You are assuming is to be a binary tree, its not. Some nodes will
share a common pour.
On Feb 25, 9:24 pm, atul anand atul.87fri...@gmail.com wrote:
i guess this would work...
n=number of nodes
h=height;
pour=quantity poured;
capacity = capacity of each cup
n=pow(2,h+1) -1;
Hi
I may not have understood your solution properly. But i think that
your solution is an implementation of brute force where you are
dealing with all cases valid in the range n/2k and 3n/2k without any
optimization with regard to DP. Am i right?
On Nov 28, 2:17 am, sourabh
plz post a samle input output..
On Nov 4, 10:24 am, sunny agrawal sunny816.i...@gmail.com wrote:
What can be a better solution than a Brute Force O(N^2* No of iteration)
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B.Tech. V year,CSI
Indian Institute Of Technology,Roorkee
fb1_input.txt
1KViewDownload
is ur brute force O(1) for space?
On Nov 4, 10:24 am, sunny agrawal sunny816.i...@gmail.com wrote:
What can be a better solution than a Brute Force O(N^2* No of iteration)
--
Sunny Aggrawal
B.Tech. V year,CSI
Indian Institute Of Technology,Roorkee
fb1_input.txt
1KViewDownload
I think Dan's solution is the best one here. TC O(n log n) and SC O(1)
where n is the maximum no. of elements in any array
Instead of starting with K given arrays, just take the first 2.
Sort both of them - time is nlogn
Now run two pointers on each array to save the common elements as they
are
How are we going to write a program which if fed the data gives the
answer?
Any ideas for the algorithm?
My approach:
We have a graph here.
We have vertices with indegree 0 and outdegree 1. - call it set A
(start points) (m vertices)
We have vertices with indegree 1 and outdegree 0. - call it
@bharat:
for the second part where u multiplied (160+5) with 999, it should be
160*999 because it is max of (150,120,160,140,5). Correct me if i am
wrong.
On Sep 26, 4:02 pm, bharatkumar bagana bagana.bharatku...@gmail.com
wrote:
for the first instruction : 150+5+120+5+160+5+140=585 ns
for the
@Victor:
Instead of 0 and 1, shouldn't it be like DP[i-1][j-1] + 0 and DP[i-1]
[j-1] + 1
On Sep 7, 1:10 am, Victor Manuel Grijalva Altamirano
kavic1.mar...@gmail.com wrote:
Try with DP, a little modicated of Edit Distance algorithm
State i=the begin of the word , j=the end of the word
how to check if intermediate words are dictionary words or not?? Also
what would be the approach for that?
On Aug 31, 3:10 am, icy` vipe...@gmail.com wrote:
Yea, I hope the intermediate words are dictionary words; it's more
fun that way. I played such a game before, where you and a friend
Level Order traversal if you are ok with the Nulls being stored.
Otherwise its pre order traversal.
On Aug 30, 12:34 pm, Vidit Sinha vidit.it...@gmail.com wrote:
whats the final answer guys? Level order traversal??
On Tue, Aug 30, 2011 at 12:56 AM, Don dondod...@gmail.com wrote:
I'm
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
Any help regarding the preprocessing required for O(1) operation for
LCA??? I read the wiki's article involving eulers traversal.. but its
nt clear.
On Aug 24, 1:12 am, DK divyekap...@gmail.com wrote:
Hmm.. Interesting question.
*Here's a solution:*
There are n elements in the tree and
So, suggest some approach plz for this ques.
On Aug 22, 12:59 am, Dave dave_and_da...@juno.com wrote:
@Sanjay: Will method 1 really work? What would it return on the string
abraxyzarba? The longest common substring between this string and
its reverse is abra, but how does that relate to the
first option is correct. Because we are not allowed to use any extra
space.
On Aug 21, 6:36 pm, priya ramesh love.for.programm...@gmail.com
wrote:
A Most efficient data structure is designed to optimize the following
operations. Pop, Push, Min. The best possible time-complexities with no
extra
We can't use a heap. Balanced BST is correct because Deletion of the
smallest element Insertion of an
element if it is not already present in the set - for this we need
to search for the element and searching in heap is O(n).
