Following code works for [A-Za-z], can be extended for whole character-set :
> #include
>
> int main()
> {
> unsigned long long int a = 0;
> char str[50];
> int i;
>
> scanf ("%s", str);
>
> for ( i = 0; str[i]; i++ ) {
> if ( str[i] >= 'A' && str[i] <= 'Z' ) {
>
In Linux, there is no difference between threads and process. It is
like, a thread is merely a process that shares certain resources with
other processes. Each thread has its own process descriptor and
appears to kernel as a normal process - threads just happen to share
resources, such as an addres
I think he means to edit the comparison function to get the order. so
at a time only 2 elements are compared.
On May 28, 7:51 am, Logic King wrote:
> @sunny it will work fine if you have 2 numbers only...but what about the
> list...3..4 or 5..or morethen the possible number of combinations wi
Given a n by n matrix. Suggest an algorithm to verify its correctness
given a configuration. User can enter numbers only between 1 to n.
I need this in 2 ways -
1. for the n by n matrix, suggest an elegant way for validating it.
2. suggest a data structure for this sudoku so that the structure aids
string getStringWithoutDuplicateChars(string input)
{
create_empty_trie_ds (say trie)
integer count = 0;
for_each_char_in_string (say ch)
{
if(trie->contains(ch)) //if ch not there in ds then add it and return
false otherwise return true
{
input.remove(count)
}
count++
Design an algorithm and write code to remove the duplicate characters in a
string without using any additional buffer.
NOTE: One or two additional variables are fine.
An extra copy of the array is not.
--
Thank You
Rajeev Kumar
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@Dave: I have suggested another solution in previous threads. Please go
through that. That is without maps. It uses array for mapping.
On Fri, May 27, 2011 at 10:09 PM, Dave wrote:
> If map insertion is O(log n), then the algorithms that insert each
> element into the map will be O(n log n), but
If map insertion is O(log n), then the algorithms that insert each
element into the map will be O(n log n), but the problem statement
insists that we find two elements of the array that sum to a given
number in O(n) time. Thus, Aakash's solution (http://groups.google.com/
group/algogeeks/msg/541180
@ross
i can do it in O(n) time using an extra array of the same size as the given
array
first sort both the even indexed terms and odd indexed terms by modifying
quick sort...it can be done in one traversal.
and then using extra array and then move the even indexed terms followed by
odd index
ya .. it can work for negative indexes also if the bound is known.. like if
the range is from -10 to +10 then declare an array of 20 and then refer
a[10] as a[0] and use negative indexes to do the same procedure !!
On Fri, May 27, 2011 at 2:32 PM, bhavana wrote:
> hey...bt this one works only in
Yup...I think thats the way to go...modify heap sort...'ignoring the odd
childs' for ascending and ignoring the even child for decending.
If i am wrong in this then quick sort has to be the solution.
On Sat, May 28, 2011 at 8:44 AM, subramania jeeva wrote:
> I hope merge sort is not an inplace so
http://samples.msdn.microsoft.com/ietestcenter/
On Fri, May 27, 2011 at 9:40 AM, radha krishnan <
radhakrishnance...@gmail.com> wrote:
> Check whether this is storing Google Search Results ? ? ?? ? ? ? ? ??
> [HONEY POTTING]
>
>
> On Fri, May 27, 2011 at 10:07 PM, UTKARSH SRIVASTAV <
> usrivas
I hope merge sort is not an inplace sorting.. It'll take extra space (in
function merge) and since it uses recursion it'll also take stack space..
I think this sol will work.
1. partition the array by holding the even elements on first space(n/2 or
n/2-1 depends on odd sized and even sized array)
@sunny it will work fine if you have 2 numbers only...but what about the
list...3..4 or 5..or morethen the possible number of combinations will
be 'N!'...where n is the number of digits...the code will work quite
slowly for larger 'n'.
On Fri, May 27, 2011 at 3:33 PM, Dave wrote:
> @Shu
@Shubham: Try 8, 89, 7. Your algorithm gives 8897, but the right
answer is 8987.
