constants, so the complexity is O(row*col).
Don
On Apr 26, 8:19 am, rahul sharma wrote:
> got the islands...but first we scan each element then also dfs for them if
> all are 1..then how it can be o(row*col)...plz explain me complexity ofr
> this
>
>
>
>
>
>
>
&g
ood solution, but if there are very many mines, finding
the best solution becomes very expensive.
Don
On Apr 29, 6:42 am, sreekanth guru wrote:
> Problem:
>
> We have m * n grids. Each grid can have one of earth/water/mines. You can
> go to top/bottom/left/right adjacent grids. You can go to
time complexity is
O(N).
Don
On May 2, 12:49 pm, Nishant Pandey
wrote:
> Guys i am looking for optimize algorithm for this, please help.
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, you can't do that in O(n) time and constant space.
Don
On May 7, 8:49 am, Nishant Pandey
wrote:
> i am little confused with the solution, like if i say there are duplicate
> words, will this algo produce the words with equal
> probability?
>
>
>
>
>
>
>
> On Mo
I don't think that there is a shortcut, if the array contains
arbitrary data.
You'll want to create an array of bit counts for either 8 or 16 bits.
I would go with 16.
int bitCount[65536];
// Do this once at initialization time
void init()
{
bitCount[0] = 0;
for(int i = 1; i < 65536; ++i) b
Counting the set bits in one integer is not the problem which was
asked.
However, I think that something like this is both faster and more easy
to understand than what you have written:
int bitCount(unsigned int x)
{
int result = 0;
while(x)
{
if (x & 1) ++result;
x >>= 1;
http://en.wikipedia.org/wiki/Rope_(data_structure)
I don't know if this will make it easy, but it should help.
On May 17, 4:28 am, Nishant Pandey
wrote:
> I want to implement rope data stucture from scratch in c , is there any
> good material that can help me implement this with ease.
>
> Thanks
My program works with any numbers.
Don
On May 22, 3:45 am, Pramida Tumma wrote:
> This above program works only if the array contains consecutive numbers
> starting from 1 to n. What to do if the array contains random numbers?
>
>
>
>
>
>
>
> On Fri, May 1
Do you mean the middle position or median value?
If you implement the stack in an array, finding the middle position
should be easy.
Don
On May 22, 11:45 am, MAC wrote:
> Saw this on a seperate group .. Since answer not known to me sharing it
> here
>
> you have a sta
it as a stack , but as an array ..
>
> thanks
> --mac
>
>
>
>
>
>
>
> On Wed, May 22, 2013 at 9:34 PM, Don wrote:
> > Do you mean the middle position or median value?
>
> > If you implement the stack in an array, finding the middle position
> > should be
This is not necessarily possible.
If you have teams A, B, and C, it is possible that
A beat B
B beat C
C beat A.
Based on the first two games the ranking should be A,B,C. But based on
the third game, C should be ranked higher than A.
Don
On May 23, 11:06 am, Nishant Pandey
wrote:
> I hav
If you create a directed graph where each node is a team and an edge
exists from A->B if A lost to B, then find a Hamiltonian Path in the
graph. That path will be the sequence you need.
Don
On May 23, 12:02 pm, bharat b wrote:
> http://www.careercup.com/question?id=14804702
>
> On
coins.
Don
#include
int main()
{
const int numCoins = 6;
const int maxS = 1;
int coins[numCoins] = {1,5,10,25,50,100}; // You can change this
if you want different sets of coins
int min[maxS];
int i, j;
for(i = 1; i < maxS; ++i) min[i] = 10;
min[0]
void findWords(trie *root, char *filter)
{
if (!root) return;
if (*filter == 0) // When you reach the end of the filter at the
end of a valid word, add the word.
{
if (root->words) words.add(root->word);
}
else if (*filter == '.') // Search for words with any letter
for(int i = 'a'; i <= 'z' ; ++i)
findWords(root->link[i], filter+1, word+i);
}
else // Search for words with the required letter
{
findWords(root->link[*filter], filter+1, word+*filter);
}
}
On May 28, 11:36 pm, avinesh saini wrot
It won't use a lot of stack space because the stack space is related
to the depth of the trie which is only as deep as the length of the
longest word. But I'm all for doing it iteratively.
