Re: [algogeeks] ADOBE SOLUTIONS PARTNER Consulting Services -Increase your Digital Business with Infinate 360

[sanjiv yadav]

Hi Mark,
I will be available between 10 am to 4 pm tomorrow. So you can connect me
any time in between.

On Tue, 16 Jun 2020, 19:02 Pavan_DigitalTechnocryne, <
pavan.mtech1...@gmail.com> wrote:

> Hi ,
>
>
>
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>
> Mark Anthony
>
> CEO – Infiante 360
>
> Website will be back by 12 June 2020
>
> Linkedin : https://www.linkedin.com/company/13734744/admin/
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Re: [algogeeks] binary search tree over btree

[sanjiv yadav]

In BST the height can be made as bad as u can but in case of btree the
height can not be more than log n base 2 because for each internal node it
is necessary to have at least 2 child and here all the leaf nodes must be
at the same label.

On Sun, Apr 1, 2012 at 8:34 PM, arun kumar  wrote:

> hi i just like to know when you will go for binary search tree over
> btree. advantage and disadvantage, application of both of them.
> thank you in advance
> Regards,
> Arun kumar
>
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MobNo.-  8050142693

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Re: [algogeeks] Re: m*n matrix, sorted rows, sorted columns, WAP to find an element efficiently

[sanjiv yadav]

we can also apply binary search on rows and columns separately by which we
will decide that in which 1/4th part we need to search.So in this case
complexity will be O(log m + log n).

just check and let me know...

On Wed, Apr 4, 2012 at 5:27 AM, Ashish Goel  wrote:

> why not start from middle(m/2, n/2)
> Best Regards
> Ashish Goel
> "Think positive and find fuel in failure"
> +919985813081
> +919966006652
>
>
>
> On Wed, Apr 4, 2012 at 12:29 AM, Karthikeyan V.B wrote:
>
>> Hi,
>>
>> Start from right top corner
>> If matrix element > key move to previous column
>> else if matrix element < key move to next row
>>
>> int search(int** a,int key,int m,int n)
>> {
>> if (key < a[0][0] || key > a[m-1][n-1]) return 0;
>> int min=a[0][n-1],i=0,j=n-1;
>> while(i<=m-1 && j>=0)
>> {
>> if(a[i][j] > key)
>> j--;
>> else if(a[i][j] < key)
>> i++;
>> else
>> return 1;
>> }
>> return 0;
>> }
>>
>> Regards,
>> Karthikeyan.V.B
>>
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Re: [algogeeks] Re: determining if frequency is greater than n/2

[sanjiv yadav]

if all values x are known to be contiguous then I think it will take o(1)
time because it will be majority element if it occurs more than n/2 times
as x must be at the middle.

so if mid=(low+high)/2 and x==a[mid]then x is majority element otherwise
not.

just checklet me know.

On Thu, Feb 9, 2012 at 11:40 PM, Don  wrote:

> It can't be done in O(log n) unless the array is sorted or there is
> some other special property, for example, if all values x are known to
> be contiguous, allowing you to use a binary search to find the first
> and last location of x.
>
> In the general case it is impossible in O(log n) because you have to
> examine at least n/2 elements to reach a conclusion. In some cases you
> must examine all n elements.
>
> Don
>
> On Feb 8, 1:45 pm, Prakhar Jain  wrote:
> > Hello everyone,
> >
> > suppose we have an array of size n and a number, say x, and we have to
> > determine whether the number x is present in the array more then n/2
> times
> > or not?
> > can we have an O(log n) algo for determining it..?..pls help...!!!
> >
> > --
> > --
> > Prakhar Jain
> > IIIT Allahabad
> > B.Tech IT 3rd Year
> > Mob no: +91 9454992196
> > E-mail: rit2009...@iiita.ac.in
> >   jprakha...@gmail.com
>
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Re: [algogeeks] Array problem

[sanjiv yadav]

u r right.

