@ shashank and all
can we an approach that take less time than O(N^2). I mean not in O(N ) or
nlogn but n^2 some optimiation
On Thu, Aug 25, 2011 at 8:43 PM, WgpShashan)
k shashank7andr...@gmail.com wrote:
@Sharvan ..Yes We Can do That and yes i forgot that intead of
locals.add(a[i][j]) , we
Hey guys..
I know in this group all geeky stuffs goes on... but following lines of code
i foind it intresting and thah y wants ot sahre it with you all. Its very
basic but some may of you may not know,
int main()
{
int i=0;
if(i++)
printf(Hello);
else
printf(Bye);
return 0;
}
/*
ER-diagram is given and asking for min and max table. Can anyone
Explain me ..
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O\P-Bye I think so.plz explain how its hello
On Fri, Aug 26, 2011 at 12:12 PM, Suraj Fale surajfa...@gmail.com wrote:
Hey guys..
I know in this group all geeky stuffs goes on... but following lines of
code i foind it intresting and thah y wants ot sahre it with you all. Its
very
in java a tring object is immutable
but in following code
String s=java;
s+=c c++;
System.out.print(s);
the output is javac c++
why this is so
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think in 2d and I assume that both the sphere are touching each other.
it should be simple 2d maths now :)
On Aug 25, 8:32 pm, rakesh kumar rockey.rav...@gmail.com wrote:
Could you explain the solution for shere problem
On Thu, Aug 25, 2011 at 3:49 PM, vikas vikas.rastogi2...@gmail.com
Its true that in Java String are immutable. Even in the code given its
immutable.. Its actually the code executed like this..
String s=java; // S holds to java.
s+=c c++; //
Here S = S + c c++
Java + c c++ which is now assigned to the new variable which
here is s itself.
includestdio.h
void fun(int);
typedef int (*pf) (int, int);
int proc(pf, int, int);
int main()
{
int a=3;
fun(a);
return 0;
}
void fun(int n)
{
if(n 0)
{
fun(--n);
printf(%d,, n);
fun(--n);
}
}
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its been a while tat i had gave test bt i remebr quant was full of geometry
application so clear ur concept for geometry. and in logical they give this
kinda alphanumeric prob. my advice in quant and logical correct 3-4
questions out of 20 each u gonna get a gud percentile. for sure.
On Fri, Aug
Thats correct it will print 'Bye'... i just said Hello so everyone will
try to compile it and ask questions like you.
Thanks for showing intrest. Btw you can share such aptitude questions if u
have
On Fri, Aug 26, 2011 at 12:15 PM, Vijay Khandar vijaykhand...@gmail.comwrote:
O\P-Bye I
O\P= 1 2 1 3 1 2 1 is it correct?
On Fri, Aug 26, 2011 at 12:23 PM, SAMMM somnath.nit...@gmail.com wrote:
includestdio.h
void fun(int);
typedef int (*pf) (int, int);
int proc(pf, int, int);
int main()
{
int a=3;
fun(a);
return 0;
}
void fun(int n)
{
if(n 0)
{
as per my knowledge, in java
String s=java; // S holds to java.
s+=c c++; // now s is pointing to string javac c++ not the
appended string (java+c c++)
// strcat and append syntactically(o/p strings) are same but wrt to memory
they differ
Thank you,
Siddharam
On Fri, Aug 26, 2011 at
ar[0]=a;
ar[1]=b;
printf(largest %d,ar[((a-b)31)0x1]);
On Wed, Aug 24, 2011 at 11:20 PM, kranthi kumar damarlakran...@gmail.comwrote:
@priyanka raju, Good solution...
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Computer Science Engg.
IIT Madras.
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So, in short you are using a BST and a FIFO linked list. Whereas, a
Stack is actually a LIFO linked list. Am i right?
On Aug 25, 11:37 pm, Don dondod...@gmail.com wrote:
You will have to keep two pointers, one to the root of the tree and
one to the head of the FIFO linked list.
To push, insert
The O/P is 0,1,2,0,
But how
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For
o/p:- 0,1,2,0..
iIts simple stack concepts... during each recursive call it will goes on to
store 2,1,0 and then *it will print 0* . call it agin with -1 then
... its false then* it will print 1* and again it will become 0... its false
so it takes value from stack which is 2.. so *it will
Its O(n*n) algo.. so you have to just iterate n times.. please take an
exampla and solve it.. u'll get to know urself :)
On Aug 19, 3:14 am, Prakash D cegprak...@gmail.com wrote:
hey, thanks..
but if it needs many iteration, then we've to check each time whether the
array is sorted.. is there
abs function itself has a condition??