On Aug 21, 6:16 pm, priya ramesh love.for.programm...@gmail.com
wrote:
Check interpolation which is a substitute for binary search for a
uniformly sorted array. It has complexity of this order.
On Aug 3, 7:31 pm, Ajai Sathyan ajaisath...@gmail.com wrote:
Can you suggest a sample program which has complexity O(log log n)?
Regards
Ajai Sathyan
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Given an array of length n of integer numbers. Output 1 or 0 depending
on whether or not there exists a sub sequence in the array with sum S.
Suggest a fastest algorithm plz.
e.g. say given an array with numbers as 10 20 30 40 50 60 70
Sum = 110
Output is 1 because 20+30+50 is 110.
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You
Use multilevel indexing
On Jul 27, 11:07 pm, himanshu kansal himanshukansal...@gmail.com
wrote:
if u hv say 20 million records and u have to create a b+ tree then you
might be storing 20 million pointers at the leaf levelhow can u
optimize this(using b+ tree only)???
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check the above solution.. urs is O(logn) mine is O(log k).
On Jul 19, 11:35 pm, sagar pareek sagarpar...@gmail.com wrote:
well i dont think so...any better solution will be there compare mine one
On Wed, Jul 20, 2011 at 12:00 AM, Dumanshu duman...@gmail.com wrote:
heres the code
Given an infinite length list. u got to find index of an element k.
use this approach-
initially, take length as 2^x where x increases from 1 to ...
while (still not found)
{
now if arr[2^x-1] k,
increment x
else
binarysearch on length 2^(x-1) to 2^(x)
}
Please help me to find the
Describe
an
algorithm
that
takes an
unsorted
array
of
axis‐
aligned
rectangles
and returns
any
pair
of
rectangles
that
overlaps,
if
there
is such
a
pair.
Axis‐aligned means
that
all
the
rectangle
sides
are
either
parallel
or
perpendicular
to
the
x
and y‐axis.
You
that is why m confused. how would u rate this algo? in what order?
On Jul 20, 10:44 pm, Ankur Khurana ankur.kkhur...@gmail.com wrote:
when n is not defined , you want complexity in terms of ?
On Wed, Jul 20, 2011 at 3:16 PM, Dumanshu duman...@gmail.com wrote:
Given an infinite length
it should involve the exponential increment! somebody plz help
On Jul 20, 10:56 pm, Ankur Khurana ankur.kkhur...@gmail.com wrote:
in my opinion , it is log(indexof(k))
On Wed, Jul 20, 2011 at 11:23 PM, Dumanshu duman...@gmail.com wrote:
that is why m confused. how would u rate this algo
..)
- suppose you found a certain node at position n whose node-data
k , then now u only have to search for k between index n/2 to n (i.e.
if you found 16th element k then search for k between 8th and 16ht
element)..
correct me if any flaws.
On Jul 19, 3:38 am, Dumanshu duman...@gmail.com
What is the answer to the first one?
On Jul 19, 10:07 am, sagar pareek sagarpar...@gmail.com wrote:
1. Let's say
S=D=N
repeat
D=(floor)D/2;
S=S-D;
till D=0
From the above logic, figure out which algorithm is followed by 'S'
2. How can you divide a triangle, to 5 equal triangles?
surender
On Tue, Jul 19, 2011 at 4:06 AM, Dumanshu duman...@gmail.com wrote:
@Gaurav: The best solution would be to manipulate the given BTree in
place and get the BST. We don't need a separate tree but convert the
existing one using the same space occupied by nodes of BT already
heres the code
http://ideone.com/rX130
On Jul 19, 9:35 pm, swetha rahul swetharahu...@gmail.com wrote:
Hi,
Find the kth smallest element in O(logk) given 2 sorted arrays.
Merging the arrays is not allowed. I can do it in O(k).. How to do in
O(logk)..
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Given billions of unsorted numbers and we need to find first 1 million
numbers if all the numbers were sorted in non-descending order.
approach is to split up the numbers to multiple machines and sort
them.
