@senthilnathan : very nice indeed
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Rohit Saraf
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Dept. of Computer Science and Engineering
IIT Bombay
http://www.cse.iitb.ac.in/~rohitfeb14
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On Jun 7, 8:49 am, Senthilnathan Maadasamy
senthilnathan.maadas...@gmail.com wrote:
We can do it in O(n * log n) by individually binary-searching for zero on
each of the rows. Once we get the index of the first position where zero
appears, counting the number of negative number is
@ dheerraj...u cant measure 8 litre...u hve no additional instrument
@mohit...what do u mean by n th stageplzz elaborate
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@jitendra u got it right but parent and child are using the same text region
that's why control is transferring back and forth..
try running this code and see that line numbers are repeating...because of
same text region it is working like a loop...
#includestdio.h
int main() {
int id,i;
How will it be 12 12 5 6 6?? I can understand that the first number in
the list is place first so
it could be 12 12 6 6 5. How will it be 12 12 5 6 6?
On Jun 6, 7:47 am, divya jain sweetdivya@gmail.com wrote:
output willl be 12 12 5 6 6
On 6 June 2010 18:27, souravsain souravs...@gmail.com
@sharad..
sorry bt i dint get how to use bellman ford or topological sort here...
can u plz explain...
On 8 June 2010 05:53, sharad kumar aryansmit3...@gmail.com wrote:
for placing police man we can use bellman ford bfs.or even make use of
topological sort.
On Mon, Jun 7, 2010 at 9:59 PM,
What is the most efficient way of combining 2 separate ordered list
into a single ordered list ?
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Why do you want to try a brute force approach, when you have some better
algorithms and heuristics available?
On Mon, Jun 7, 2010 at 10:07 PM, Jean Carlo Mendes jean.men...@gmail.comwrote:
Hello Guys
Anyone have a implementation of knapsack 0-1 using brute force approach ?
Or… Do you
What do you mean by a stack or a queue in which each item is a
variable number number of integers?
Is it a queue of a queue, queue of a stack etc
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Actually, this might not always be the best approach. Example:
-1 1 2 3 4
1 2 3 4 5
1 2 3 4 5
1 2 3 4 5
1 2 3 4 5
2*N = 10 steps.
With my algo, you'll go:
Step 1: top-left: negative, count++,
Step2: [0][1] non negative, set limitRow=0 (or 1 depending on how you
code)
Step3: for([i][j]
@Divya: That was the question i previously asked. If n=3 000,001,010,100,101
are valid. So the solution for this is fib(n+2).
If n=4 no. of sequences will be fib(6) i.e 8
On Tue, Jun 8, 2010 at 9:31 AM, Rohit Saraf rohit.kumar.sa...@gmail.comwrote:
getting fibonacci nos is trivial using matrix
@saurabh: Hey u're right this is interesting. I checked for n=5 its giving
13 i.e fib(7)
On Mon, Jun 7, 2010 at 9:19 PM, saurabh gupta sgup...@gmail.com wrote:
it might be referring to no of sequences (say T(n) ) with no consecutive
1's
for n = 3, ans would be 5 viz.
000, 001, 010, 100, 101
yes even i dont think its possible as there is no n-1th state..ie there is
no state from whr u can come to 2 8 5..so plz check
On 7 June 2010 20:23, mohit ranjan shoonya.mo...@gmail.com wrote:
is it possible ?
if we say nth state is [2, 8, 5]
I could not find possible (n-1)th state
Mohit
i think both statements shd give error. as u r trying to change int to const
int in 2 and const int to int in 1..
On 7 June 2010 19:59, mohit ranjan shoonya.mo...@gmail.com wrote:
@Raj,
no they are not same
case 1: i is const
case 2: ptr is const
and whatever is const cann't be modified
Can you expain edge of the tree. I could not get that.
Anurag Sharma
On Tue, Jun 8, 2010 at 5:53 AM, sharad kumar aryansmit3...@gmail.comwrote:
for placing police man we can use bellman ford bfs.or even make use of
topological sort.