Don
On May 30, 7:57 am, avinesh saini wrote:
> Don, I'm trying to get all the words fro
Sorry, that was not really clear. I was just adding the newly found
word to a list of words. That list will be the output. I treated it as
if it was declared globally. It would be better form to pass the list
in as a reference, or to have the function return the list.
Don
On May 29, 12:14 am
// Flip a region including location (x,y) from "from" to "to".
// Returns true if region is surrounded.
bool flip(char board[][], int n, int x, int y, char from, char to)
{
if ((x >= 0) && (y >= 0) && (x < n) && (y < n)) return false;
if (board[x][y] != from) return true;
board[x][y] = to;
It touches an edge of the region, and is therefore not surrounded.
On Jun 12, 2:39 am, Nishant Pandey
wrote:
> juat want to ask why the last region was not flipped from o to x?
>
>
>
>
>
>
>
> On Tue, Jun 11, 2013 at 11:57 PM, Don wrote:
> > // Flip a region in
I should mention that this process is file I/O bound, so having four
threads will not help significantly unless each thread can use a
different disk device.
Don
On Jun 13, 3:31 pm, Don wrote:
> I would suggest splitting the data into four pieces and having each
> thread do an external s
I would suggest splitting the data into four pieces and having each
thread do an external sort on one piece. Essentially that means
breaking up the data into chunks which can be stored in memory,
sorting them, and then merging the chunks back into one piece. After
the 4 threads are done sorting the
It the four threads have to share the bottleneck resource, you won't
get anything close to a 4x speedup. That's all I'm saying.
Don
On Jun 13, 11:17 pm, nick wrote:
> Yes the process is IO bound but you have to read them anyways!
> Let's say you are unable to do pa
int bitCount[N];
bitCount[0] = 0;
for(int i = 1; i < N; ++i)
bitCount[i] = bitCount[i>>1] + (i&1);
On Thursday, June 20, 2013 11:03:35 PM UTC-4, shubham saini wrote:
>
> How to count no of set bits for all numbers from 1 to n for a given n...
>
> i knew brute force any better solution ??
>
-
int bitCount(int n)
{
if (n < 3) return n;
int x=31-__builtin_clz(n);
n -= 1<
> How to count no of set bits for all numbers from 1 to n for a given n...
>
> i knew brute force any better solution ??
>
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"Algorit
is set
plus the number of bits in 0..n-2^x. That is what the last line computes.
Don
On Friday, June 21, 2013 1:42:08 PM UTC-4, shubham saini wrote:
>
> thanx Don for your algos, bt i m not able to understand your second
> approach can you please explain it a liitle
>
>
> On Fri,
ould save it and try
for N+1 using only the N+1 smallest letters.
Don
On Tuesday, June 25, 2013 7:14:53 AM UTC-4, jagannath wrote:
>
> Is this not similar to knapsack problem?
>
>
> On Fri, Jun 21, 2013 at 11:24 AM, Ravi Ranjan
>
> > wrote:
>
>> There is a Bl
Get the NTL library: http://www.shoup.net/ntl/
ZZ is the extended precision integer class.
Don
On Saturday, July 6, 2013 2:31:02 PM UTC-4, subrat kumar prasad wrote:
>
> can anyone provide working code for biginteger in c++?
> please upload!!
>
--
You received this message
The trees would be the same if
1. The root is the same
2. The left and right subtrees are both the same
bool sameBstFormed(intVector t1, intVector t2)
{
if (t1.size() != t2.size()) return false; // Trees with different
number of nodes can't be the same
if (t1.size() == 0) return true;
Yes, you are right. Good catch.
On Tuesday, July 9, 2013 2:13:47 PM UTC-4, Sambhavna Singh wrote:
>
> Amazing solution Don, thank you :) You missed a small check in the code:
> if(t1[0] != t2[0]) return 0;
>
>
> On Tue, Jul 9, 2013 at 11:27 PM, Don >wrote:
>
>>
right1.push_back(t1[i]);
if (t2[i] < t2[0]) left2.push_back(t2[i]);
else right2.push_back(t2[i]);
}
result = sameBstFormed(left1, left2) && sameBstFormed(right1, right2);
}
return result;
}
On Tuesday, July 9, 2013 5:41:57 PM UTC-4, Don wrote:
>
> Y
partition algorithms are not stable. That's really the question I was
asking. Can you do that partition in place?