Dave
On May 27, 1:11 pm, shubham wrote:
> check whether these steps work:
>
> step 1:
> sort the given numbers in the decreasing order based on their first
> digit.
> step 2:
> if two numbers come ou
will it work
modifying the comparison function of the sort as follows
generate two new numbers c and d from arguments a,b as
c = ab; //concatenation
d = ba;
compare c and d
if c >= d return true means a will come before b in final ordering
else return false means b will come before in final order
how large n will be...O(n) can't grow more than O(nlogn)so in any
case the complexity is going to be O(nlogn)..so i dnt see any point of
bringing our any modification of mergersort...even if u think so,
provide a concrete algo
On 5/28/11, LALIT SHARMA wrote:
> It will give correct answer ,
>
It will give correct answer ,
but instead of doing manipulation after taking input.
as it would take some O(n) time. if n would be large , we would incur
this extra cost .
we should change the termination condition of merge-sort function and
modify the merge function of merge sort ..to reach our o
main()
{
int a[100];
int i,j,N;
printf("enter the number of elements: ");
scanf("%d",&N);
for(i=0;i wrote:
> wat about insertion sort (with some limited conditions obviously ) ??
>
> On Sat, May 28, 2011 at 12:56 AM, Piyush Sinha
> wrote:
>
>> will it be given th
wat about insertion sort (with some limited conditions obviously ) ??
On Sat, May 28, 2011 at 12:56 AM, Piyush Sinha wrote:
> will it be given that the number of elements is always even??
>
> On 5/28/11, ross wrote:
> > Hi all,
> >
> > Sort all elements in odd indices of an array in ascending o
@J,
I think you are wrong about the thread scheduling on multiple cores ,
As one of the basic mapping technique would be used to map user level
thread to kernel thread ,
one to one or many to many would serve the purpose .
the point it that , you cant control how it would be mapped to a
kernel thre
will it be given that the number of elements is always even??
On 5/28/11, ross wrote:
> Hi all,
>
> Sort all elements in odd indices of an array in ascending order and
> even indices in descending order.
> Finally, rearrange so that all even indexed elements come first.
>
> eg:
>
> input – 7 2 6
@srajan: ya , i made a mistake...the correct ans will of-course be 9978
On Fri, May 27, 2011 at 12:11 PM, srajan dongre wrote:
> @ankit sambyal.i think d rite answer will be 9978 in dat case.
>
>
>
>
> On Fri, May 27, 2011 at 11:50 PM, ankit sambyal wrote:
>
>> @shubam: won't work
Whats the point? There is infinite points to use threads, even in a
single core.
On Fri, May 27, 2011 at 4:05 PM, jagannath wrote:
> hi guys,
> i know that pthread is an user-level thread and an user-level
> can't take the advantage of SMP. Then what is the point of creating
> user-level th
@ankit sambyal.i think d rite answer will be 9978 in dat case.
On Fri, May 27, 2011 at 11:50 PM, ankit sambyal wrote:
> @shubam: won't work
>try following test case: 8,89,9
>
> Ankit Sambyal
>
>
>
> On Fri, May 27, 2011 at 11:11 AM, shubham wrote:
>
>> check whether these steps wo
hi guys,
i know that pthread is an user-level thread and an user-level
can't take the advantage of SMP. Then what is the point of creating
user-level threads if they can't be scheduled on multiple cores?Please
clear my doubt which has been hitting me for long...
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You received this me
To solve this, look at an 8x8 grid representing the games played. The
diagonal is not used, because teams do not play themselves. Below the
diagonal is the first game between each team and above the diagonal is
the second game. Assume that teams 1-4 are the ones who will go to the
semi-finals. This
Hi all,
Sort all elements in odd indices of an array in ascending order and
even indices in descending order.