On Mon, Jun 7, 2010 at 9:59 PM, divya
how do u come to this formula T(n)=fib(n+2).. plz explain
On 7 June 2010 21:19, saurabh gupta sgup...@gmail.com wrote:
it might be referring to no of sequences (say T(n) ) with no consecutive
1's
for n = 3, ans would be 5 viz.
000, 001, 010, 100, 101
T(n) = fib(n+2)
where fib = Fibonacci
What is the time complexity of insertion and deletion in
1. Linked list
2. Queue
3. Queue implemented using a linked list
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@Mohit: If u're saying that in case 2 ptr is const then what is int *const
ptr. I thought this is a constant pointer. Constant pointer is one which
can't be made to point to any other address rit? How is *ptr++ coming into
the way of constant pointer ?
On Mon, Jun 7, 2010 at 7:59 PM, mohit ranjan
Two binary trees T1 and T2 are isomorphic if T1 can be transformed
into T2 swapping left and right children of the nodes in T1.Give an
algorithm to report whether 2 given binary trees are isomorphic.
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Actually the first statement i gave const int i=5; int *ptr=i is itself
giving an error on gcc and a warning on borland. We have to modify it as
const int *ptr=i otherwise it gives illegal pointer conversion error.
On Tue, Jun 8, 2010 at 12:11 PM, divya jain sweetdivya@gmail.comwrote:
i
I just checked out with so many possibilities and it is indeed matching. I
may not be correct though.
On Tue, Jun 8, 2010 at 7:50 PM, divya jain sweetdivya@gmail.com wrote:
how do u come to this formula T(n)=fib(n+2).. plz explain
On 7 June 2010 21:19, saurabh gupta sgup...@gmail.com
Use a n x n array.
initialize with -1 (don't care = no constraints). If there is an
equality constraint, set to 1, if
explicity non-equality constraint is expected, replace with 0. While
doing either of these if
you come across a situation where 0 has to be overwritten by 1 or 1
has to be
Check whether the first element in the array is greater than the last
element in the array. If it is true, it means the array is rotated after
sorting.
Now, take the middle element of the array and check whether it is greater
than the last element of the array. If true, it means the first half of
merging as in merge sort.
On 8 June 2010 19:59, Raj N rajn...@gmail.com wrote:
What is the most efficient way of combining 2 separate ordered list
into a single ordered list ?
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I found a nice explanation (from some other site) on how to get this formula
T(n) = fib(n+2).
Consider the set of all binary numbers of length n that end with 0. The
first n-1 positions can be anything (0 or 1). So if we take the set of all
binary numbers of length n-1 and append 0 to each of
Actually the solution is best thought of with recursion.
See, for a simple 4 element array: a1 a2 b1 b2, you can get the result
by swapping only the two elements in the middle.
Now think, if you had an 8 element array: a1 a2 a3 a4 b1 b2 b3 b4,
group these into pairs of 2 like: a1 a2 - b1 b2 / a3
Linked List: Insertion O(n) deletionO(n)
Queue using linked list : Insertion O(1) deletion O(n).
On Tue, Jun 8, 2010 at 1:38 AM, Raj N rajn...@gmail.com wrote:
What is the time complexity of insertion and deletion in
1. Linked list
2. Queue
3. Queue implemented using a linked list
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*What if there are no zero elements at all?*
*
*
*...@minotauraus Very valid question. The terminating condition for the
binary search can be modified to return the count of negative numbers in the
array instead. *
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One approach is to parse the element of the two tree in two arrays and
compare the arrays to fiind if they are isomorphic.
parsing takes O(n).
Comparision takes O(n/2)
for (i=0;in;i++)
{
if(2i+1 n)
break;
if(arr_1[2i] != arr_2[2i+1])
{
break;
}
}
if(i == n/2)
@Sharad
let's say that it will take n steps to reach from [15,0,0] to [2,8,5] then
after nth state will be 2,8,5
and (n-1)th state will be say [x,y,z] from which one transfer will lead to
o/p [2,8,5]
hope it's clear
Mohit Ranjan
On Tue, Jun 8, 2010 at 6:54 AM, sharad kumar
I dont think so
This approach is better than O(nlogn)
On Tue, Jun 8, 2010 at 9:10 AM, harit agarwal agarwalha...@gmail.comwrote:
@vadivel selvaraj
your approach is O(nlogn)
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You have a Numerator and Denominator. After division you might get a
recurring decimal points float as the answer.