Don
On Sunday, July 14, 2013 2:48:46 PM UTC-4, sourabh wrote:
>
> You don't need to take vector for this you can have input in array also.
> You just need to check t
x = *temp;
printf("%d\n", x);
You know that x=*temp; will crash, but I'm guessing that it will not reach
the first printf.
Don
On Thursday, July 18, 2013 9:33:51 AM UTC-4, phoenix wrote:
>
>
> I tried the following on ideone
>
> #include
> int mai
Just mail me a check for my drinks.
Thanks!
Don
On Thursday, July 18, 2013 2:43:46 PM UTC-4, .bashrc wrote:
>
> Hi all,
>
> Party at my house on 28 July, 5 pm.
>
> Unlimited drinks.
>
> All are welcome. :)
>
> Regards,
> Dinesh Sohu
>
>
> On Thu, Jul 18
?
Don
On Thursday, August 8, 2013 1:02:08 AM UTC-4, payel roy wrote:
>
> There is a stream of numbers coming in and you have to find K largest
> numbers out of the numbers received so far at any given time. Next problem
> is that a constraint is added. memory is limited to m. m < k
Use Kadane's algorithm.
http://en.wikipedia.org/wiki/Maximum_subarray_problem
Don
On Thursday, August 29, 2013 11:09:59 PM UTC-4, yash wrote:
>
> 1. find largest sum contiguous subarray
> 2. find the contiguous subarray which produce largest sum
>
> Ex:
> a[7] = {2,1,
Provide a function which returns a value randomly and uniformly selected
from the range 0..N-1 excluding values in the array a[S] containing sorted
values in the same range. A rejection algorithm would work well enough if S
is much smaller than N, but what if N is large and S is slightly smaller
You can do a lot better than N log(N).
Don
On Wednesday, October 9, 2013 2:03:02 PM UTC-4, hppygrry wrote:
>
> What's wrong with just a simple approach with checking the values in the
> array[s] ? Since its already sorted, it takes log(s) time. Assuming S
> approaches N,
Describe in more detail how you would use reservoir sampling.
What would be the memory requirement?
There is a solution to this problem which is O(log S) time and constant
memory.
Don
On Tuesday, October 8, 2013 2:24:54 PM UTC-4, atul007 wrote:
>
> can't we use idea of reservoir sa
ult, so the xth value is x+min.
return x+min;
}
You can prove that it works by trying all possible values of x and
verifying that it produces the set of valid responses.
Don
On Thursday, October 10, 2013 12:41:10 AM UTC-4, Dave wrote:
>
> @Don: Note that a[j] - j is the number of n
t need one and the
statistical properties are not important, you can do something really
simple like:
int x = (time() * getpid()) % N;
Don
On Thursday, October 10, 2013 5:51:42 AM UTC-4, atul007 wrote:
>
> Hello,
>Given integer N , How to generate random number from 0 to N wi
. Once you know that, you can solve a big problem
easily.
Don
On Monday, October 28, 2013 3:22:54 AM UTC-4, robin wrote:
>
> We have a game of matchboxes with the following rules:
> * It is a 2-player game.
> * There are n matchboxes with varying number of matchsticks in them. The n
Atul, the depth is the important factor, not the number of nodes.
If you always pass the message to the deeper subtree first, followed by the
other subtree, you get the minimum number of rounds.
The problem is to compute the required number of rounds. This can be done
in a way similar to comput
Atul, the depth is the important factor, not the number of nodes.
If you always pass the message to the deeper subtree first, followed by the
other subtree, you get the minimum number of rounds.
The problem is to compute the required number of rounds. This can be done
in a way similar to comput
Save it in pre-order.
Rebuild by inserting nodes in the order they occur in the file.