Finally, rearrange so that all even indexed elements come first.
eg:
input – 7 2 6 4 8 3 1
even indexed : 7 6 8 1 => sort 8 7 6 1
odd indexed: 2 4 3 => sort 2 3 4
output – 8 7 6 1 2 3
@shubam: won't work
try following test case: 8,89,9
Ankit Sambyal
On Fri, May 27, 2011 at 11:11 AM, shubham wrote:
> check whether these steps work:
>
> step 1:
> sort the given numbers in the decreasing order based on their first
> digit.
> step 2:
> if two numbers come o
@shubham, 10 101 ??
On Fri, May 27, 2011 at 11:41 PM, shubham wrote:
> check whether these steps work:
>
> step 1:
> sort the given numbers in the decreasing order based on their first
> digit.
> step 2:
> if two numbers come out to be equal in the above case & both of
> their ne
check whether these steps work:
step 1:
sort the given numbers in the decreasing order based on their first
digit.
step 2:
if two numbers come out to be equal in the above case & both of
their next digit exist then sort on the basis of their next digit, otherwise
the number
@Piyush: try 97,8,9
acc. to ur algo, adding 0s: 97,80,90
then sorting : 97,90,80
so final ans acc. to ur algo: 9798
whereas the correct ans is : 9897
Ankit
BITS Pilani
On Fri, May 27, 2011 at 6:58 AM, Piyush Sinha wrote:
> how about adding zeroes at the end of integers to make to equal to
Ah! sorry.
This combination is not possible.
It will be 10,10,10,10,10,4,2,0. So, the answer is 11.
On May 27, 10:10 pm, L wrote:
> The worst case will occur when 5 teams have the same number of wins.
> As only 4 can qualify, one team with the same number of points will
> not be able to qualify.
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The worst case will occur when 5 teams have the same number of wins.
As only 4 can qualify, one team with the same number of points will
not be able to qualify.
1. 11
2. 11
3. 11
4. 11
5. 11
6. 1
7. 0
8. 0
In this scenario, a team with 11 points will not be able to qualify.
So, to ensure that i
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Check whether this is storing Google Search Results ? ? ?? ? ? ? ? ??
[HONEY POTTING]
On Fri, May 27, 2011 at 10:07 PM, UTKARSH SRIVASTAV wrote:
> test cases for internet explorer
>
> --
> *UTKARSH SRIVASTAV
> CSE-3
> B-Tech 2nd Year
> @MNNIT ALLAHABAD*
>
> --
> You received this message beca
test cases for internet explorer
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CSE-3
B-Tech 2nd Year
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No, Kadane's algorithm considers subarray sum, we are considering
concatenation ( for whole array ).
The solution with custom string comparator : http://ideone.com/doASH.
On May 27, 9:15 pm, Supraja Jayakumar
wrote:
> Hi
>
> Isnt this the Kadane's (largest subarray) problem ?
>
> Rgds
> Supraja J
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Kadane's algorithm is considers subarray sum, we are considering
concatenation here.
On Fri, May 27, 2011 at 9:45 PM, Supraja Jayakumar wrote:
> Hi
>
> Isnt this the Kadane's (largest subarray) problem ?
>
> Rgds
> Supraja J
>
> On Fri, May 27, 2011 at 9:41 AM, anshu mishra
> wrote:
>
>> @all g
Hi
Isnt this the Kadane's (largest subarray) problem ?
Rgds
Supraja J
On Fri, May 27, 2011 at 9:41 AM, anshu mishra wrote:
> @all go through this code
>
> #include
> #include
>
> using namespace std;
> bool compare(int a, int b)
> {
> string u, v;
> u = v = "";
> while (
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a king has two sons eric and bob.he wants to divide his
islands
the islands are in a queue.eric being elder gets the first
chancethey both can pick d island alternatively from beginning or
end of the queue only.design an algo so tht eric gets the max.
piece of land.
i hv sol
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@all go through this code
#include
#include
using namespace std;
bool compare(int a, int b)
{
string u, v;
u = v = "";
while (a)
{
u += (a % 10 + '0');
a/=10;
}
while (b)
{
v += (b % 10 + '0');
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@vishwakarma
thanks for rectifying me...
its clear... 12 is not posible, i was in another way :)
On Fri, May 27, 2011 at 7:46 AM, vishwakarma wrote:
> correction---it was typo mistake ...
