit Sinha!!!
On Fri, Sep 3, 2010 at 8:14 AM, kartheek muthyala
wrote:y
>
> Yeah
> The solution given by Ankit is gr8. But inorder to not destroy the array we
> need to go by the geek method where
>
> Suppose x is the duplicated element and y is the missing element in the
> array.
> Multiply all the elements in the array to prod and sum all the elements in
> the array to sum.
> Divide prod with 100! that is gonna give value of x/y
> Subtract sum from 5050 that is gonna give 5050 + x - y
> Solve the two equations to get x and y.
> It is gonna take O(N) ideally.
> Cheers
> Kartheek
>
> On Fri, Sep 3, 2010 at 11:09 AM, Ankit Sinha wrote:
>>
>> @Dhritiman, It's good algo man!!!The only thing is we are destroying
>> the array but also that's mandatory as only o(n) complexity we are
>> interested in.
>>
>> As Somebody wanted the code, here I am attaching below: -
>>
>> int a[SIZE_A] = {0,2,1,4,0};
>> int i = 0, dup = 0, pos = 0, t =0;
>> while (i< 5)
>> {
>> if (a[i] == i)
>> i++;
>> else if (a[a[i]] == a[i])
>> {
>> dup = a[i];
>>
>> printf ("\nduplicated element is [%d]", dup);
>> pos = i;
>> i++;
>> }
>> else
>> {
>> t= a[i];
>> a[i] = a[a[i]];
>> a[t] = t;
>> }
>> }
>> printf ("\nmissing element is [%d]", pos);
>>
>>
>> Cheers,
>> Ankit Sinha
>>
>> On Thu, Sep 2, 2010 at 7:08 AM, Dhritiman Das
>> wrote:
>> > @Dinesh,
>> > Yes, we can't apply this method, if that's not allowed.
>> >
>> > On Thu, Sep 2, 2010 at 10:31 AM, dinesh bansal
>> > wrote:
>> >>
>> >>
>> >> On Wed, Sep 1, 2010 at 11:08 AM, Dhritiman Das
>> >> wrote:
>> >>>
>> >>> Given a array, A[1..n], do the following.
>> >>> Start from i=1 and try placing each number in its correct position in
>> >>> the
>> >>> array.
>> >>> If the number is already in its correct position, go ahead. if (A[i]
>> >>> ==
>> >>> i) , i++
>> >>> else if the number was already restored to its correct position, then
>> >>> it
>> >>> is
>> >>> a duplicate , so remember it and move ahead if (A[A[i]] == A[i]), dup
>> >>> =
>> >>> A[i], i++ ;
>> >>> else
>> >>> swap the elements at the current index i and that at A[i]'s correct
>> >>> position in the array.
>> >>> continue this until all numbers are placed in their correct position
>> >>> in
>> >>> the array
>> >>> Finally, A[missing] will be dup.
>> >>> O(n)
>> >>
>> >>
>> >> @Dharitiman, good solution man. No expensive computation, no extra
>> >> memory
>> >> required. Only problem is it will change the input array to sorted
>> >> order
>> >> which may not be desired.
>> >>
>> >>>
>> >>> On Wed, Sep 1, 2010 at 7:14 AM, Dave wrote:
>>
>> Suppose that x is duplicated and y is omitted. Then the sum of the
>> numbers would be 5050 + x - y. Similarly, the sums of the squares of
>> the numbers would be 338,350 + x^2 - y^2. Calculate the sum and sum
>> of
>> squares of the numbers and solve the resulting equations for x and y.
>>
>> Dave
>>
>> On Aug 31, 1:57 pm, Raj Jagvanshi wrote:
>> > There is an array having distinct 100 elements from 1 to 100
>> > but by mistake some no is duplicated and a no is missed.
>> > Find the duplicate no and missing no.
>> >
>> > Thanks
>> > Raj Jagvanshi
>>
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>> >> --
>> >> Dinesh Bansal
>> >> The Law of Win says, "Let's not do it your way or my way; let's do it
>> >> the
>> >> best way."
>> >>
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