Then make a tournament tree in such a way that each machine feeds the
lowest number it has,
Hi
sry for not being so clear with the problem statement.
The list of numbers may have repetitions. Lots of numbers can be
missing from the list, you just need to output any one.
Regards
Dumanshu
BITS-Pilani
On Jul 18, 5:28 pm, ankit sambyal ankitsamb...@gmail.com wrote:
The question says
Quadratic equation solver: just solve it the way we do on paper, take
coefficients as input and apply formula x = (-b+sqrt(b^2 - 4*a*c))/
2*a...
anything wrong with this?
And for the given problem, it says atleast once and not exactly
once. So, Ankit's solution of bit vector is the best one.
On
You are given a long string and you have to print the first non
repeating character.
Solve it keeping SC and TC in mind. That is present both solutions,
one with high SC and low TC and viceversa.
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here it is -http://rosettacode.org/wiki/MD5
It has the C implementation if u need in C.
On Jul 18, 6:09 pm, Anuj kumar anonymize...@gmail.com wrote:
some body can give me code of MD5
PLZ its Urgent ..
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You have a sorted linked list of integers of some length you don't
know and it keeps on increasing. You are given a number k. Print the
position of the number k.
Basically, you have to search for number k in the ever growing sorted
list and print its position.
Please write the complexity of
Lets say DLL is sorted. We have to convert the DLL so basically its
the manipulation of pointers.
Can we do it recursively?
//n is the total number of nodes in the list.
// next corresponds to right and prev corresponds to left pointer.
Node * BuildBalBST(Node *head, int n)
{
if(!n)
return NULL;
1. //BT to BST - function used is to swap values
Node* bubbleData(Node *root)
{
if(!root)
return NULL;
if(root-right)
{
if(root-data root-right-data)
swap((root-data),(root-right-data));
root-right = bubbleData(root-right);
}
if(root-left)
{
if(root-data
heres my solution with TC O(n) and SC O(26)
input string str
int arr[26] = {0};
traverse the string, character by character and increment the
corresponding counter.
i.e. arr[str[i]]++;
Now traverse the string again and print out the first character
encountered whose arr[str[i]] == 1;
On Jul 18,
Sry, I made a mistake in my code for problem 1 to convert BT to BST
Ignore it ... its wrong
any case missing??
3. Do we have to give an algorithm for garbage collection like Mark
sweep algo or Do we have to write a code? if code, plz tell how to
write?
On Jul 18, 4:59 pm, Balaji S
);
}
}
}
On Jul 18, 7:24 pm, Dumanshu duman...@gmail.com wrote:
1. //BT to BST - function used is to swap values
Node* bubbleData(Node *root)
{
if(!root)
return NULL;
if(root-right)
{
if(root-data root-right-data)
swap((root-data),(root-right-data));
root-right
. else if (ptr2-datak) set ptr1=ptr2 goto 2
5. else traverse the ptr1 upto ptr2, if found return its position else
return fail
if anyone has more efficient solution then pls tell :)
On Mon, Jul 18, 2011 at 6:53 PM, Dumanshu duman...@gmail.com wrote:
You have a sorted linked list
:12 pm, surender sanke surend...@gmail.com wrote:
@Dumanshu it also doesn't works, as min node doesn't satisfies bst
conditions, u swap it but it again creates inconsistencies with its left
subtree.
void binarytreetobst(btree *root)
{
if(root == NULL) return;
else if(root
@Balaji: for third question, were u asked to write the code?
On Jul 18, 10:04 pm, Balaji S balaji.ceg...@gmail.com wrote:
wats the mistake..
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You have to use parallel computing to find the first repeating number
in O(n) time if theres nothing special about the 2-D array
Use n +1 processes, n processes to scan each row and 1 process to scan
the rows last and next rows first element to check for repetition.
Each process uses hash table to
traverse using recursion and instead of printing it
just pass to function which is making BST??
and is this right as someone above said first sort it and then make BST...
dont we want that root of both Tree to be same or something like that...??
