it is rotated to get 14-1 O(log(n)) how can
you find rotation point in log(n) ?
On Fri, Jun 8, 2012 at 6:57 PM, partha sarathi Mohanty
partha.mohanty2...@gmail.com wrote:
It is easy.. find the point where it is rotated to get 14-1 O(log(n))
since 214 that means u have to find
@saurabh: why would u count all??? just see while counting if the bitmap is
set.. then return the char.
On Fri, Jun 8, 2012 at 4:33 PM, Saurabh Yadav saurabh...@gmail.com wrote:
order of hashing and counting is important
eg. abba
if we do hashing by characters 'a' is stored before 'b'
and
It is easy.. find the point where it is rotated to get 14-1 O(log(n))
since 214 that means u have to find it in subarray [123].. do a binary
search here o(long(n))
final 2*O(log(n))...
On Fri, Jun 8, 2012 at 7:44 PM, Dave dave_and_da...@juno.com wrote:
@Hassan: This is not possible without
no
if ta/om/to/am/ba/batot/amoma are words then it should be the one. no
5*2 3*3
On Tue, Jun 5, 2012 at 7:45 PM, Gene gene.ress...@gmail.com wrote:
Does this sufficae?
Suppose you were using a dictionary from the frapplewonk language,
which has only 5 words:
tab
oma
to
am
ba
@ashish : couldnt get u.. can u give an example??
On Tue, May 22, 2012 at 5:45 PM, Prem Krishna Chettri hprem...@gmail.comwrote:
What I Could possibly think of is
For each string S1 that is an anagram of some string S, use Map and Store
the Key Value as (S1,S). Now there is a trick here abt
Java has something call treeMap. it stores strings lexographically.. u can
always do permutations and store them in a treeMap. and get the rank
then... just the idea.. will post the solution once i am done.. what do u
guys think.abt the idea???
On Tue, May 22, 2012 at 9:46 AM, atul anand
recurse on left and right will do it and keep comparing the values
On Tue, May 22, 2012 at 2:50 PM, atul anand atul.87fri...@gmail.com wrote:
no need of creating another mirror tree
you just need to call the function func(root-left,root-right);
now left sub tree and right sub tree will be
);
permute(st + chars.charAt(i), newString);
} catch (Exception e) {
e.printStackTrace();
}
}
}
}
On Tue, May 22, 2012 at 2:19 PM, partha sarathi Mohanty
partha.mohanty2...@gmail.com wrote
@ashish.. it wont be constant space then.. surely it will be o(n) though
On Mon, May 21, 2012 at 7:23 PM, Ashish Goel ashg...@gmail.com wrote:
Dave,
Cant we have a hash table with the item as key and its count as value
(walk over array A and build HT).
For permutation check, walk over
a[] = [-1,-3,4,0,7,0,36,2,-3]
b[] = [0,0,6,2,-1,9,28,1,6]
b1[] = [0,7,0,36,4,-6,3,0,0]
b2[] =[-1,-3,11,0,0,0,35,0,0]
suma = 42 proda = -84*72*3
sumb = 51 prodb = -84*72*3
sumb1 = 44 prodb1 = -84*72*3
sumb2 = 42 prodb2 = 33*35
do the sum and prod operation w/o 0s and compare the values..
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