is O(long n!)
n! is O(n^n) by sterling approximation
hence it is O(log n^n) = O(nlogn)
On Sun, Oct 28, 2012 at 11:14 PM, Pralay Biswas
pralaybiswas2...@gmail.comwrote:
@ Siddharth :
Well, here is how you may understand it:
1) There is an outer loop that runs n times. (k)
2) Then there is an
use pascal's triangle
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string str;
while(cin str)
{
cout str endl;
}
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@bharat: When each step of nishant's algo is O(n) how can it sum up to
O(nlogn)???
On Tue, Sep 13, 2011 at 9:18 AM, bharatkumar bagana
bagana.bharatku...@gmail.com wrote:
@nishant : your algo takes O(nlogn) time with higher constant behind
this. can't we write better than this ?
@sairam
With hashing..
make a hash table of elements from 0 to sum/2
if an element k is in sum/2 to sum then check if sum-k is in the hashtable.
take care of the case when sum is even and sum/2 occurs only once.
TC: O(n) Space complexity: O(sum)
Method2: Sort the elements.
Now maintain to pointers one at
@Neha take 42, 21 and 1
42 ^ 1 =43
while 42 ^21 =63
On Fri, Aug 26, 2011 at 10:28 PM, Neha Singh neha.ndelhi.1...@gmail.comwrote:
Sort the nos., which can be done in O(nlogn)
Now the 1st and the last integers are the required integers.
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Hi hary,
when do you think that there can be problem? The sq is never made to store a
value 10^9 (for previous sq) +9*10^9 (for new sum)
hence the sum never exceeds 10^10 which is under the range of int.
Correct me if i go wrong somewhere.
On Thu, Apr 28, 2011 at 12:26 PM, hary rathor
sort two of the list and pick one number in C(unsorted) to find if there
exists a pair with the sum.
This can be done in linear time by maintaining two pointers.
Hence O(n^2).
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amazom needs 7*4=28 bytes
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@Dave: Sorry It was a typo but for the probability figures,
When C shoots B then if he is successful then A will shoot C
Hence he must be unsuccessful and then
if B is unsuccessful then A must will Kill B and then C must kill A
if B is successful then C must kill B now
Hence P(when C shoots at
@swayambhoo:
ofcourse a cubical room must me symmetrical at allcorners, Hence, neway it
will reach in
min_dis=sqrt((4+3)^2+5^2)=8.6
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For n x m take first element of all the rows and insert in a min heap.
Now take the smallest element and and if we have a track of from which row
it belongs, we can take an element out of that row
and insert in the heap.
this will be done n x m times. Hence we have a time complexity of
O(nmlog(m))
for each element in the second array apply binary search in first array. ie,
For X[1] find 1 in first array with binary search.
Time Complexity=O(nlogn)
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make another array(B) from (A) with all elements negated
now find one element from B and one from A whose sum is x or -x. This can
ofcourse be done in O(n).
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Algo:
In first traverse find the min and the max values.
if (max-min) not equals (N-1)
return false
In next traverse map each in a hashtable of size N where index=key-min. Now
in case of collision return false
return true
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rem=1;
for(j=3;j=N+1;j++)
rem=(rem*j)%n;
return rem;
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@coolfrog
the question never asked u to find thm in order of thier distances.
Correct me if i m wrong!
find the distances in O(n)
now using the partitioning process of quicksort.
Dividing the array into two parts:
Now if the Length of the first part is less than or equal to i we have to
now make
#includestdio.h
int main()
{
int arr[11]={0},sum2[1001]={0};
int type[1001]={0};//0 for tails
int N,Q,i,j,sum;
int a,b,c;
scanf(%d,N);
scanf(%d,Q);
for(i=0;iQ;i++)
{
scanf(%d%d%d,a,b,c);
if(a==0)
{
j=b;
int
@fenghuang: the last step will take O(n logn ) . Or there is some better
way?
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will O(n) work or u wish something lesser dependent on k or log(n) ?
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result=((n ((128)/3))1)|((n ((129)/3))1);
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