On Mon, Mar 12, 2012 at 11:17 AM, atul anand wrote:

> @sanjiv : wont work for this test case :-
>
> {1,5,3,6,2,7,8};
>
>
> On Mon, Mar 12, 2012 at 10:54 AM, sanjiv yadav wrote:
>
>> @atul anand- It will still work as follows---
>>
>> (3,0)
>> /  \(5,0+3)
>>   (1,0) \(6,0+3+5)
>> \(2,0+1)\(7,0+3+5+6)
>>   \(8,0+3+5+6+7)
>>
>> here, my logic is that if number is grater than its parent,then add the
>> parent in the current sum,else keep it as such.
>>
>> check it and made correction in my logic if i m wrong.
>>
>>
>> On Mon, Mar 12, 2012 at 10:33 AM, atul anand wrote:
>>
>>> @piyush : i dont think so BIT would work over here , we are not just
>>> reporting cumulative sum tilll index i.
>>>
>>> On Mon, Mar 12, 2012 at 12:58 AM, Piyush Kapoor wrote:
>>>
>>>> This can be done very easily with the help of a Binary Indexed Tree,and
>>>> it is very short to code as well.Simply process the numbers in order,and
>>>> for each number output the cumulative frequency of the index of the number
>>>> you are processing.
>>>>
>>>>
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>>
>> Sanjiv Yadav
>>
>> MobNo.-  8050142693
>>
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Re: [algogeeks] Array problem

[sanjiv yadav]

@atul anand- It will still work as follows---

(3,0)
/  \(5,0+3)
  (1,0) \(6,0+3+5)
\(2,0+1)\(7,0+3+5+6)
  \(8,0+3+5+6+7)

here, my logic is that if number is grater than its parent,then add the
parent in the current sum,else keep it as such.

check it and made correction in my logic if i m wrong.

On Mon, Mar 12, 2012 at 10:33 AM, atul anand wrote:

> @piyush : i dont think so BIT would work over here , we are not just
> reporting cumulative sum tilll index i.
>
> On Mon, Mar 12, 2012 at 12:58 AM, Piyush Kapoor wrote:
>
>> This can be done very easily with the help of a Binary Indexed Tree,and
>> it is very short to code as well.Simply process the numbers in order,and
>> for each number output the cumulative frequency of the index of the number
>> you are processing.
>>
>>
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Re: [algogeeks] Array problem

[sanjiv yadav]

u r right payal but
can u expln o(n) time complexity..

On Sun, Mar 11, 2012 at 6:10 PM, payal gupta  wrote:

> By Augmented BST-
> TC-O(n)
>
>
> On Sun, Mar 11, 2012 at 3:08 PM, Gaurav Popli wrote:
>
>> given an array of size n...
>> create an array of size n such that ai where ai is the element in the
>> new array at index location i is equal to sum of all elements in
>> original array which are smaller than element at posn i...
>>
>> e.g
>> ar[]={3,5,1,6,7,8};
>>
>> ar1[]={0,3,0,9,15,22};
>>
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Re: [algogeeks] Re: Doubt in removing loop from linked list

[sanjiv yadav]

actually how u r moving from 7 to 4?

can tell me in the detail...

On Sat, Mar 10, 2012 at 4:37 AM, Dave  wrote:

> @Rahul: The definitions of m and p are in the first sentence of the
> posting.
>
> Dave
>
> On Friday, March 9, 2012 12:46:43 PM UTC-6, rahul sharma wrote:
>
>> @dave..
>>
>> plz tell use of variable...m and p??
>> p i thnik no. of nodes in cycle..???
>>
>> plz tell what is p and m
>>
>> On Fri, Mar 9, 2012 at 10:16 PM, Dave  wrote:
>>
>>> @Rahul: You have to figure out where the fast and slow pointers meet,
>>> which will depend on the distance m from the head to the cycle and the
>>> period p of the cycle. It is awkward to explain in words, so draw a picture
>>> and label it according to this discussion.
>>>
>>> Suppose the two pointers meet k nodes from the beginning of the cycle.
>>> Then the slow pointer has moved m + t*p + k nodes, where t is the number of
>>> times the slow pointer traverses the cycle, and the fast pointer has moved
>>> m + u*p + k nodes, where u is the number of times the fast pointer has
>>> traversed the cycle. The number of double steps of the fast pointer is (m +
>>> u*p +k)/2, and this equals the number of single steps the slow pointer has
>>> taken. Thus, m + u*p + k = 2(m + t*p + k), so k = (2*t - u)*p - m, i.e.,
>>> some number of times around the cycle minus m. Now start the fast pointer
>>> at the head and take m single steps with both pointers. The fast pointer is
>>> at the beginning of the cycle, and the slow pointer has traversed the cycle
>>> (2*t  - u) times and is back at the beginning of the cycle. The pointers
>>> couldn't be equal sooner since the fast pointer is not in the cycle after
>>> less than m steps and the slow pointer is in the cycle.
>>>
>>> Dave
>>>
>>> On Friday, March 9, 2012 4:18:14 AM UTC-6, rahul sharma wrote:
>>>
>>>> i have 2 pointers fast and slow.now if tehy meet there is a loop...
>>>>
>>>> now keep one ptr at meeting point and take other one to the begining of
>>>> listmove both at speed of one..they will meet at start of loophow
>>>> this happens???why they meet at start..plz tell logic behind this???thnx in
>>>> advance
>>>>
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Re: [algogeeks] Doubt in removing loop from linked list