> Team C loses to(one to A and one to B)
>
> On May 27, 7:44 pm, vishwakarma wrote:
> > so here we go
correction-->>"Then C loses two of its matches to(one to A and one to
C). " to "Then C loses two of its matches to(one to A and one to B)
".
On May 27, 7:44 pm, vishwakarma wrote:
> so here we go
>
> Let A loses two of its matches to (one to B and one to C).
> Let B loses two of its matches
correction---it was typo mistake ...
Team C loses to(one to A and one to B)
On May 27, 7:44 pm, vishwakarma wrote:
> so here we go
>
> Let A loses two of its matches to (one to B and one to C).
> Let B loses two of its matches to(one to A and one to C)
> Then C loses two of its matches to(o
so here we go
Let A loses two of its matches to (one to B and one to C).
Let B loses two of its matches to(one to A and one to C)
Then C loses two of its matches to(one to A and one to C).
Now.
team D, if it ever plays with (A,B,C) will loses..hence minimum number
o matches it is going to
@Aakash, yeah missed that, overriding default string comparator while
sorting will do that.
On Fri, May 27, 2011 at 8:11 PM, Arpit Mittal wrote:
> @piyush :
>
> for 1, 100
>
> you did
> 100,100
> then sort
> result 100,100
>
> so u said ans is 1100 but it could also be 1001 as 100=100...
>
> co
@piyush :
for 1, 100
you did
100,100
then sort
result 100,100
so u said ans is 1100 but it could also be 1001 as 100=100...
correct me if i am wrong?
On Fri, May 27, 2011 at 7:39 AM, Aakash Johari wrote:
> @vipul: try for 100 and 10
>
>
> 2011/5/27 • » νιρυℓ « •
>
>> @Piyush Sinha,
>> what
@vipul: try for 100 and 10
2011/5/27 • » νιρυℓ « •
> @Piyush Sinha,
> what about 9, 801
>
> 2011/5/27 • » νιρυℓ « •
>
>> Take input as vector of string or array of string
>> sort the vector
>> print from end to beginning
>>
>>
>> On Fri, May 27, 2011 at 7:51 PM, Logic King
>> wrote:
>>
>>> i
@D.N. I Think if you removes default constructor and give only parametrized
constructor...then it i going to give any errorit should work fine..
On Fri, May 27, 2011 at 4:40 AM, Aakash Johari wrote:
> Provide some default value to the parameterized constructor.
> *
> A(int m = 0) {
>
@Piyush Sinha,
what about 9, 801
2011/5/27 • » νιρυℓ « •
> Take input as vector of string or array of string
> sort the vector
> print from end to beginning
>
>
> On Fri, May 27, 2011 at 7:51 PM, Logic King wrote:
>
>> i agree with piyush...can't find the countercase...satisfied with the
>> alg
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Participate in
Take input as vector of string or array of string
sort the vector
print from end to beginning
On Fri, May 27, 2011 at 7:51 PM, Logic King wrote:
> i agree with piyush...can't find the countercase...satisfied with the algo.
>
>
> On Fri, May 27, 2011 at 6:58 AM, Piyush Sinha wrote:
>
>> how about
@Vishwakarma
it is now ok that 11 should be the answer, but why any 4 teams cannot win 12
matches in total...
for that they have to score 12*4 = 48 points out of 56. then wats the
problem.
i know how it is coming 11 now, but i am replying back for the doubt i have
in a line u just mentioned in y
@rishabh : now i understand it better... thanks :)
On Fri, May 27, 2011 at 7:22 AM, Rishabh Maurya wrote:
> because we want upper 4 teams to win maximum matches altogether so
> to satisfy this criteria .. last team should win 0 , and team 7 must have
> lost all its matches except from t
@Arpit
Any four team cannot win 12 matches in total.