I think it is not that good solution.
On Wed, Aug 24, 2011 at 6:33 PM, priyanka raju priyark...@gmail.com wrote:
int a,b,max,min;
max=(a+b+abs(a-b))/2;
min=(a+b-abs(a-b))/2;
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cheers
priyanka
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Here is a simpler way to understand.
fun(0) = prints nothing
fun(1) = f(0) + print 0 + fun(-1) = 0
fun(2) = fun(1) + print 1 + fun (0) = 0 1
fun(3) = fun(2) + print 2 + fun(1) = 0 1 2 0
Karthik R,
RD Engineer,
Tejas Networks.
On Fri, Aug 26, 2011 at 1:02 PM, Suraj Fale surajfa...@gmail.com
i figured out algo to find the inorder predecessor of a bst without
using parent pointer... just wanna confirm if its missing any case
if the left child(subtree) of node exist, then predecessor ll be the
max value in the left subtree.
else predecessor ll be one of the ancestor in this
@Rohit.
Could you please tell wht was the marking scheme of the test?
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A book by 'Yashwant Kanetkar'
On Fri, Aug 26, 2011 at 9:03 AM, Navneet navneetn...@gmail.com wrote:
Ankit, i would also like to mention Mark Allan Weiss book on Data
Structures. Available in both C and C++ (different books)
On Aug 25, 3:59 pm, Abhishek mailatabhishekgu...@gmail.com wrote:
Posting few questions which Playdom asked a friend of mine in telephonic
interviews. He had to write code in all cases.
1. Print m*n matrix in spiral form. Modify to make it iterative.
2. Print level order traversal or tree along with level information. Tree
need not be complete
a 1
b c
Thank you,
Siddharam
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For more options,
1.
int top,bottom,left,right,size;
void printTop(int arr[][size])
{
int i=top,j;
for(j=left;j=right;j++)
printf(%d ,arr[i][j]);
}
void printBottom(int arr[][size])
{
int i=bottom,j;
for(j=right;j=left;j--)
printf(%d ,arr[i][j]);
}
void printLeft(int arr[][size])
{
for ques 3:
]1) If right subtree of node is not NULL, then succ lies in right subtree.
Do following.
Go to right subtree and return the node with minimum key value in right
subtree.
2) If right sbtree of node is NULL, then succ is one of the ancestors. Do
following.
Travel up using the parent
One thing i should mention in ques3, you have the parent pointer of node.
(so left, right and parent pointers can be used)
On Fri, Aug 26, 2011 at 3:22 PM, Navneet Gupta navneetn...@gmail.comwrote:
Posting few questions which Playdom asked a friend of mine in telephonic
interviews. He had to
For ques 2:
./*Function to print level order traversal of tree*/
printLevelorder(tree)
for d = 1 to height(tree)
{
printGivenLevel(tree, d);
printf( %d\n,d);
}
/*Function to print all nodes at a given level*/
printGivenLevel(tree, level)
if tree is NULL then return;
if level is
@tech coder:
When u execute : s+=c c++;
a new string object is created with the value of s appended with c c++.
Now s is made to hold the reference of this new string object. Thus the
older string object is not modified, but a new string string object is
created.
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is it not possible to traverse tree in order and store in array. then figure
out the element and print the previous element?
On Fri, Aug 26, 2011 at 2:04 PM, Vikram Singh singhvikram...@gmail.comwrote:
i figured out algo to find the inorder predecessor of a bst without
using parent pointer...
count no of 0s,1s and 2s.fill first 0s followed by 1s den 2s :)
On Thu, Aug 25, 2011 at 11:37 PM, icy` vipe...@gmail.com wrote:
not enough information, imo. Tell me more about the given string...
is the string made up of consecutive integers/characters ? Are there
always the same number of
ya thats one option but that gives ans in O(n), requires additional
memory... and unnecessarily finds for all which is not required...
my sol doesnt require any extra space i.e. in O(1) space... and also
in O(log n) time...
tell if dere is any missing case
On Aug 26, 4:58 pm, sukran dhawan
Vikram : will u plz elaborate more on ur solution ?
