it wud give u continuous...subarray...if u want non continuous..question
shud be subsequence...and for that u need to all combination O(n^20..which
is offcourse bruteforce...
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make a sum array..such that..sum[i]=sum of number from 0 to i-1make a
node which has 2 data..sum[i] and ith index itself..sort this node array
according to sum[i]...now check for consecutive value which have same
value...corresponding index i,j to that sum nodes will be start and end of
Code:-
#includeiostream
#includevector
using namespace std;
void recursion(int sum,int i,vectorint v,int size)
{
vectorint v1=v;
int size1=size;
if(sum==0)
{
for(int k=0;ksize;k++)
coutv[k] ;
coutendl;
}
else
{
for(int j=i+1;j10;j++)
@atul-I think This Should work for n dimension-
complexity O(n^no.of dimesions)
:-
have N-dimension check for which Tile can contain which Tile.i.e (3,3,4)
can contain (2,3,4) .
Now
1.Take the titles which cannot contain any-other tile set no-of tile if it
is base =1;
2.now take tiles which can
Hope This will Work..
int Recursion(node *root)
{
if(root-left==Nullroot-right==Null)
{
return root-data;
}
int a=0,b=0;
a=recursion(root-left);
b=recursion(root-right);
if(a+b+root-datamax)
max=a+b+root-data;
return max(a,b)+root-data;
}
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@all-This is almost the same question asked in facebook..
Dave sir is correct!
Example-file contain only a b a c;
Then step1:
save a;
step2:
save b with probability of value 0 by rand()%2;
therefore pr[1]=1/2;pr[2]=1/2;
step3;
save a with probability of value 0 by rand()%3;
therefore
@vaibhav-yup :-)
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@Dave Sir-Got ur point..thnks!!
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@Carl-At each step you are calculating lcp between the text and the last
entry probably to compare ... lcp[i+1]..is it linear still
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@Atul- can you explain ur approach a little more..What is A and B string wd
reference to question..for each A and B string it will take O(n) time to
make the table...ur approach is O(n)..dont think so..please explain may be
i m wrong
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@carl-Sorry was making suffix array by naive approx in the link
http://en.wikipedia.org/wiki/Suffix_array .there are more optimized algos
for it...!! thanks
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Thanks.carl and atul.!!.@carl-i got lgn due to string comparison sort while
making the suffix array..:)
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@Carl- I didnt got ur point completely abt Lcp array..can you demonstrate
on the below example...
Example for ababaa
answer shud be -11
suffix array wud be:-
a
aa
abaa
ababaa
baa
and Lcp array would be then
0
1
1
3
0
..correct if wrong..whats next...
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@carl- got ur point..but complexity is more..suffix array takes
o(n^2lgn)..considering string comparisons. complexity to build...i already
have o(n^2)..want o(n)..
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may be Lucas theorem will be helpful...google it!!
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What is the efficient algorithm to partition a big number like 1000 into
500 parts such that each part is=0,i need to know all the partition ;ex
partition of 5 into 3,4 1 0,3 1 1,2 2 1,i know brute force methods..please
provide some algorithm which is feasible to calculate all the partition
upto
@Dave- really nice , that will do,thnks a lot!
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I want to calculate all binomial coeffficient upto a particular number say
1000.,like 1000C1 100C2
1000C3...999C1,999C2..998C12C1 1C1.
all possible binomial coefficient out of 1000*1000 modulo 17;I want
a maximum complexity of o(n^2);for 1000,i had some code of mine
@jai-thanks!!! , it worked...,is there any method for big numbers like
10,in my case 1000 so storing in array[10^3][10^3] worked but what if
10^5,storing will not work then??,anyway thanks a lot again
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@amrit- thnks!!
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@atul-its nice one, but its complexity is probably greater than o(n^2)
just to calculate one mCk,i want a sequence of all mCk in a
less complexity
i tried for some code which calculate all mC1,mC2..mCk in less
complexity.this works good till (100)Ck ;
if while loop can be reduced i can get a
@atul-its nice one, but its complexity is probably greater than o(n^2) just
to calculate one mCk,i want a sequence of all mCk in a
less complexity
i tried for some code which calculate all mC1,mC2..mCk in less
complexity.this works good till (100)Ck ;
if while loop can be reduced i can get a
@prakash...ya i realized that and it will be sorted row and column
wise.which will become same as forming min-heapand my algo will
become same what lucifer had already specified...
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I thought of a simpler algo, i m using the property of this matrix
That is. a[i][0] is smallest of a[i][] and a[i+1][]so on,so to
decide next smallest element i will only consider
a[i][0]...
i will swap element if required and sort the row(too by swapping) if
element are changed.
on the
50 6 9---
8 7 22 45 55--- 7 8 22.and so on...
for swapping without space use
a=a+b;
b=a-b;
a=a-b;
it worked for examples i had takenpls notice me if any flaw with this..
On 1/11/12, pankajsingh psingh...@gmail.com wrote:
srry some formatting problem...i will repost later
sorryi was thinking n as number of nodes...
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yours question is slightly unclear.,if u need to count all possible
number of triangle,why i need to divide in 3 parts...,what i think u r
looking for is dividing in 3 parts which give me maximum number of possible
triangleif it is so, divide in more or less equal parts...
i want to calculate values like (100 C 1) %17,what would be
better algorithm for it,i think lucas theorem cant be used in this
case.want some efficent algorithm,actually i want to calculate
mC1,mC2mC1000 %17,such that may is about 10^6
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Thanks Dave,
but i want some more efficient in my case, something like O(k) to calculate
all mC1 mC2...mCk,
i already had a worst time O(k^2),
i.e
for (long long int i1=1;i1=k;i1++)
{
while((result*(m-i1+1))%i1)
{
result=result+17;
}
result=(result*( m-i1+1));
result /= i1;
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