On Friday, November 8, 2013 8:33:19 AM UTC-5, kumar raja wrote:
>
> What is the effective way to save and restore the binary trees to files
> effectively?
>
>
>
>
>
>
>
>
>
>
>
>
>
> Regards,
> Kumar Raja.
>
--
}
On Friday, November 8, 2013 1:00:35 PM UTC-5, atul007 wrote:
>
> @don : it is not BST , it is binary tree ...so your approach will
> not work in this case.
>
> @kumar : save pre-order and in-order traversal with some delimiter in
> b/w traversals.
>
> pre-order : a
is a leaf, 'B'
if it has only a left subtree, 'C' if it has only a right subtree, and 'D'
if it has both left and right subtrees. Then you read the line, store the
string minus the first character, and recursively build the left and then
right subtree, as appropria
Printf will print one value per item in the format string.
If you don't supply that many, it will look at the undefined memory where
that value should have been, so you will get gibberish.
Don
On Friday, December 6, 2013 9:17:36 AM UTC-5, segfault wrote:
>
> Hi All,
>
> I&
http://www.geeksforgeeks.org/median-of-two-sorted-arrays-of-different-sizes/
This gives a reasonable solution. However, I would use iteration instead of
tail recursion.
Don
On Monday, January 6, 2014 4:15:41 AM UTC-5, atul007 wrote:
>
> @dave : could you provide pseudo code for ur ap
Sort the input string and remove duplicates, keeping a count of the number
of occurrences of each character.
They you can build the permutations easily.
For your example, you would start with
char *str = "aba";
int len = strlen(str);
Which would be converted to
char *str "ab";
int rep[N] = {2,
It is no longer a binary tree. The node you pick may have three children.
The path from the original root to that node will now be a third branch.
Don
On Wednesday, January 15, 2014 12:17:47 AM UTC-5, atul007 wrote:
>
> Imagine a binary tree lying on the floor with nodes as balls and edges
Good point, Vikas.
What if the balls all have a negative charge and repel each other?
On Thursday, January 16, 2014 6:36:06 AM UTC-5, vikas wrote:
>
> it will all collapse and form a single monolithic straight line with bunch
> of balls hanging in between :D
>
> On Wednesday, 15 January 2014 10:
This works, and I think is O(N*log(N)) which is similar to sorting and
scanning.
An unordered map will be faster, in general. It could be made faster in
most cases by looping over items left in the map, to avoid processing the
same number more than once. Also, when the number of items left in t
No. If you start at any number in a sequence it will find the entire
sequence. There is no need to start again at some other number in that
sequence.
Don
On Wednesday, January 29, 2014 12:19:21 AM UTC-5, Varun Sachdeva wrote:
>
> If we don't process the same number more than once,
// Using George Marsaglia's Multiply With Carry generator
int rnd10()
{
static unsigned int x = time(0);
x = 63663 * (x&65535) + (x>>16);
return (x & 65535) % 10;
}
On Sunday, February 2, 2014 2:51:47 AM UTC-5, atul007 wrote:
>
> Generate random number form 1 - 10 with probability of 1/10
Just noticed that you asked for 1-10, and my code will clearly generate
0-9. Add one for the desired range.
Don
On Wednesday, February 5, 2014 4:29:26 PM UTC-5, Don wrote:
>
> // Using George Marsaglia's Multiply With Carry generator
> int rnd10()
> {
>static unsigned int
gt; On Thursday, February 6, 2014 1:47:15 PM UTC-8, Don wrote:
>>
>> Just noticed that you asked for 1-10, and my code will clearly generate
>> 0-9. Add one for the desired range.
>> Don
>>
>> On Wednesday, February 5, 2014 4:29:26 PM UTC-5, Don wrote:
>>>
&
mp; 65535) % (i+1);
n[i] = n[j];
n[j] = i;
}
// n now contains a permutation of the values 1-10.
On Thursday, February 6, 2014 11:33:53 PM UTC-5, atul007 wrote:
>
> @don: algo look fine..i tested it ,but it did not generate 1-10 with
> probabilty of 1/10 everytime.
> actu
64-bit multiply
with carry generator, while what I posted was a 32-bit multiply with carry
generator.