Sanju
:)
On Fri, Aug 26, 2011 at 5:24 AM, Vikram Singh singhvikram...@gmail.comwrote:
ya thats one option but that gives ans in O(n), requires additional
memory... and unnecessarily finds for all which is not required...
my sol doesnt
hmmm k
On Fri, Aug 26, 2011 at 5:54 PM, Vikram Singh singhvikram...@gmail.comwrote:
ya thats one option but that gives ans in O(n), requires additional
memory... and unnecessarily finds for all which is not required...
my sol doesnt require any extra space i.e. in O(1) space... and also
in
i m writing just a pseudocode...
// root is the root of treeand node is the node whose
predecessor is to be found
predecessor(root, node)
{
parent=NULL;
if(root==NULL)
return ;
if(node-left!=NULL)
{
// find max value in left subtree...
}
else
{
while(root!=NULL
got it Vikram :)
Sanju
:)
On Fri, Aug 26, 2011 at 5:43 AM, Vikram Singh singhvikram...@gmail.comwrote:
i m writing just a pseudocode...
// root is the root of treeand node is the node whose
predecessor is to be found
predecessor(root, node)
{
parent=NULL;
if(root==NULL)
@Neha,
For your solution of 1, you needed to take m*n matrix into account.
Though i think your approach should work fine even in that case.
Function signature will change.
For 3rd solution, the corner cases actually little non-trivial while
writing code(need to consider root, root with no right
Also for ques 1, instead of having four auxiliary functions, you can
have a variable flag for rows and cols functions being passed as param
and reduce the number to two.
On Aug 26, 6:07 pm, Navneet navneetn...@gmail.com wrote:
@Neha,
For your solution of 1, you needed to take m*n matrix into
lol :P
On Wed, Aug 10, 2011 at 11:35 PM, $hr! k@nth srithb...@gmail.com wrote:
Tie the rope at the top of the tower
Climb down with the help of the rope up to 100 mt peg possItion
Tie the rope to that peg, Climb up to the top of the tower with that rope.
Now release the rope at the top and
Yayashwant kanitekar and deepali srivastava are great books
If you find all the mistakes that they have made you have learned DS(and
coding style) quite well.:p
On Fri, Aug 26, 2011 at 2:44 PM, Suraj Fale surajfa...@gmail.com wrote:
A book by 'Yashwant Kanetkar'
On Fri, Aug 26, 2011
varun: can u explain it little further..
On Wed, Aug 10, 2011 at 7:49 PM, varun pahwa varunpahwa2...@gmail.comwrote:
make two ropes 50m and 100 meter. make a loop kind of thing with that now
you have two 50 mtr ropes so get down to 100 mtr point and tie loop rope in
downward now cut the loop
no it results in segmentation fault
On Fri, Aug 26, 2011 at 3:48 PM, siddharam suresh
siddharam@gmail.comwrote:
Thank you,
Siddharam
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but its its valid address right
Thank you,
Sid.
On Fri, Aug 26, 2011 at 8:16 PM, SANDEEP CHUGH sandeep.aa...@gmail.comwrote:
no it results in segmentation fault
On Fri, Aug 26, 2011 at 3:48 PM, siddharam suresh siddharam@gmail.com
wrote:
Thank you,
Siddharam
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wat are u tying to say :P ? is that a good book or wat ?
On Fri, Aug 26, 2011 at 7:55 PM, saurabh singh saurab...@gmail.com wrote:
Yayashwant kanitekar and deepali srivastava are great books
If you find all the mistakes that they have made you have learned DS(and
coding style) quite
i think it is reserved for system functions and you are not allowed to work
as your program is not alloted that area if somehow your program is allowed
that area there will be no segmentation fault
On Fri, Aug 26, 2011 at 7:57 AM, siddharam suresh
siddharam@gmail.comwrote:
but its its valid
thanks to all got it
On Fri, Aug 26, 2011 at 4:54 AM, Neha Singh neha.ndelhi.1...@gmail.comwrote:
@tech coder:
When u execute : s+=c c++;
a new string object is created with the value of s appended with c c++.
Now s is made to hold the reference of this new string object. Thus the
older
Please share Process and experience of tejas Network for software
profile.
Thanks
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Usually that is done before run time and hard coded into the hash
function.
But for numbers in the range you are talking about, a sieve would
work. However trial division should not take long either, if you just
need to do it once when the hash table is created.