...Rishabh is wid right answer that is ( " 11 " ).
Hence any team winning its any 11 out of 14 matches ensures its entry
to semis.
But not below 11 its entry to semi will depend on other team
performance.
On May 27, 7:11 pm, Arpit Mittal
i agree with piyush...can't find the countercase...satisfied with the algo.
On Fri, May 27, 2011 at 6:58 AM, Piyush Sinha wrote:
> how about adding zeroes at the end of integers to make to equal to the
> integer with maximum number of digits and sort them...
>
> ex-
> 101 10
>
> adding zeroes..
>
because we want upper 4 teams to win maximum matches altogether so to
satisfy this criteria .. last team should win 0 , and team 7 must have
lost all its matches except from team 8 , so it wins 2 and similarly team 6
wins 4 and team 5 wins 6 .
don't forget to watch MI vs RCB .. :)
--
sorry !!! correction-->>..i misread the problem
My solution gives "what is the lowest possible number of matches won
by a qualifying team ".
On May 27, 6:37 pm, vishwakarma wrote:
> Correct me if i m wrong !!!
>
> Number of matches of each team = 14.
> Let team A,B,C,D qualify for semifinal.
> 1
@rishabh :
in your solution u have taken scores of last 4 teams as 6 4 2 0. what if i
take 2 2 2 2 then the ans would be 56-(2+2+2+2)/4 = 12...!!!
and i can also take the scores of last 4 teams as 6 4 4 2 then the ans would
be
56-(6+4+4+2)/4 = 10!!!
so how you can say it would be 11?
On Fri,
how about adding zeroes at the end of integers to make to equal to the
integer with maximum number of digits and sort them...
ex-
101 10
adding zeroes..
101 100
sort 100 101
therefore make number as 10110
100 1
adding zeroes
100 100
therefore number is 1100
I am not sure of the method.i
No , you are wrong .. problem statement says how many matches should a
teams win to ensure its qualification , their no word like minimum or
maximum ...
8 gets wrong if following situation arises
1 -> 9
2 -> 9
3 -> 9
4 -> 9
5 -> 8
6 -> 6
7 -> 4
8 -> 2
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@radha, i think your solution is wrong.
for this case: 101,10
in your solution , the ans is 10101,but the max ans is 10110.
2011/5/27 radha krishnan
> 10100 is max ans
> okay
> convert the numbers to strings and sort based on the first character
> !!!
> if equal do that recursively and the
how about this case:
9, 100 -> 9100
100 9
9100
2, 3, 9, 78 -->
78 9 3 2
9 78 3 2
I guess solution should be:-
sort the array of numbers in an ascending order and then check for the first
element in the array, if there is any other element greater than it, shift
all the elements one right and
10100 is max ans
okay
convert the numbers to strings and sort based on the first character !!!
if equal do that recursively and then if length is less give that
preference !!
i think this solution ..
may be this is wrong !!
On Fri, May 27, 2011 at 7:07 PM, wujin chen wrote:
> @Piyush， how t
@Piyush， how to deal with this case :100 , 10
2011/5/27 Piyush Sinha
> we can work out if we sort according to the leftmost integer
>
> On 5/27/11, adityasir...@gmail.com wrote:
> > are you kidding me. Just simple sort wont work.
> >
> > On Fri, May 27, 2011 at 9:31 AM, radha krishnan <
> > rad
Correct me if i m wrong !!!
Number of matches of each team = 14.
Let team A,B,C,D qualify for semifinal.
1.maximum number of matches A can win = 14 (all played )
2.maximum number of matches B can win = 12 (all played except played
with team A)
3.maximum number of matches C can win = 10 (all play
Haha !! Any counter case against sort ? ?? ? :P
On Fri, May 27, 2011 at 7:02 PM, wrote:
> are you kidding me. Just simple sort wont work.