Don
On Saturday, February 8, 2014 2:26:34 AM UTC-5, atul007 wrote:
>
> @don : awesome +1 . I was not aware of George Marsaglia's Multiply
> With Carry generator. But it i
generate a number in the range 0..n-1 is to divide the output by
65535/n. Then check the result and if it is >= n, repeat the process until
you get a valid result.
Don
On Tuesday, February 11, 2014 1:18:26 AM UTC-5, SVIX wrote:
>
> :) true... that was silly of me.. I was too quick to
I like that Python code, and for three digits numbers it is just fine.
If the number could be in a larger range, factoring out digits starting
with larger digits should give the correct sequence, in reverse. If the
number has a prime factor larger than 9, there is no solution.
// Returns the sm
No, that is not the same.
It won't find a square smaller than s.
Don
On Thursday, March 27, 2014 2:56:29 AM UTC-4, atul007 wrote:
>
> @Don : your algo time complexity is O(n^2)
>
> It can be reduced to this :-
>
> // Now look for largest square with top left corner at
00
00
010100
011100
01
00
In this case, when i and j are 1, your algorithm will set s = 3. The if
statement will fail, and it will never notice that it could have formed a
square with s=2.
Don
On Sunday, March 30, 2014 9:49:21 AM UTC-4, atul007 wrote:
>
> @don : Accord
might not give a result exactly equal to S. But if A
contains positive real numbers I think that it will work.
Don
On Sunday, March 30, 2014 10:02:08 AM UTC-4, atul007 wrote:
>
> Hello,
> i found this question online and its solution tooBut i am not able
> to fully understand
The only square is found when s=2. Your program will look at s=3 and not
find a square, so it will never find the square.
Try it and you will see what I mean..
Don
On Monday, March 31, 2014 2:42:13 PM UTC-4, atul007 wrote:
>
> @Don: what is the point of considering s=2 when we have a
Use the fact that a*b*c*d = exp(log a + log b + log c + log d)
double sum = 0.0;
double x;
for(x = n-r+1.0; x <= n; x += 1.0)
sum += log(x);
for(x = 2; x <= r; x += 1.0)
sum -= log(x);
result = exp(sum);
Don
On Tuesday, April 8, 2014 2:06:47 PM UTC-4, kumar raja wrote:
>
Note that my example is for C(n,r), but you can use the same method for
your second formula. Just add up the logs of whatever you multiply in the
numerator and subtract the logs of the factors of the denominator. Then the
result is exp of the resulting sum.
Don
On Tuesday, April 8, 2014 2:15
Very nice solution.
Don
On Wednesday, April 9, 2014 12:55:43 AM UTC-4, Dave wrote:
>
> What you want to do is automate the process of cancelling common factors
> that you would do if you were calculating nCr by hand. Fill an array a[]
> with the numerator terms, n, n-1, n-2, ...,
.
Don
On Sunday, April 20, 2014 11:52:44 AM UTC-4, bujji wrote:
>
> Consider a city whose streets are defined by an X ×Y grid. We are
> interested in
> walking from the upper left-hand corner of the grid to the lower
> right-hand corner.
> Unfortunately, the city has bad ne
// The result will always include the smallest value of A[i] where i is
less than j,
// so just keep track of that minimum value as you scan through the array.
int findPair(int *A, int N)
{
min = A[0];
max = 0;
for(j = 1; j < N; ++j)
{
if ((A[j] - min) > max) max = A[j] -
+1] == distance[i][j]-1) ++j;
else if (distance[i][j-1] == distance[i][j]-1) --j;
printf("Move to (%d, %d)\n", i,j);
}
}
On Monday, April 21, 2014 11:08:55 AM UTC-4, Don wrote:
>
> Create a matrix distance[x][y] and set all values to a large number.
> This will rep
Good catch. A function called "markShortestPath" ought to mark the shortest
path, huh?