Don
On Aug 25, 10:18 pm, Navneet
0 is reserved as the address which can't be referenccd
On Fri, Aug 26, 2011 at 9:08 PM, UTKARSH SRIVASTAV
usrivastav...@gmail.comwrote:
i think it is reserved for system functions and you are not allowed to work
as your program is not alloted that area if somehow your program is allowed
After working on it quite a bit I got an O(log n) algorithm working.
For small cases (size 10) it sequentially finds the solution.
For larger cases it uses a binary search:
Starting at the midpoint, find the left and the right ends of the
region with values equal to the value at the midpoint.
Do anyone have ideas regarding Sandisk technical and interview
questions...?? or any link to refer..??
Thanks in advance
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To
it can be done in O(N) by using XOR ing the elements
1: Xor all the elemnts since those elemnts that even freq will nullify each
other we get number taht will tell in which the two required number differ.
2: divide the array in two sets on the basis of bit in which numbers
differ
3:1 element
how to divide an integer array into 2 sub-arrays and make their averages
equal?
array is unsorted and we can also take any numbers and the numbers in the
array need not be contiguous in the original array.
how many total such array's are possible. Output them
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Given n unsigned integer, output 2 integers which has the maximum result
after XOR
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hi guys...microsoft is coming to our campus..plz nyone tell their
recruitment procedure..n give me their previous xams question if nyone
has...but please tell me their selection procedure n
roundscuming after 2 days.plz reply soon.n tell me the
subjects to crack
yeah...just clear concept of recursion...its simple form test ur c
skilss..(function.)
On Aug 26, 1:22 pm, kARTHIK R k4rth...@gmail.com wrote:
Here is a simpler way to understand.
fun(0) = prints nothing
fun(1) = f(0) + print 0 + fun(-1) = 0
fun(2) = fun(1) + print 1 + fun (0) = 0 1
fun(3)
Sort the nos., which can be done in O(nlogn)
Now the 1st and the last integers are the required integers.
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college?
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@Neha take 42, 21 and 1
42 ^ 1 =43
while 42 ^21 =63
On Fri, Aug 26, 2011 at 10:28 PM, Neha Singh neha.ndelhi.1...@gmail.comwrote:
Sort the nos., which can be done in O(nlogn)
Now the 1st and the last integers are the required integers.
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thaparit is 100% yet sure.but it is 99% sure.i wana
prepare early...
On Aug 26, 10:00 pm, Neha Gupta nehagup...@gmail.com wrote:
college?
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@Tech: I'm not sure I understand your algorithm. Let's try it on
{1,1,2,2,3,4,5,5,6,6,7,7}. The two number occurring an odd number of
times are 3 and 4. We xor the numbers getting 7 = 111 in binary. Now
how do we divide the numbers into two groups?
Dave
On Aug 26, 11:09 am, tech coder
The function of that flag would be to decide direction (bottom up or
top down/ left to right or otherwise)
On Aug 26, 6:15 pm, Navneet navneetn...@gmail.com wrote:
Also for ques 1, instead of having four auxiliary functions, you can
have a variable flag for rows and cols functions being passed
@Rajesh: abs() can be done without conditional operations. There
probably are many ways. The first two that come to mind are:
abs(x) = (x 31) x | ~(x 31) ~x
abs(x) = (x 31) ^ x + (x 31) 1
Dave
On Aug 26, 2:56 am, rajesh singarapu rajesh0...@gmail.com wrote:
abs function itself has a
And in my college , samsung is giving 6.25lpa whereas winshuttle 7lpa with
dream status.. thats y a bit confused :O
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Here is a problem:
Given an array of size n. Find all the MAXIMAL subsets whose sum is = K.
The solution is not a concern, the optimization is required.
-piyush
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I believe this is what techcoder is saying:
int a[N];
// Find the bitwise xor of all the array values.
// These are the bits which are different between the two results.
int xor = 0;
for(i = 0; i N; ++i)
xor ^= a[N];
// Find the low order bit of xor
int bit = 1;
while(!(xor bit))
bit = 1;
take h as array/has table as max element of a as size.
just a psecudeocde.
for(i=0;in;i++)
{
sub=k-a[i];
if(h[a[i]]==1)
return (a[i],sub);
h[a[i]]=1;
}
log n complexity for all subsets of sum=k;;;
just add one more condition in loop for checkeing also...
coreect if m wrng??