On Tue, Jul 19, 2011 at 2:17 AM, Dumanshu
, Dumanshu duman...@gmail.com wrote:
@Swathi: plz give the TC of ur algo (exponential plus log)
On Jul 18, 10:14 pm, Swathi chukka.swa...@gmail.com wrote:
The solution to this problem will be a combination of exponential increase
and the binary search..
start = 0;
end = 0;
index =0
@Ankit: this particular problem can be solved with min heap
implementation for tournament tree concept.
But i want to code the actual tournament tree, where we can also find
out the winner over all and all other contestants that lost to the
winner.
how to code that?
Regards
Dumanshu
BITS-Pilani
Given billions of integers in a file. Memory Constraints exist so need
to parallelize. One solution can be to split the file over a 100
machines and each machine calculates median of their part using
quickselect and sends the result to host machine. Now the host
calculates median of these medians
plz give some example..sry but what do you mean by child sibling
version??
On Jul 16, 3:46 pm, Reynald reynaldsus...@gmail.com wrote:
Given a Parent -Child binary tree ,build the child -sibling version of
it?
Minimize the space requirements wherever possible.
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Well, for the third case, you can elaborate on the implementation. We
need to compare the first half with the later half. So, in case of
even no. of characters it splits perfectly and in case of odd number
of characters, just ignore the middle character.
On Jul 17, 8:28 pm, swetha rahul
+919966006652
On Sun, Jul 17, 2011 at 8:57 PM, Dumanshu duman...@gmail.com wrote:
Given billions of integers in a file. Memory Constraints exist so need
to parallelize. One solution can be to split the file over a 100
machines and each machine calculates median of their part using
I want to have an in-memory dictionary. One word can have multiple
meanings.
1. If you want to go with hash tables then do Suggest some good
choices for hash functions.
2. Also, how can you modify this dictionary so that it can be used to
solve a crossword puzzle?
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@Ashish: if i got ur algo correct, contrary to all the above examples,
u r forming a linked list of level order traversal of the tree. m i
right?
On Jul 17, 8:49 pm, Ashish Goel ashg...@gmail.com wrote:
1. PUSH ROOT IN Q
2. PUSH DUMMY NODE IN Q, DEFINE PREVIOUS NODE AS NULL
3. WHILE Q IS NOT
, Jul 18, 2011 at 1:43 AM, Dumanshu duman...@gmail.com wrote:
I want to have an in-memory dictionary. One word can have multiple
meanings.
1. If you want to go with hash tables then do Suggest some good
choices for hash functions.
2. Also, how can you modify this dictionary so that it can
are u sure that this code works??? because last time i checked it
didn't.
On Jul 18, 1:22 am, hary rathor harry.rat...@gmail.com wrote:
int dividend,divisor,remainder;
int division(int p,int q){
int quotient=1;
/*if divisor and diviend are equal then quotient=1*/
if(p==q){
remainder=0;
Ashish Goel
Think positive and find fuel in failure
+919985813081
+919966006652
On Sun, Jul 17, 2011 at 8:57 PM, Dumanshu duman...@gmail.com wrote:
Given billions of integers in a file. Memory Constraints exist so need
to parallelize. One solution can be to split the file over a 100
given 4 billion integers i.e 2^32 integers. find the integer that is
not there in the list.
Memory constraints - 2^16 bytes can be used at max.
Does the presence of -ve numbers affect the solution??
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@Hary: thanks, ur code works. Could someone tell me the complexity of
the above mentioned code???
heres the working version of the same http://ideone.com/fDTfj
Its increasing exponentially.. so can we say log_2(n)???
On Jul 18, 1:57 am, Dumanshu duman...@gmail.com wrote:
are u sure
how about using voters algorithm? TC O(n) and SC O(1)
On Jul 16, 6:28 am, Anand Shastri anand.shastr...@gmail.com wrote:
Given a file containing 4,300,000,000 integers, how
can you *find **one* that *appears* at *least **twice*
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heres another approach.