Don
On Monday, April 21, 2014 4:38:32 PM UTC-4, bujji wrote:
>
>
> Nice solution. good. Looks like in your markShortestPath(int
> i, int i, int d) function
> you
How about this:
const int N = 6;
unsigned int coins[N] = {100,50,25,10,5,1};
unsigned int count[N] = {2, 1, 1, 5, 4, 10};
int best = 10;
int minCoins(int s, int f=0, int d=0)
{
if (d == 0) best = s; // It can never take more than s coins
if (s < 1) return 0; // No coins required
Sid,
Your proposal is to try removing every possible combination of nodes and
for each case, do a DFS to determine if the graph is still connected?
What is the time complexity of that approach?
Don
On Friday, May 9, 2014 4:58:15 AM UTC-4, siddharam wrote:
>
> use DFS
>
>
> re
If you find a way to do that for more than a few coins I'd be interested in
seeing it too.
Don
On Thursday, May 15, 2014 3:00:04 PM UTC-4, atul007 wrote:
>
> @Don : i am intersted in DP bottom up approach with less time complexity.
> solving it recursively is a simpler approach.
In my solution, the line which reads:
if (sum < result) sum = result;
Should read
if (sum < result) result = sum;
Don
On Friday, May 16, 2014 11:51:52 PM UTC-4, Shashwat Anand wrote:
>
> @Don
> int coins [] = {1, 3, 5};
> int cnt [] = {7, 3, 1};
> int S = 9;
> Your
Shashwat, very nice.
Don
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BFS might be faster, though.
Don
On Tuesday, April 22, 2014 1:11:00 AM UTC-4, bujji wrote:
>
> Hi Don,
> At most we can reach a point from 4 adjacent points. So,
> time complexity will be O(XY) only.
>
> -Thanks,
> Bujji.
>
>
> On Mon, Apr 21,
from the list which are not consistent with that answer. Repeat the
process until you have only one order remaining.
Don
On Tuesday, July 1, 2014 12:15:14 PM UTC-4, vikas wrote:
>
> One of my friend posted this question to me and I am unable to get
> anywhere. Could you guys please he
was to
find the minimum number of hints required, not to minimize the computation
time to find those hints. A faster algorithm is to ask for hints randomly,
but that will require more hints.
Don
On Wednesday, July 2, 2014 11:20:32 AM UTC-4, vikas wrote:
>
> Don, if you are talking about
card by one of the hints which you do request.
There are only 120 combinations of 7 hints. If you try each one of those to
determine if it distinguishes each pair of cards selected, you can
determine if only 7 hints will suffice. Keep trying with fewer hints until
you find the lower limit.
Don
not distinguish those cards. So a valid
combination of hints has a non-zero bitwise AND with all values in the
pairs list. Find the value of i with the fewest bits which meets this
condition.
Don
On Tuesday, July 1, 2014 12:15:14 PM UTC-4, vikas wrote:
>
> One of my friend posted this ques
// This function returns the number of hints required to
// determine the order of the cards.
// Parameter int c[n] specifies n cards which are provided
// in an unknown order. Each card is represented as a 10-bit field
// with two bits set, one representing the color of the card and the
// other r
determine the correct
response to a guess. The algorithm should require a feasible amount of
memory for P<1000 and C<1000 (meaning you can't have a table of size C^P).
Don
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You received this message because you are subscribed to the Google Groups
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To
void change()
{
printf("10"); while(1) {}
}
On Feb 15, 10:17 am, Balaji S wrote:
> Insert only one line in the function change() so that the output of the
> program is 10.
> You are not allowed to use exit(). You are not allowed to edit the function
> main() or to
> pass the parameter to change
A semicolon is valid in the middle of a line in C or C++.
For instance, no one says that
for(i = 0; i < 10; ++i)
is three lines of code.
Don
On Feb 15, 11:31 am, jalaj jaiswal wrote:
> after termination of semicolon , that will be considered a separate line i
> guess
>
>
>
Mike Johnson wrote:
> Plesae rite a program for me to find prime nummers. It should be recursive
> prorgram. What duz that mean?
> If u type a nummer like 10 it should say "1 is prime, 2 is prime, 3 is prime,
> 4 is not prime" up to 10.
> This iz not homewurk I just thout of it myself. Lol.