[sanjiv yadav]

I think u r making mistake i.e. after 7,5 will com not 4 because loop
starts from 5.check it again

On 3/9/12, rahul sharma  wrote:
> @sanjiv...how can u take q at c.they will meet at some position ...
> suppose u have
>
> 1-2-3-4-5-6-7-4
>
> now initialy
>
> slow and fast both at 1
> start incrementing slow by 1 and fast by 2
>
> slow @2  ...fast @3
> slow @3fast @5
> s...@4...fast@7
> slow @5...fast@5
>
> both are equalmeans dere is loop.
>
> take anyone of the slow/fast say fast @ start i.e @1
>
> now we have s...@5...fast@1...
> incrementing bith by 1
>
> s...@6...fast@2
> s...@7fast@3
> s...@4..fast@4...
> they are equal..means we are at start of loop.
>
> i got the algo...but didnt get the logichow they are meeting @start.
>
>
> On Fri, Mar 9, 2012 at 7:42 PM, sanjiv yadav wrote:
>
>> suppose linked list is
>>
>> a->b->c->d->e
>>
>> and suppose loop starts from 'c'
>>
>> according to u let one pointer be at 'c' say *q and another be at 'a' say
>> *p. Now if we move both at the speed of one then
>>
>> After first pass
>>
>> p will be at b
>>
>> q will be at d
>>
>> After second pass
>>
>> p will be at c
>>
>> q will be at e
>>
>> After third pass
>>
>> p will be at d
>>
>> q will be at c
>>
>> After fourth pass
>>
>> p will be at e
>>
>> q will be at d
>>
>> After fifth pass
>>
>> p will be at c
>>
>> q will be at e
>>
>>
>> and so on.
>>
>> Correct me if i am wrong.
>>
>>
>> On Fri, Mar 9, 2012 at 7:28 PM, rahul sharma
>> wrote:
>>
>>> @terencei cant get..can u eleboratethnx for the sol..but plz
>>> elaborate...
>>>
>>>
>>> On Fri, Mar 9, 2012 at 5:59 PM, Terence  wrote:
>>>
>>>>  @ rahul sharma:
>>>> the linked list is a combination of a list a->b->...->p->q and a cycle
>>>> q->r->...->z->q. (z != p).
>>>> noting that the start of cycle q is the only node with 2 predecessor: p
>>>> and z.
>>>> if 2 pointers meet at some node x, different from q, in last step they
>>>> must have met at x', the predecessor of x.
>>>> the above logic holds for all nodes in cycle except q.
>>>>
>>>> @ sanjiv yadav:
>>>> They will meet at the start of loop.
>>>> ex.  a->b->c->d->e->c->d->e...
>>>> First round:
>>>> A: a->b->c->d
>>>> B: a->c->e->d
>>>> meet at d.
>>>> Second round:
>>>> A: a->b->c
>>>> B: d->e->c
>>>> meet at c.
>>>>
>>>>
>>>> On 2012-3-9 18:39, sanjiv yadav wrote:
>>>>
>>>> No They will not meet at the start in a case containing 5 nods and
>>>> having loop at the third node. once check this
>>>>
>>>> On Fri, Mar 9, 2012 at 3:48 PM, rahul sharma
>>>> wrote:
>>>>
>>>>> i have 2 pointers fast and slow.now if tehy meet there is a loop...
>>>>>
>>>>>  now keep one ptr at meeting point and take other one to the begining
>>>>> of listmove both at speed of one..they will meet at start of
>>>>> loophow this happens???why they meet at start..plz tell logic
>>>>> behind
>>>>> this???thnx in advance
>>>>>  --
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>>>>
>>>>
>>>>
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>>>> Regards
>>>>
>>>>  Sanjiv Yadav
>>>>
>>>>  MobNo.-  8050142693
>>>>
>>>>  Email Id-  sanjiv2009...@gmail.com
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Re: [algogeeks] A strange doubt with cpp class. Why compile error ?