For given n sets, all permutations have a { in the beginning and } in
the end. So, we need to permute the middle string with (n-1) sets. I
have generated all the permutations of n sets i.e. say n ==3 then
generate all permutations of (n-1==2 sets) {}{} string. Now before
how about this?
take pointer p1 set to end of array A and pointer p2 set to end of
array B.
while(you get n values)
{
print A(p1),B(p2)
now if( A(p1-1)+B(p2)A(p1)+B(p2-1) ) then print A(p1-1),B(p2)
followed by print A(p1),B(p2-1)
decrement p2 and p1 both
}
m i missing something?
On Jul 9,
@oppilas:
light the first string at both ends and in the middle. So it will burn
completely in 15 min as the fire is advancing in 4 ways irrespective
of the burn rate.
when the first string is completely burnt, light the second string at
both ends to get another 30 min. So, overall 45 min.
On
an hour to burn, but the burn rate is not
constant
Can't we have a string which take 45 minutes to burn till half length.
0-L/2. And 15 min from L/2 to L.
On Sat, Jul 9, 2011 at 8:57 PM, Dumanshu duman...@gmail.com wrote:
@oppilas:
light the first string at both ends and in the middle
the string burns in an hour if lit from one end and string burns in 30
min if lit from both ends.
This fact is based on - light up the string from either end it burns
in 60 min.
now heres the solution:
Light up the first string from both ends and the second string from
one end at the same time.
There are K pegs. Each peg can hold discs in decreasing order of
radius when looked from bottom to top of the peg. There are N discs
who have radius 1 to N; Given the initial configuration of the pegs
and the final configuration of the pegs, output the moves required to
transform from the initial
given an array of intergers. find the any integer that occurs only
once. Others might occur any no. of times.
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no constraints but try to give the optimized solution and plz no
hashing.
On Jul 8, 7:02 pm, Dumanshu duman...@gmail.com wrote:
given an array of intergers. find the any integer that occurs only
once. Others might occur any no. of times.
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how about this?
given n milestones
1.find max number from the array and delete it.
2.now find any (x,y) both from the array such that x+y == max.
3. put min(x,y) in the front of output array, delete y from array and
if(sizeof(output array)==(n-2))
put x also in front of output array and
Initially we can sort the array in O(nlogn) and then given a max
value, find a pair (x,y) in O(n). here n is also of quadratic order if
taken in terms of no. of milestones.
On Jul 7, 5:52 pm, Dumanshu duman...@gmail.com wrote:
how about this?
given n milestones
1.find max number from
given two sorted arrays a[m] b[2*m], each contains m elements only.
You
need to merge those two arrays into second array b[2*m].
anything better than O(m^2)
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@Radha: could u plz elaborate on getting the first n elements???
On Jul 8, 12:55 am, radha krishnan radhakrishnance...@gmail.com
wrote:
ok ! i got a O(n lgn) finally
i don know exact complexity
Let N = size of first array
Find the first N smallest elements using one pointer in each array
now
ans1. use counting sort for character array (0 to 255) then check for
the second string if same or not.
ans2. send 1 and 2, 1 comes back, send 10 and 5, 2 comes back, send 2
and 1
ans3. As vikas said, sum of digits should b 8. In that case the number
must have a zero digit because 1st digit the
@Azhar: could u plz explain the q8. problem.
On Jul 5, 12:13 pm, Azhar Hussain azhar...@gmail.com wrote:
Q8 is a DP problem, here is the solution
#include stdio.h
#define MAX 125
#define VAL 6
int Min[MAX];
int N[VAL] = {5, 10, 20, 25, 50, 100};
int main(void)
{
int i, j;
will count as one
traversal?
Correct me if I am wrong?
My approach:
traverse storing each char in a string.when space encountered push the
string in stack
I am not sure how my solution is but it doesnt appears gud ryt now.
On Tue, Jul 5, 2011 4:34 PM, Dumanshu duman...@gmail.com wrote
xor all the elements of both arrays ==0
sum of 1st array == sum of 2nd array
no. of elements in 1st == no. of elements in 2nd
if the above conditions are met, they have the same set.
m i missin sth?
On Jul 3, 1:23 am, mittal mohitm.1...@gmail.com wrote:
Given two arrays of numbers, find if each
..
if the value at prev center is small then next center won't skip
positions it may be just the next one.
so overall, it gives linear complexity.
