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
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 cards
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 help
A game
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, 2014 at 1:38 PM, bujji jajala jajal
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 code returns 9
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...
On 15 May 2014
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
remove 1( and all
.
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 neighborhoods, whose
// 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] -
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 represent the distance to travel to the destination on the
shortest route.
Now set the distance at the destination to zero.
Set
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, ..., n-r+1
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:
Hi all,
I know
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
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 : According
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 solution.Need your
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 already
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 (i,j)
for(i = 0; i n; ++i
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
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 post... anyway, the
point
for a 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 is soo clll
, 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:
// Using George Marsaglia's Multiply With Carry generator
int rnd10
[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.
actually question was asked in an intrview
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 x = time(0);
x
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, does
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
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
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
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] =
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 approach
On 6
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'm not able to get output
, '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 appropriate.
Don
On Monday, November 11, 2013 1
;
}
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 b c d e
in-order : c b e a d
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.
--
You received
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
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
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
matchboxes are numbered from 1
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 without
using rand() function
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 non-excluded responses a[j].
Here
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, we are still only
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 sampling here
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
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,-7,5,8,-1,-3}
Ans is: a[3
?
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. How would you
;
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 stdio.h
int main(){
int *temp;
printf
http://www.opengroup.org
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, 2013 at 9:19 PM, Don dond...@gmail.com javascript:wrote
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 the elements in each
.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:
Yes, you are right. Good catch
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
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 dond...@gmail.com javascript:wrote:
The trees
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 because you
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
ravi.c...@gmail.comjavascript:
wrote:
There is a Blank Paper Sheet, Given a list
int bitCount[N];
bitCount[0] = 0;
for(int i = 1; i N; ++i)
bitCount[i] = bitCount[i1] + (i1);
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 ??
--
You
int bitCount(int n)
{
if (n 3) return n;
int x=31-__builtin_clz(n);
n -= 1x;
return x*(1(x-1)) + bitCount(n) + n + 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
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, Jun 21, 2013 at 9:50 PM
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 tarunguptaa...@gmail.com wrote:
Yes the process is IO bound but you have to read them anyways!
Let's say you are unable to do parallel IO
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 dondod...@gmail.com wrote:
I would suggest splitting the data into four pieces and having each
thread do
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
It touches an edge of the region, and is therefore not surrounded.
On Jun 12, 2:39 am, Nishant Pandey nishant.bits.me...@gmail.com
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 dondod...@gmail.com wrote:
// Flip a region
// 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;
return
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 avinesh.sa...@gmail.com wrote:
Don, I'm trying to get all the words
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
)
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 avinesh.sa...@gmail.com wrote:
Thank you Don, I was also trying in similar
coins.
Don
#include stdio.h
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
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 nishant.bits.me
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 bagana.bharatku...@gmail.com wrote:
http://www.careercup.com/question?id
My program works with any numbers.
Don
On May 22, 3:45 am, Pramida Tumma pramida.tu...@gmail.com 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 17, 2013 at 6:55 PM, Don
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 macatad...@gmail.com wrote:
Saw this on a seperate group .. Since answer not known to me sharing it
here
you have a stack
, but as an array ..
thanks
--mac
On Wed, May 22, 2013 at 9:34 PM, Don dondod...@gmail.com wrote:
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 macatad...@gmail.com wrote
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 nishant.bits.me...@gmail.com
wrote:
I want to implement rope data stucture from scratch in c , is there any
good material that can help me implement
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)
, you can't do that in O(n) time and constant space.
Don
On May 7, 8:49 am, Nishant Pandey nishant.bits.me...@gmail.com
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 Mon, May 6, 2013 at 7
is
O(N).
Don
On May 2, 12:49 pm, Nishant Pandey nishant.bits.me...@gmail.com
wrote:
Guys i am looking for optimize algorithm for this, please help.
--
You received this message because you are subscribed to the Google Groups
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To unsubscribe from this group and stop receiving
, but if there are very many mines, finding
the best solution becomes very expensive.
Don
On Apr 29, 6:42 am, sreekanth guru sreekanth.i...@gmail.com 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
the complexity is O(row*col).
Don
On Apr 26, 8:19 am, rahul sharma rahul23111...@gmail.com 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
On Fri, Apr 26, 2013 at 2:07 PM
if (B[j+1] A[i]) max = i-1;
else break;
}
printf(Median is %d\n, (A[i] B[j]) ? A[i] : B[j]);
Don
On Apr 24, 1:19 pm, rahul sharma rahul23111...@gmail.com wrote:
IS this code correct?
float findMedianUtil( int A[], int N, int B[], int M )
{
// If the smaller array has only one
There is no need to use recursion here. Tail recursion is essentially
using recursion as an expensive loop.
int leastCommonAncestor(struct node *root, int n1, int n2)
{
while(root)
{
if ((root-data == n1) || (root-data == n2) || ((root-data
n1) != (root-data n2)))
return
.
Don
On Apr 18, 12:04 am, marti amritsa...@gmail.com wrote:
Given a list of words wordlist on 1st line (no of words = 100) and a
string qstr of len = 1 million on 2nd line, print the index at qstr where
a continuous substring exists that contains all the given words in wordlist.
--
You
Yes, you can use mid+2 there.