On Aug
yeah can be done in poly tym also...but we dnt knw whether we have
unsorted arryit is possible in sorted array.
On Aug 26, 10:52 pm, Don dondod...@gmail.com wrote:
This is the knapsack problem.
Find a polynomial-time solution and you will be a hero.
Don
On Aug 26, 12:43 pm, Piyush Grover
XOR all the elements in the array, the result will be the XOR of the two
numbers occuring odd number of times.
Now take any set bit of th result(u can determine the position of any bit
set in the number). Divide the array such that for the numbers for which at
this location(where the bit is set
@rahul...I'm unsure if your algo returns all the subsets.
On Fri, Aug 26, 2011 at 11:24 PM, rahul sharma rahul23111...@gmail.comwrote:
yeah can be done in poly tym also...but we dnt knw whether we have
unsorted arryit is possible in sorted array.
On Aug 26, 10:52 pm, Don
@Tech, Don: How about this: given n and array a[n]:
int x = 0, result[2] = {0};
for( i = 0 ; i n ; ++i )
x ^= a[i];
x |= x(x-1); // low order 1-bit of xor
for( i = 0 ; i n ; ++i )
result[a[i]x?0:1] ^= a[i];
Dave
On Aug 26, 12:49 pm, Don dondod...@gmail.com wrote:
I believe this is
yes it will
return in c return 1 value at tym...
ijust given the code snipetjust modify it..store trhm in some
other array like thebut it will
On Aug 26, 11:02 pm, Piyush Grover piyush4u.iit...@gmail.com wrote:
@rahul...I'm unsure if your algo returns all the subsets.
On Fri,
A certain number of men can finish a piece of work in 10 days. If however
there were 10 men less it will take 10 days more for the work to be
finished. How
many men were there originally?
(a) 110 men
(b) 130 men
(c) 100 men
(d) none of these
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20?
Yanan Cao
On Fri, Aug 26, 2011 at 1:12 PM, priya ramesh
love.for.programm...@gmail.com wrote:
A certain number of men can finish a piece of work in 10 days. If however
there were 10 men less it will take 10 days more for the work to be
finished. How
many men were there originally?
@rahul
Your code will only find pairs which sum to k. The problem is to find
a subset of as many elements in the array as required to sum as close
as possible to k.
It is a well-known problem and after years of study, no polynomial
solution is known. There are reasonably fast solutions for small
i too got the same ans.
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20 men ?
(x-10)20=10x
Sanju
:)
On Fri, Aug 26, 2011 at 11:17 AM, gmagog...@gmail.com
gmagog...@gmail.comwrote:
20?
Yanan Cao
On Fri, Aug 26, 2011 at 1:12 PM, priya ramesh
love.for.programm...@gmail.com wrote:
A certain number of men can finish a piece of work in 10 days. If however
@yanan how it is 20.
Rahul Verma
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the answer is 130 itseems. i too got 20
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For
@Rahul
Assume the productivity of each man is the same
let original number of man be x
The total workload= x*10*p
also workload = (x-10)(10+10)*p
solve it
so x=20
Yanan Cao
On Fri, Aug 26, 2011 at 1:21 PM, Rahul Verma rahul08k...@gmail.com wrote:
@yanan how it is 20.
Rahul Verma
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Sry for prrevious post .
It is question of R S Aggarwal, and solved using conecpt of direct and
indirect proportions.
Originally,let there be x men.
Less men,more days(indirect proposition)
(x-10) : x :: 100 : 110
(x-10) * 110 = x * 100
ANS : x = 110
Sanju
:)
On Fri, Aug 26, 2011 at
In this question, instead of 10, it is 100.
Check it again.
Sanju
:)
On Fri, Aug 26, 2011 at 11:26 AM, Sanjay Rajpal srn...@gmail.com wrote:
Sry for prrevious post .
It is question of R S Aggarwal, and solved using conecpt of direct and
indirect proportions.
Originally,let there be x
If it is 10 days originally, answer is 20 for sure.
Sanju
:)
On Fri, Aug 26, 2011 at 11:27 AM, Sanjay Rajpal srn...@gmail.com wrote:
In this question, instead of 10, it is 100.
Check it again.