>
> On Fri, May 27, 2011 at 9:31 AM, radha krishnan <
> radhakrishnance...@gmail.com> wrote:
>
>> sort :)
>>
>>
>> On Fri, May 27, 2011 at 6:57 PM, ross wro
we can work out if we sort according to the leftmost integer
On 5/27/11, adityasir...@gmail.com wrote:
> are you kidding me. Just simple sort wont work.
>
> On Fri, May 27, 2011 at 9:31 AM, radha krishnan <
> radhakrishnance...@gmail.com> wrote:
>
>> sort :)
>>
>>
>> On Fri, May 27, 2011 at 6:57
are you kidding me. Just simple sort wont work.
On Fri, May 27, 2011 at 9:31 AM, radha krishnan <
radhakrishnance...@gmail.com> wrote:
> sort :)
>
>
> On Fri, May 27, 2011 at 6:57 PM, ross wrote:
>
>> Hi all,
>>
>> Given an array of elements find the largest possible number that can
>> be formed
sort :)
On Fri, May 27, 2011 at 6:57 PM, ross wrote:
> Hi all,
>
> Given an array of elements find the largest possible number that can
> be formed by using the elements of the array.
>
> eg: 10 9
> ans: 910
>
> 2 3 5 78
>
> ans: 78532
>
> 100 9
>
> ans: 9100
>
> --
> You received this message b
Hi all,
Given an array of elements find the largest possible number that can
be formed by using the elements of the array.
eg: 10 9
ans: 910
2 3 5 78
ans: 78532
100 9
ans: 9100
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Is it the minimum required matches to ensure for semifinals?
On Fri, May 27, 2011 at 6:06 AM, Rishabh Maurya wrote:
> suppose bottom 4 teams have won least matches and upper 4 teams have won
> equal number of matches ...
>
> 1 -> x
> 2 -> x
> 3 -> x
> 4 -> x
>
> 5 -> 6
> 6 -> 4
> 7 -> 2
> 8 ->
suppose bottom 4 teams have won least matches and upper 4 teams have won
equal number of matches ...
1 -> x
2 -> x
3 -> x
4 -> x
5 -> 6
6 -> 4
7 -> 2
8 -> 0
total matches are 56
and let upper four teams have won x matches each
so x = (56-(6+4+2+0))/4
x = 11
so in this way to ensure qualific
could you please explain how?
On Fri, May 27, 2011 at 3:45 AM, varun pahwa wrote:
> i think 11.
>
> On Fri, May 27, 2011 at 3:06 PM, Arpit Mittal wrote:
>
>> 8?
>>
>>
>> On Fri, May 27, 2011 at 2:26 AM, anil chopra
>> wrote:
>>
>>> 11
>>>
>>> On Fri, May 13, 2011 at 12:14 AM, amit wrote:
>>>
>>
Provide some default value to the parameterized constructor.
*
A(int m = 0) {
a = m;
} *
On Fri, May 27, 2011 at 4:36 AM, D.N.Vishwakarma@IITR wrote:
> without default constructor what will be solution
>
> On Wed, May 25, 2011 at 10:56 AM, Aakash Johari wrote:
>
>> This way you can do:
>>
without default constructor what will be solution
On Wed, May 25, 2011 at 10:56 AM, Aakash Johari wrote:
> This way you can do:
>
> #include
>
> using namespace std;
>
> class A {
> public:
> int a;
>
> A(int m) {
> a = m;
>
Can be solved in this way also :
> #include
> #include
>
> using namespace std;
>
> string a, b, c;
> int memo[51][51][51];
>
> int interleave(int ai, int bi, int ci)
> {
> int r1, r2;
>
> r1 = r2 = 0;
>
> if ( ai == a.size() && bi == b.size() && ci == c.size() ) {
> return 1
i think 11.
On Fri, May 27, 2011 at 3:06 PM, Arpit Mittal wrote:
> 8?