On Apr 17, 1:23 pm, rahul sharma rahul23111...@gmail.com wrote:
following is logn soln
int ceilSearch(int arr[], int low, int high, int x)
{
int mid;
/* If x is smaller than or equal to the first element,
then return the first element */
if(x =
is known at node A ,whether it
is up or down.
On Tue, Apr 16, 2013 at 12:23 AM, Don dondod...@gmail.com wrote:
I think that is called a load leveler. It needs to keep track of the
status of the resources, and the capacity of each one, when means that
it must be able to detect when
I have not tested this, but it should get you started:
struct item
{
int val;
int freq;
int index;
};
// Compare function for qsort.
// Returns 1 if a occurs with greater frequency than b
// Returns -1 if b occurs with greater frequency than a
// Otherwise returns 1 if a is encountered
for more information.
Don
On Apr 14, 1:16 pm, payel roy smithpa...@gmail.com wrote:
You are to test whether a number if prime or not.
Digit of that number can be as large as 1000. How do you do that?
I was thinking of basic idea.
a) First I shall calculate all primes which is less than
For numbers larger than what can be stored in your computer's native
integer types, you should use an extended precision math library. NTL
is one such option. It provides a type called ZZ which supports
integer operations where the size is limited only by your computer's
memory and speed.
Don
Reformatting to make it easier to see:
11000
01001
10011
0
10101
In this case an island is any set of 1's which are connected
vertically, horizontally, or diagonally.
So the five islands are
11000
01002
10022
0
30405
Don
On Apr 10, 4:34 pm, rahul sharma rahul23111...@gmail.com wrote
M is a matrix, not a number. M is NxN, so the algorithm is O(N^3) as
stated in the text.
On Apr 10, 4:19 pm, rahul sharma rahul23111...@gmail.com wrote:
isnt the complexity should be o(m*n*n) instead of (n*n*n) as m can be
greater than n..plz comment
On Wed, Apr 10, 2013 at 10:11 PM, rahul
head is not even declared, so I doubt that it would compile.
I believe that you want to free head, not next.
On Apr 9, 11:31 am, rahul sharma rahul23111...@gmail.com wrote:
Is the following code correct for linked list deletion or i need to copy
head in some tem. pointer and dlete and
is pointing to
a node which has already been deallocated.
Don
On Apr 9, 2:22 pm, rahul sharma rahul23111...@gmail.com wrote:
sory..following is correct code
void deleteList(struct node** head)
{
/* deref head_ref to get the real head */
struct node* next;
while (*head != NULL
, at which
point it will crash.
Don
On Apr 9, 2:29 pm, rahul sharma rahul23111...@gmail.com wrote:
A = {5, 3, 8, 9, 16}
After one iteration A = {3-5,8-3,9-8,16-9}={-2,5,1,7}
After second iteration A = {5-(-2),1-5,7-1} sum =7+(-4)+6=9
Given an array, return sum after n iterations
my sol/
void
.
For bonus points, can you determine how this problem relates to
Pascal's Triangle.
Don
On Apr 9, 2:43 pm, rahul sharma rahul23111...@gmail.com wrote:
If you have any other solution ..please post that...i thnik recursion is ok
with base case...we need to scan again after first iteration
Use a backtracking search to build a prefix expression. If there are
two more operands than operators in the expression, the next item
could be either an operand or an operator. Otherwise, it must be an
operand.
In very loose pseudocode:
search(int target, list operands, stack opStack, string
Simplify the expression by evaluating expressions inside parenthesis
first. Follow the order of evaluation, doing multiplications first and
then addition and subtraction. It should be possible to reduce any
expression to the form
ax+b=0. Then x=-b/a.
Don
On Apr 4, 11:18 am, arun kumar kumar0
I meant postfix, of course.
Don
On Apr 4, 10:32 am, Don dondod...@gmail.com wrote:
Use a backtracking search to build a prefix expression. If there are
two more operands than operators in the expression, the next item
could be either an operand or an operator. Otherwise, it must be an
operand
I like that solution better than the one I suggested.
Don
On Apr 4, 4:45 pm, Dave dave_and_da...@juno.com wrote:
@Kumar0746: Technically, you can't solve an _expression_; you can solve an
_equation_, which is a statement of the form expression = expression, which
is what you have.
Don's
they are asking for, you have to find the median
location for each distinct value in the array and find the sum of the
differences between each value and the median.
I would consider using a map to store the locations indexed by value.
Then iterate over the map and process each set of locations one by
one.
Don
I don't see how a statement like 3 coins together weigh x kg
provides any new information.
Using a binary search algorithm you should be able to find any coins
which weigh the same in 17 comparisons in the worse case.
On Mar 2, 12:42 am, Shubham Sandeep s.shubhamsand...@gmail.com
wrote:
@dave
precisely is very important.
BFS is not a workable solution. For any matrix larger than about 10x10
it will take longer than you want to wait.
Don
On Feb 21, 2:56 am, shady sinv...@gmail.com wrote:
Given a matrix of size mXn, find the number of paths from the top left cell
to the bottom right cell
that in your input array and you'll find that you
don't get the desired output.
I don't know of a solution better than sorting and scanning the array.
Don
On Feb 12, 3:14 pm, prakhar singh prakharsngh.1...@gmail.com wrote:
#includestdio.h
int main()
{
int a[] = {2,2,3,3,3,1,1,4,4
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