Sanju
:)
On Fri, Aug 26, 2011 at 11:26 AM, Sanjay Rajpal srn...@gmail.com wrote:
Sry
it's basic unitary method problem:
x men do work in 10 days
1 man will do-in 10*x days
x-10 men do it in 10*x/(x-10) = (10+10)
Solve it and x = 20
it can't be 130
On Fri, Aug 26, 2011 at 11:53 PM, gmagog...@gmail.com
gmagog...@gmail.comwrote:
@Rahul
Assume the productivity of each man
oh thanks :) It's given as 10 only. Printing mistake i guess.
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@Tech: I'm not sure I understand your algorithm. Let's try it on
{1,1,2,2,3,4,5,5,6,6,7,7}. The two number occurring an odd number of
times are 3 and 4. We xor the numbers getting 7 = 111 in binary. Now
how do we divide the numbers into two groups?
see we come to know that both number differ at
@ Don exactly waht u write i wanted to say
On Fri, Aug 26, 2011 at 11:52 AM, tech coder techcoderonw...@gmail.comwrote:
@Tech: I'm not sure I understand your algorithm. Let's try it on
{1,1,2,2,3,4,5,5,6,6,7,7}. The two number occurring an odd number of
times are 3 and 4. We xor the numbers
i am getting repeatdly WA in this question link is
https://www.spoj.pl/problems/ACODE/
plzz provide me some test cases where my code
fails.
my code is
# includecstdio
# includecstring
int main()
{
while(1)
{
char a[50100];
scanf(%s,a);
I hope you dont mind that I respond to the original question about the
6x6 matrix. As I understand it, all elements have to be either 1 or
-1, and product of *every* row and *every* column is 1 = how many
arrangements?
Now a bunch of you seem to think(nxn) = 2^((n-1)^2) gives the
answer,
The 0-1 knapsack problem is still the knapsack problem.
Don
On Aug 26, 1:55 pm, Piyush Grover piyush4u.iit...@gmail.com wrote:
it's similar to knapsack but not the same. In knapsack, types of items are
limited and we play on the quantity of each item.
Here each element will come once in the
@Umesh I really appreciate your solution and thinking to understand the
complexity of the program.
Actually I don't have that much idea about the *how to calculate complexity
of any program*. So could you please show some light on the evaluation
procedure of complexity.
Rahul Verma
--
You
Cut the rope in 50mtrs and 100mtrs length.
Make a small loop(of negligible length at one end of the 50 mtrs rope)
Tie the other end of the rope at the top and from the loop end side pass the
100mtrs rope
such that you have both the ends of 100mtrs rope in your end.
now get down at 100mtrs peg
Does anyone know about the recruitment procedure for Research Intern at MS
Redmond or MS Bangalore ? What are the type of questions do they ask and
what should be paid attention to ?
There are lot of posts on MS SDE and SDET positions but none about research.
If anyone knows then kindly reply.
Other than making little loops and risking the fall on the first trip
down, I dont think the rope question has an answer. NVIDIA just
wanted to see if you were suicidal =D
On Aug 26, 3:36 pm, Piyush Grover piyush4u.iit...@gmail.com wrote:
Cut the rope in 50mtrs and 100mtrs length.
Make a
i hope now it clear:
[image: Screenshot.png]
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If the seeker wants to do entire self study , then I would suggest CS
106B Reader , it's in C ++ , BUT ideal for people who rely only on
internet. ALso that Data STructure USing C ,Book by Y. KAnitikar , I
never saw any math in any chapters for analysis of algorithms they
write
. lol. Google it
Check this out:
Tie it at the 200th meter mark. Throw the 150mt rope down.
Climb down to the 100th meter pole. Tie the rope there from the middle, and
not the end.
So what you have is a 150 mt rope that is tied at 200 mt mark, 100 mt mark
and 50 mts of the rope from 100 mt marks is hanging.
It's not knapsack in knapsack we find max or min subset here we have to find
all subsets =k not just one which is min or max , so I guess we have to
form all subsets and check their sum hence the algoritm will be 0(2^n) where
n is number of elements in the set ,
please correct me if i am wrong or
@don really cool algo man
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For more options,
thanks everybody for so many suggestions... :)
On Sat, Aug 27, 2011 at 6:40 AM, Rahul raikra...@gmail.com wrote:
If the seeker wants to do entire self study , then I would suggest CS
106B Reader , it's in C ++ , BUT ideal for people who rely only on
internet. ALso that Data STructure USing C
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