[sanjiv yadav]

write..

Abc a=Abc()

it will execute...

On Fri, Mar 9, 2012 at 8:15 PM, ~*~VICKY~*~  wrote:

> #include
> using namespace std;
> class Abc
> {
>
> public :
> int i;
> Abc(){ i = 0;}
>
> };
> int main()
> { Abc a = new Abc();
> cout< }
>
>
> --
> Cheers,
>
>   Vicky
>
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Sanjiv Yadav

MobNo.-  8050142693

Email Id-  sanjiv2009...@gmail.com

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Re: [algogeeks] Doubt in removing loop from linked list

[sanjiv yadav]

suppose linked list is

a->b->c->d->e

and suppose loop starts from 'c'

according to u let one pointer be at 'c' say *q and another be at 'a' say
*p. Now if we move both at the speed of one then

After first pass

p will be at b

q will be at d

After second pass

p will be at c

q will be at e

After third pass

p will be at d

q will be at c

After fourth pass

p will be at e

q will be at d

After fifth pass

p will be at c

q will be at e


and so on.

Correct me if i am wrong.

On Fri, Mar 9, 2012 at 7:28 PM, rahul sharma wrote:

> @terencei cant get..can u eleboratethnx for the sol..but plz
> elaborate...
>
>
> On Fri, Mar 9, 2012 at 5:59 PM, Terence  wrote:
>
>>  @ rahul sharma:
>> the linked list is a combination of a list a->b->...->p->q and a cycle
>> q->r->...->z->q. (z != p).
>> noting that the start of cycle q is the only node with 2 predecessor: p
>> and z.
>> if 2 pointers meet at some node x, different from q, in last step they
>> must have met at x', the predecessor of x.
>> the above logic holds for all nodes in cycle except q.
>>
>> @ sanjiv yadav:
>> They will meet at the start of loop.
>> ex.  a->b->c->d->e->c->d->e...
>> First round:
>> A: a->b->c->d
>> B: a->c->e->d
>> meet at d.
>> Second round:
>> A: a->b->c
>> B: d->e->c
>> meet at c.
>>
>>
>> On 2012-3-9 18:39, sanjiv yadav wrote:
>>
>> No They will not meet at the start in a case containing 5 nods and having
>> loop at the third node. once check this
>>
>> On Fri, Mar 9, 2012 at 3:48 PM, rahul sharma wrote:
>>
>>> i have 2 pointers fast and slow.now if tehy meet there is a loop...
>>>
>>>  now keep one ptr at meeting point and take other one to the begining
>>> of listmove both at speed of one..they will meet at start of
>>> loophow this happens???why they meet at start..plz tell logic behind
>>> this???thnx in advance
>>>  --
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>>
>>
>>
>>  --
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>>
>>  Sanjiv Yadav
>>
>>  MobNo.-  8050142693
>>
>>  Email Id-  sanjiv2009...@gmail.com
>>
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Email Id-  sanjiv2009...@gmail.com

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Re: [algogeeks] Doubt in removing loop from linked list

[sanjiv yadav]

No They will not meet at the start in a case containing 5 nods and having
loop at the third node. once check this

On Fri, Mar 9, 2012 at 3:48 PM, rahul sharma wrote:

> i have 2 pointers fast and slow.now if tehy meet there is a loop...
>
> now keep one ptr at meeting point and take other one to the begining of
> listmove both at speed of one..they will meet at start of loophow
> this happens???why they meet at start..plz tell logic behind this???thnx in
> advance
>
> --
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Sanjiv Yadav

MobNo.-  8050142693

Email Id-  sanjiv2009...@gmail.com

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Re: [algogeeks] Re: Exchanging bit values in a number

[sanjiv yadav]

suppose we have to swap bits  b1 and b2  in a number n.Then

if(b1==b2)  do nothing

else take a hexadecimal number whose all bits are zero except bits b1 and
b2.bit b1=1 and bit b2=1

now  xor of original no with this no will give the desired result.