Problem is: You need to identify the recurring part for a given
decimal no?
For example 23.34563456 ...
return 3456
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Min number of police man could be placed on the nodes which does not have
leaf nodes.
Only things needs to find out is the height of the tree.O(logn).
For any given tree: 1. calculate the leaf node: 2^h (h is the height of
the tree)
2. Subtract leaft nodes from the
bool isIsomorphic(List* h1,List* h2)
{
if(!h1 !h2) return true;
if(h1 h2)
return(h1-data == h2-data isIsomorphic(h1-left,h2-right)
isIsomorphic(h1-right,h2-left));
else return false;
}
On Tue, Jun 8, 2010 at 8:31 PM, divya sweetdivya@gmail.com wrote:
Two
I would say in most situations it is a stack of linked lists (queues).
I'd suggest for the sake of simplicity and problem solving, don't
think of it that way.
Think of it as a stack of 'objects'. These objects can be of any type.
Type can be an integer, a float, a pointer or an entire array.
Makes
This question was asked in a Microsoft interview 2 years back.
On Mon, Jun 7, 2010 at 8:05 PM, divya jain sweetdivya@gmail.com wrote:
let a[n][n] be the input array nd b[][] be the output
for(i=0;in;i++)
for(j=0;jn;j++)
b[i][j]=a[n-j-1][n-i-1]
On 7 June 2010 08:26,
Divide and Conquer approach works. Look up merge sort.
I don't know how the most efficient way. Actually I'm not sure if
there are many ways to do this.
It will take O(n) time either way (unless there are other specifics
given which we can make use of).
On Jun 8, 7:29 am, Raj N rajn...@gmail.com
edge is the link connecting two nodes of a tree
On 8 June 2010 11:23, Anurag Sharma anuragvic...@gmail.com wrote:
Can you expain edge of the tree. I could not get that.
Anurag Sharma
On Tue, Jun 8, 2010 at 5:53 AM, sharad kumar aryansmit3...@gmail.comwrote:
for placing police man we
Yeah, I didn't get the question either. Do you want to generate these
numbers or do you want to be able to tell if
the number input is valid or not? And how does Fibo. fit in the
picture either way?
On Jun 7, 8:13 pm, Dave dave_and_da...@juno.com wrote:
Hmmm. The first few Fibonacci numbers
@Raj,
Sorry for the confusion
yes, you are right that 1st one is giving warning/error
though for 2nd case
int i=5;
const int *ptr=i;
*ptr++;
i am nt getting any error/warning (gcc) and i remains 5
but
int i=5;
const int *ptr=i;
(*ptr)++;
is giving error
Mohit Ranjan
On Tue, Jun 8, 2010
recursion.
for length n if the count is T(n) then
T(n)
= 1st digit 0 + 1st digit 1
= T(n-1) + 1st digit 1 2nd digit 0
= T(n-1) + T(n-2)
QED.
On Tue, Jun 8, 2010 at 9:03 PM, Raj N rajn...@gmail.com wrote:
I just checked out with so many possibilities and it is indeed matching. I
may not be
@senthilnathan: Indeed a nice logic. Complexity is O(logn) worst case(if
all elements are negative).
http://codepad.org/hRbYV287
On Tue, Jun 8, 2010 at 8:18 AM, Minotauraus anike...@gmail.com wrote:
Actually, this might not always be the best approach. Example:
-1 1 2 3 4
1 2 3 4
Your time complexity is not O(c) but O(n^2)
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Second Year Undergraduate,
Dept. of Computer Science and Engineering
IIT Bombay
http://www.cse.iitb.ac.in/~rohitfeb14
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Fib comes because she wants the number of such sequences
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Second Year Undergraduate,
Dept. of Computer Science and Engineering
IIT Bombay
http://www.cse.iitb.ac.in/~rohitfeb14
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which is just the recursive version of the abovementioned iterative
solution.
P.S. -Please remove this quoted text when you are composing
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Second Year Undergraduate,
Dept. of Computer Science and Engineering
IIT Bombay
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