/* S
template
void sort(Iterator first, Iterator last, Compare cmp);
On Feb 16, 5:57 am, bittu wrote:
> write header file for sorting general element. it can sort any type of
> object(int,float,complex number, objects)
>
> Thanks
> Shashank
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ed of 1 space every third turn until you
find the other robot's starting spot
SKIP NEXT INSTRUCTION IF OIL
GOTO Search
Fast: RIGHT // Move at speed of 1 space every other turn
GOTO Fast
Both robots will move to the right, but the one on the left will
eventually find the oil left by the other rob
class coord
{
public:
int x;
int y;
};
class SolveMaze
{
public:
SolveMaze() { location.x = location.y = 0; }
bool Explore();
private:
list visited;
coord location;
};
bool Explore()
{
// Determine if we've already been here
if (visited.found(location))
return false;
// Bas
Did he get frostbite on his toes?
On Mar 2, 1:54 am, Lavesh Rawat wrote:
> *WalK The River Problem Solution*
> *
> *A whole village crowded together at the edge of the lake, they all came for
> a common purpose.
> To see the man who could walk on water.
> Many bets were placed, many a dreamer ima
Game 2 is better if p > 0.5.
P(win game 1) = p
P(win game 2) = p^3 + 3(p^2 * (1-p))
To find the solution, solve for p when
p^3 + 3(p^2 * (1-p)) > p
Which simplifies to
3p - 2p^2 > 1
Which is true when p > 0.5
On Mar 1, 5:15 pm, bittu wrote:
> You have a basketball hoop and someone says that
be to make the total odd or even, as indicated
by the previous prisoner.
Each prisoner in line must keep track of the current state, odd or
even, which changes each time someone behind him indicates having a
red hat.
Don
On Mar 3, 2:18 am, freecoder wrote:
> You are one of 20 prisoners on death
The problem with BFS is that you have to remember how to get from
where you are to every other point you have explored.
Of course, the problem with a depth-first search is that in an
unbounded maze, you may never reach a point even if it is very close
to your starting point.
Don
On Mar 3, 1:31
I'm not sure what you mean.
You might use an priority queue to store scheduled events and process
them in time order.
On Mar 4, 10:27 am, Luciano Junior wrote:
> Hello,
>
> I need a clock algorithm to use with in a simulation system that I be
> creating.
>
> Someone have any Idea ?
> --
> --
Another alternative to a void pointer is a union. Typically you would
also want an enumerated type to indicate the actual type stored in the
union.
On Mar 16, 9:57 am, himanshu kansal
wrote:
> can nodes of linked list in c/c++ contain different types of
> datameans suppose 1st node of the lis
This only works if the file is sorted. If the file starts out with
values 5,7,6,... and never contains another 7, the result will be 7,
which is in the file.
On Mar 17, 12:19 pm, "arpit.gupta" wrote:
> read the first no. .
> now ans= first no +1;
> if now ans is encountered while reading the next
bool uniform()
{
static bool state = true;
state = !state;
return state ^ nonuniform();
}
On Mar 17, 9:24 am, saurabh agrawal wrote:
> Given a function which returns true 60% time and false 40% time.
>
> Using this function you have to write a function which returns true 50% of
> the time.
int n = printf("Hello\n");
int main()
{
}
On Mar 17, 8:08 am, himanshu kansal
wrote:
> is there any way to print hello in c also wdout writing anythng in
> main()
>
> On Thu, Mar 17, 2011 at 6:34 PM, kunal srivastav
> > wrote:
> > n...c does not support classes
>
> > On Thu, Mar 17, 20
Which is why stating the requirements precisely is important.
If modeling a coin toss is the goal, this should work.
bool uniform()
{
static unsigned int x = 1234567;
x = 63663*(x&65535)+(x>>16);
return ((x&64)==0) ^ nonUniform();
}
On Mar 17, 3:29 pm, Dave
There is no upper bound on the runtime of this function.
On Mar 20, 4:33 pm, Jammy wrote:
> clearly the prob. of getting true|false and false|true are equal to
> 0.6*0.4. Therefore the following code works,
>
> bool uniform(){
> bool f1;
> bool f2;
> do{
> f1 = non_uniform();
>
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