On Sun, Oct 30, 2011 at 10:32 AM, shiva@Algo  wrote:

> we need to swap only if both the bits are not same
>
> if((n^(1< n^=(1<
> On 10/29/11, praveen raj  wrote:
> > int func(int x)
> > {
> > int  y=(1< >  int z=x&y;// if after bitwise and  ..we get power of 2 then
> ...
> > we have to flip the bits..
> > if((z&(z-1))==0)
> >return(x^y);
> > else
> >   return x;
> > }
> >
> > With regards,
> >
> > Praveen Raj
> > DCE-IT
> > 735993
> > praveen0...@gmail.com
> >
> >>
> >>
> >
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Re: [algogeeks] Re: Exchanging bit values in a number

[sanjiv yadav]

suppose we have to swap bits  b1 and b2  in a number n.Then

if(b1==b2)  do nothing

else take a hexadecimal number whose all bits are zero except bits b1 and
b2.bit b1=1 and bit b2=1

now  xor of original no with this no will give the desired result.suppose
we have to swap bits  b1 and b2  in a number n.Then

if(b1==b2) do not do any thing

else take a hexadecimal number whose all bits are zero except bits b1 and
b2.bit b1=1 and bit b2=1

now  xor of original no with this no will give the desired result.

On Sun, Oct 30, 2011 at 10:32 AM, shiva@Algo  wrote:

> we need to swap only if both the bits are not same
>
> if((n^(1< n^=(1<
> On 10/29/11, praveen raj  wrote:
> > int func(int x)
> > {
> > int  y=(1< >  int z=x&y;// if after bitwise and  ..we get power of 2 then
> ...
> > we have to flip the bits..
> > if((z&(z-1))==0)
> >return(x^y);
> > else
> >   return x;
> > }
> >
> > With regards,
> >
> > Praveen Raj
> > DCE-IT
> > 735993
> > praveen0...@gmail.com
> >
> >>
> >>
> >
> > --
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Re: [algogeeks] Re: Exchanging bit values in a number

[sanjiv yadav]

suppose we have to swap bits  b1 and b2  in a number n.Then

if(b1==b2) do not do any thing

else take a hexadecimal number whose all bits are zero except bits b1 and
b2.bit b1=1 and bit b2=1

now  xor of original no with this no will give the desired result.

On Sun, Oct 30, 2011 at 10:32 AM, shiva@Algo  wrote:

> we need to swap only if both the bits are not same
>
> if((n^(1< n^=(1<
> On 10/29/11, praveen raj  wrote:
> > int func(int x)
> > {
> > int  y=(1< >  int z=x&y;// if after bitwise and  ..we get power of 2 then
> ...
> > we have to flip the bits..
> > if((z&(z-1))==0)
> >return(x^y);
> > else
> >   return x;
> > }
> >
> > With regards,
> >
> > Praveen Raj
> > DCE-IT
> > 735993
> > praveen0...@gmail.com
> >
> >>
> >>
> >
> > --
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Re: [algogeeks] Re: Missing elements

[sanjiv yadav]

it is really good algorithm

On 8/2/11, Deepak giggs  wrote:
> @varun : clear :) thanks...
>
> On Mon, Aug 1, 2011 at 11:44 PM, kartik sachan
> wrote:
>
>> similar type algo is there in geeksforgeeks.org in array section from
>> there u can understand it better way
>>
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> http://www.kurukshetra.org.in
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Re: [algogeeks] Amazon Written test Q-1

[sanjiv yadav]

node sort(node root)
{
node first;
int count=0;count1=0;count2=0;
for(first=root;first !=null;first=first->next)
{
if(first->data==1) count++;
if(first->data==2) count1++;
if(first->data==3) count2++;
}
first=root;
for(first;first !=null;first=first->next)
{
if(count>0)
{
first->data=1;
count--;
}
else if(count1>0)
{
first->data=2;
count1--;
}
else if(count2 >0)
{
first->data=3;
count2--;
}
return root;
}

On 8/2/11, lokesh  wrote:
> any other solution other than counting no. of 1s, 2s and 3s?
>
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