On Jul 1, 12:35 pm, hary rathor harry.rat...@gmail.com wrote:
@Dumanshu; worst case O(n3) hai iska
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@sunny: if a[i]==0 then 0,i is the solution
suppose a[i]==a[j] ==0 now 0,i and 0,j is the solution. is i,j also
the solution??
On Jul 1, 4:08 pm, sunny agrawal sunny816.i...@gmail.com wrote:
String = 1 0 1 1 0 1 1 1.
1. make the array = 1 -1 1 1 -1 1 1 1
2. after second operation
array = 1 0
, himanshu kansal himanshukansal...@gmail.com
wrote:
If i do #if i==0 thn it vl cmpl it..means its tkng it as 0..bt why...
On 7/1/11, Dumanshu duman...@gmail.com wrote:
i think it will take a garbage value for i because these are
preprocessor directives.
On Jun 30, 4:46 pm, himanshu
In ur second code change int i=2 and run it. It would still print bye.
check the result of preprocessor using -E flag.
On Jun 30, 4:46 pm, himanshu kansal himanshukansal...@gmail.com
wrote:
1 more thngif its possible to eval. variable expression lyk abv...
thn...
int i=2;
#if i==2
@Jitendra: where r u using k in the first call?
On Jun 30, 8:56 am, Jitendra singh jsinghrath...@gmail.com wrote:
kthlargest(a[],b[],lefta,righta,leftb,rightb,k)
{
mida=lefta+(righta-lefta)/2;
midb=leftb+(rightb-leftb)/2;
if(a[mida]bmidb])
heres the code: http://ideone.com/aiG1m
using the algo from
http://www.akalin.cx/2007/11/28/finding-the-longest-palindromic-substring-in-linear-time/
On Jun 21, 11:31 pm, Swathi chukka.swa...@gmail.com wrote:
Does any one know how to return the Longest palindrome in a string in
O(n).
From
check this code http://ideone.com/rX130
compares k/2 values of each array taking care of extremes.
On Jun 24, 9:48 pm, Decipher ankurseth...@gmail.com wrote:
Can anybody please explain how to solve this question with logarithmic
time complexity ?
Write the code/algorithm to find the k-th
i think it will take a garbage value for i because these are
preprocessor directives.
On Jun 30, 4:46 pm, himanshu kansal himanshukansal...@gmail.com
wrote:
1 more thngif its possible to eval. variable expression lyk abv...
thn...
int i=2;
#if i==2
printf(hi);
#else
printf(bye);
I have used suffix arrays. Its easier to implement than trees.
code here- http://ideone.com/kX8MV
In case u find a test case givin wrong output plz do tell.
On Jun 29, 9:54 pm, Swathi chukka.swa...@gmail.com wrote:
Given a string (assume there no spaces or punctuations), write a code that
for the input doomdoomdoom
output is 4 or 8?
On Jun 30, 2:04 am, Dumanshu duman...@gmail.com wrote:
I have used suffix arrays. Its easier to implement than trees.
code here-http://ideone.com/kX8MV
In case u find a test case givin wrong output plz do tell.
On Jun 29, 9:54 pm, Swathi
if the output for the input doomdoomdoom is 8 then refer to
http://ideone.com/kX8MV
else if the output is 4 then refer to http://ideone.com/3tAKN
the second one is having a higher complexity.
ny suggestions?
On Jun 29, 9:54 pm, Swathi chukka.swa...@gmail.com wrote:
Given a string (assume there
O(n^2) and O(n^3) respectively.
On Jun 30, 3:47 am, Dumanshu duman...@gmail.com wrote:
if the output for the input doomdoomdoom is 8 then refer
tohttp://ideone.com/kX8MV
else if the output is 4 then refer tohttp://ideone.com/3tAKN
the second one is having a higher complexity.
ny suggestions
, Dumanshu duman...@gmail.com wrote:
finally got it in 0.01 sec using all the optimizations i am aware of
including what Wladimir had suggested.
