any one give a good link to study Dynamic Programming concepts??
--Regards
Vikram
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Given an array say A=(4,3,1,2). An array B is formed out of this in
such a way that B[i] = no. of elements in A, occuring on rhs of A[i],
which are less then A[i].
eg.for the A given, B is (3,2,0,0).
Here A of length n only contains elements from 1 to n that too
distinct..
Now the problem is:
1).
The naivest solution is O(n2). Could you give a hint for
this problem?
On Oct 3, 4:39 pm, Vikram Singh singhvikram...@gmail.com wrote:
Given an array say A=(4,3,1,2). An array B is formed out of this in
such a way that B[i] = no. of elements in A, occuring on rhs of A[i],
which are less then A[i
@anshu: pls elaborate... give an example...
On Tue, Oct 4, 2011 at 9:51 AM, anshu mishra anshumishra6...@gmail.comwrote:
use segment tree or binary index tree to solve O(nlogn)
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ok.. thanks guys...
one doubt now... if this ques is asked in an interview(its already been
asked in ms interview)... then u cant just write the code... u hv to explain
the whole approach like why u r choosing that way to narrowing dowm the
range...
so pls explain how this sol is derived...
On
any one suggest an algo to write own sqrt func...??
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how abt this:
(i hv used some variables)
a=x;
b=-x;
k1=a7; (if we assume 8 bit number)
k2=b7;
n=k1 | k2;
n=n-1;
k=n7;
return (y-k*(y-z));
Explanation:
if(x0 or x0)
a and b are of opp signs,
one of k1 and k2 is 0and other is 1...
n=1;
n becomes 0;
k =0;
y is returned
if(x==0)
a=b=0;
i m asking again, cos no one replied earlier:
in example given by mohit... is the leftmost right cousin of 8 and 9,
NULL or
12??
On Aug 27, 6:42 pm, aditi garg aditi.garg.6...@gmail.com wrote:
This grp is full of MS interview ques..search the archives...
On Sat, Aug 27, 2011 at 6:55 PM, teja
how abt this:
if(!a[0][0])
{
first traverse the 1st row till we find any 1...
On Sep 2, 12:32 pm, kranthi raj kranthi...@gmail.com wrote:
oops missed Space Complexity
On Fri, Sep 2, 2011 at 12:55 PM, kranthi raj kranthi...@gmail.com wrote:
for( i = 0 ; i n ; ++i )
for( j = 0 ;
how abt this:
if(!a[0][0])
{
first traverse the 1st row till we find 1.
if dere is 1 do a[0][0]+=2;
then traverse the first column till 1..
if dere is 1... do a[0][0]+=3;
}
apply dave's method
if(a[0][0]==2)
make 1st row 1
else if(a[0][0]==3)
in above example... is the leftmost right cousin of 8 and 9, NULL or
12??
On Aug 10, 4:36 pm, Brijesh Upadhyay brijeshupadhyay...@gmail.com
wrote:
thank u , i couldnot have answered this :P
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i figured out algo to find the inorder predecessor of a bst without
using parent pointer... just wanna confirm if its missing any case
if the left child(subtree) of node exist, then predecessor ll be the
max value in the left subtree.
else predecessor ll be one of the ancestor in this
sukrandha...@gmail.com wrote:
is it not possible to traverse tree in order and store in array. then figure
out the element and print the previous element?
On Fri, Aug 26, 2011 at 2:04 PM, Vikram Singh singhvikram...@gmail.comwrote:
i figured out algo to find the inorder predecessor
;
}
}
return parent;
}
}
i hope it makes u understand@sanjay...
On Aug 26, 5:27 pm, Sanjay Rajpal srn...@gmail.com wrote:
Vikram : will u plz elaborate more on ur solution ?
Sanju
:)
On Fri, Aug 26, 2011 at 5:24 AM, Vikram Singh singhvikram...@gmail.comwrote
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