>
>
> On Fri, May 27, 2011 at 2:26 AM, anil chopra wrote:
>
>> 11
>>
>> On Fri, May 13, 2011 at 12:14 AM, amit wrote:
>>
>>> Consider a series in which 8 teams are participating. each team plays
>>> twice with all other teams.
what if c=9.. 9%4=1 ryt?
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In case of trie tree, you need not to traverse through the whole tree. Just
look at how trie tree works, you will come to know.
On Fri, May 27, 2011 at 2:27 AM, bhavana wrote:
> sorry guys fr my last post...bcoz dis doesnt provide any benefit..in terms
> of either space or time .
>
>
> On Fri, M
8?
On Fri, May 27, 2011 at 2:26 AM, anil chopra wrote:
> 11
>
> On Fri, May 13, 2011 at 12:14 AM, amit wrote:
>
>> Consider a series in which 8 teams are participating. each team plays
>> twice with all other teams. 4 of them will go to the semi final.How
>> many matches should a team win, so th
sorry guys fr my last post...bcoz dis doesnt provide any benefit..in terms
of either space or time .
On Fri, May 27, 2011 at 2:55 PM, bhavana wrote:
> @himanshu : den do map the wrong words as u go on finding them in the doc.
>
>
> On Fri, May 27, 2011 at 2:51 PM, himanshu kansal <
> himanshukan
11
On Fri, May 13, 2011 at 12:14 AM, amit wrote:
> Consider a series in which 8 teams are participating. each team plays
> twice with all other teams. 4 of them will go to the semi final.How
> many matches should a team win, so that it will ensure that it will go
> to semi finals.?
>
> --
> You
@himanshu : den do map the wrong words as u go on finding them in the doc.
On Fri, May 27, 2011 at 2:51 PM, himanshu kansal <
himanshukansal...@gmail.com> wrote:
> thnks shashankbt i hv used trie for building the dictionary..bt
> since d wrong words does nt hv a common prefix...so it is
thnks shashankbt i hv used trie for building the dictionary..bt
since d wrong words does nt hv a common prefix...so it is nt suitable for
storing wrong words(worst case scenario is common)...
so i need a ds tht wd make d search of wrong words faster.nd v dont hv
to scan d entire diction
In the second approach I wrote to use array for mapping
> "you can simply map the existence/non-existence of any particular element
> in an array. that will be in constant time (for query purposes) and O(n)
> time for preprocessing."
>
> On Fri, May 27, 2011 at 2:18 AM, sukhmeet singh wrote:
> a
actly i did.. but bhavana didn;t used STL ..!!
My question to you was regarding Dave 's query which i didn't understand
what he meant by saying : "@Aakash: And tell me how map works. Is making an
entry O(1) regardless
of the value of the entry? For example, is it O(n) to map the sequence
1, 4, 9,
@sukhmeet: could you get my approach? it was same as Bhavana explained.
On Fri, May 27, 2011 at 2:12 AM, bhavana wrote:
> hehe...d difference is regarding time complexity...bcoz map takes 0(logN)
> for insertion while array can b accessed in constant time through index.
>
>
> On Fri, May 27, 201
hehe...d difference is regarding time complexity...bcoz map takes 0(logN)
for insertion while array can b accessed in constant time through index.
On Fri, May 27, 2011 at 2:39 PM, sukhmeet singh wrote:
> k.. got it .. but it was same as putting them into a map ..if the bound 'b'
> is not known..
Newline Character '\n' => ascii 10
On Fri, May 27, 2011 at 2:36 PM, Bhavesh agrawal wrote:
> #include
> /* copy input to output; 2nd version */
> main()
> {
> int c;
> while ((c = getchar()) != EOF && printf("%d\n",c))
> {putchar(c);
> printf("\n");}
> }
>
> it's output is like
> a
> 97
> a
> 1
k.. got it .. but it was same as putting them into a map ..if the bound 'b'
is not known.. as said be Akash ..
But if STL is not allowed your approach is mch better..Noogle..
also a simple change that can be done if the each number is that we can
check if (sum - a[i] )!= i then getting same can b
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