Now wat to do for 0.00???
heres my code-
http://ideone.com/lsK8n
please suggest if any further optimizations possible.
On Jun 26, 2:21 am
@Dan: see, that is not the point. We are just looking for a better
solution not just an algorithm which fetches us 0.00 time given the
SPOJ conditions. Actually we are not worried about the compiler stuff
because its all relative. Some other person on this SPOJ platform has
submitted the code
These type of solutions require to think in binary.
First of all leave the last one because if we don't find a poisoned
bottle in first 5 then it means the last one is poisoned.
So 5 can be expressed using 3 bits.
these 3 bits will correspond to mice... 1 indicates the mice drinks
and 0 indicates
see i got 0.07 sec as the time after using counting sort.. because the
hotness scale is 0-10 so i used an array of 11 ints to count the no.
of occurrences. But i m nt able to further reduce this.
ny suggestions?
theres a comment on the problem:
On an interesting side note: the maximising
in logic.
wat do u think?
On Jun 26, 12:25 am, Dumanshu duman...@gmail.com wrote:
see i got 0.07 sec as the time after using counting sort.. because the
hotness scale is 0-10 so i used an array of 11 ints to count the no.
of occurrences. But i m nt able to further reduce this.
ny suggestions
finally got it in 0.01 sec using all the optimizations i am aware of
including what Wladimir had suggested.
Now wat to do for 0.00???
heres my code-
http://ideone.com/lsK8n
please suggest if any further optimizations possible.
On Jun 26, 2:21 am, Dumanshu duman...@gmail.com wrote:
Ok, it seems
@Piyush: could u plz post the link to the same?
On Jun 22, 2:15 pm, Piyush Sinha ecstasy.piy...@gmail.com wrote:
This question has been discussed over here once...It was concluded
that this can be solved in O(n) if we know there is a fixed range up
to which the elements keep on increasing and
Dynamic programming using memoization is the best one i think.
On Jun 21, 3:17 am, PRAMENDRA RATHi rathi prathi...@gmail.com wrote:
what is the shortest algo to find the value of nCr if both
n,r100...
-
PRAMENDRA RATHI
NIT ALLAHABAD
--
You
@Sunny: the value can still overflow because
Using the while loop u r dividing the res value whenever possible. so
the while checks if the current j divides res or not. What if current
res value is not divisible by that particular j ... it will multiply
by i for next res value and it will keep
new and delete are functions available in C++ and not in C. m i right?
On Jun 17, 2:34 pm, rohit rajuljain...@gmail.com wrote:
how to free memory allocated to an array with new function?
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Yes SPOJ uses the nasm assembler.
On Jun 18, 7:02 am, saurabh singh saurab...@gmail.com wrote:
I am using standard gcc 4.3.2 and the code does not requires any flag to be
required.I also checked the alias if gcc has been aliased to be used with
some option,but that was not the case.My
can use either.
Now plz give me some examples where mutex is preferred over binary
semaphore and vice-versa.
Thanks
Dumanshu
On Jun 18, 1:58 am, DK divyekap...@gmail.com wrote:
@Dumanshu/@Ankit: Of course mutexes can be made to work between processes
(it's an implementation detail
. Is this what you
are trying to do here? sorry i dont get your algo.
Thanks
Dumanshu
On Jun 18, 12:06 am, Ashish Goel ashg...@gmail.com wrote:
say the arrays are
a 6,7,9
b 3,4,5
c 1,2,8
the merged array would be
1c
2c
3a
4b
5b
6a
7a
8c
9a
1c,2c cant be compared as they are from
It does matter. suppose u write the c code and the compiler generates
assembly level code later u get machine code which runs. here in this
process many optimizations can be employed which are not done by gcc.
So including asm code in ur c code can actually make ur code to run
very fast provided u
U have got 3 sorted arrays A1 A2 and A3 having m n and p elements
respectively. A gap of 3 arrays is defined to be max distance between
3 nos if they are put on a no line say u pick three 2 12 and 7 then
the gap is 10. Now u have to find an efficient way of chosing 3 nos
from these 3 seperate
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