you can't do binary search with linked lists.
On Nov 17, 1:14 pm, Vijay Khandar vijaykhand...@gmail.com wrote:
Linked lists are not suitable data structures of which one of the
following problems?
a) Insertion sort
b) Binary search
c) Radix sort
d) Polynomial manipulation
Plz explain
On Thu, Nov 17, 2011 at 4:05 PM, shady sinv...@gmail.com wrote:
you can't do binary search with linked lists.
Yes you can do the binary search on the linked list.
But the only difference it makes from the array is that array elements can
be accessed in O(1) time and finding the mid in array is
roflmao, that's what i mean, else the whole purpose of binary search is
defeated, instead just linearly traverse the array and find the element
On Thu, Nov 17, 2011 at 4:17 PM, sumit mahamuni
sumit143smail...@gmail.comwrote:
On Thu, Nov 17, 2011 at 4:05 PM, shady sinv...@gmail.com wrote:
2n!/n!(n+1)!
catalan number
also asked by directi in written ...
wasim akram
MCA 3rd year
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Hello All,
What every algorithm mentioned above have some problem.
The Recursive swapping won’t work if you don’t have 2^n elements.
Same with getting the indexes, it will form a cycle.
Thanks
Pramod Negi
On Fri, Jun 11, 2010 at 7:09 PM, sharad kumar sharad20073...@gmail.comwrote:
a1a2a3b1b2b3
= [a1 a2] a3 [b1 b2] b3
= [a1 a2] [b1 b2] a3 b3
= a1 a2 b1 b2 a3 b3
= a1 b1 a2 b2 a3 b3
The algo. that I put forth works. And I think Sourav's will as well.
The grouping needs to be in numbers of 4.
The total number of elements need not.
-Minotauraus.
On Jun 13, 1:46 pm, Pramod Negi
excellent soln!!
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nice algo!
Anurag Sharma
On Wed, Jun 9, 2010 at 11:23 PM, souravsain souravs...@gmail.com wrote:
Guys
We can solve this in O(n) time like this:
Let me say total elements in array is 2N such that 1 to N are a's and N
+1 to 2N (which I will again refer to as 1 to N) are b's
Observation:
here is a sel explainatory algo
given:
abcd1234
abc1d234
ab1c2d34
a1b2c3d4
here is the link for the code : http://codepad.org/SZnufGc6
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Jitendra Kushwaha
MNNIT, Allahabad
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Its not O(n) time.
Anurag Sharma
On Wed, Jun 9, 2010 at 5:46 PM, Jitendra Kushwaha
jitendra.th...@gmail.comwrote:
here is a sel explainatory algo
given:
abcd1234
abc1d234
ab1c2d34
a1b2c3d4
here is the link for the code : http://codepad.org/SZnufGc6
--
Regards
Jitendra Kushwaha
Guys
We can solve this in O(n) time like this:
Let me say total elements in array is 2N such that 1 to N are a's and N
+1 to 2N (which I will again refer to as 1 to N) are b's
Observation:
If an element is on first 1 to N (an 'a') and has index i then in the
final array its position index (in
Actually the solution is best thought of with recursion.
See, for a simple 4 element array: a1 a2 b1 b2, you can get the result
by swapping only the two elements in the middle.
Now think, if you had an 8 element array: a1 a2 a3 a4 b1 b2 b3 b4,
group these into pairs of 2 like: a1 a2 - b1 b2 / a3
which is just the recursive version of the abovementioned iterative
solution.
P.S. -Please remove this quoted text when you are composing
--
Rohit Saraf
Second Year Undergraduate,
Dept. of Computer Science and Engineering
IIT Bombay
In order to make it inplace, time complexity has gone to n^2.
Rearrange(array,int N) //So array size is 2N
{
int i = 1;//points to index array[1] which has a2 since a1 is already
at the correct place;
int j = N;//array[0] to array[N-1] is a1 to aN. so j is index of b1
//it
For the first problem.
All occurrence of a duplicate character in a word.
void RemoveDuplicates(char *word) {
int count[256] = {0, 0};
char *p = word;
while ( p ) {
count[(int)(*p)] = 1;
p++;
}
p = word;
char *q = word;
while ( q ) {
if ( count[*q] ) {
*p = *q;
On Oct 13, 12:05 pm, Raghavendra Sharma raghavendra.vel...@gmail.com
wrote:
For the first problem.
All occurrence of a duplicate character in a word.
void RemoveDuplicates(char *word) {
int count[256] = {0, 0};
char *p = word;
while ( p ) {
This would be while(*p) {
OOPS sorry the code is not correct.
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On Sep 8, 10:47 am, yash yashpal.j...@gmail.com wrote:
wap a program in efficient manner to remove all occurrence of
duplicate character in the word and all occurrence of duplicate word
in the file.
i break the problem in two section( this is my approach it may be
better one )
wap to
On Sep 8, 10:47 am, yash yashpal.j...@gmail.com wrote:
wap a program in efficient manner to remove all occurrence of
duplicate character in the word and all occurrence of duplicate word
in the file.
i break the problem in two section( this is my approach it may be
better one )
wap to
@Debanjan Thanks for correcting this.
On Tue, Oct 13, 2009 at 5:12 PM, Debanjan debanjan4...@gmail.com wrote:
On Oct 13, 12:05 pm, Raghavendra Sharma raghavendra.vel...@gmail.com
wrote:
For the first problem.
All occurrence of a duplicate character in a word.
void
You can use a hash map. What do u mean by preserving order?
Keep inserting the characters into hash, whenever u find a duplicate,
delete it. The order is maintained still. Can you please elaborate on
what is mean by preserving order?
On Sep 8, 10:47 am, yash yashpal.j...@gmail.com wrote:
wap a
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