@Dave : is it a working code??
i have executed your code but getting wrong output...and value of s is
becoming -ve inside loops.
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0.5 ( G(h - 1, i - 1) + G(h - 1, i) ) should be 0.5 ( G(h - 1, i - 1) + G(h
- 1, i+1) )
i am not clear why the parents of a cup in upper row are consecutive?
Best Regards
Ashish Goel
Think positive and find fuel in failure
+919985813081
+919966006652
On Tue, Feb 28, 2012 at 10:43 AM, Gene
Dave, why the assumption that nothing is coming from left side.
Every cup gets something from cup left above and right above itself when
they have something extra?
Best Regards
Ashish Goel
Think positive and find fuel in failure
+919985813081
+919966006652
On Mon, Feb 27, 2012 at 8:17 PM, Dave
Gene, your DP is what i was thinking of but in code i could not coreleate
G(h - 1, i - 1) + G(h - 1, i) part (:
Best Regards
Ashish Goel
Think positive and find fuel in failure
+919985813081
+919966006652
On Tue, Feb 28, 2012 at 7:50 AM, Gene gene.ress...@gmail.com wrote:
G(h - 1, i - 1) +
@Dave : my code is not that complicated ...if you ignore the helper
function and check fillCup();
it just similar to preorder travesal and pour half to left and right child.
here is the explanation :-
node* fillCup(node *root,float pour,float capacity)
{
float temp;
if(root==NULL)
@Dave : yeah sorry its O(n) where n is number of nodes.
yeah as i said before its a nice approach...
On Tue, Feb 28, 2012 at 10:15 AM, Dave dave_and_da...@juno.com wrote:
@Atul: I don't have an n in my algorithm, so I'm not sure what your
assessment that my algorithm is O(n^2) means. My
@Gene , @dave : +1 +1
On Tue, Feb 28, 2012 at 10:49 AM, atul anand atul.87fri...@gmail.comwrote:
@Dave : yeah sorry its O(n) where n is number of nodes.
yeah as i said before its a nice approach...
On Tue, Feb 28, 2012 at 10:15 AM, Dave dave_and_da...@juno.com wrote:
@Atul: I don't have
@all
same doubt qstn appears to be of binary tree DS
but the diagram given in between qstn is not like Binary tree so
sharing is there
so how sharing is done plz explain??
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@Ravi: checkout this code...i have created tree where there is sharing of
nodes..
here is my code :-
please let me know is case you find any bug.
#includestdio.h
typedef struct tree{
int idx;
float data;
struct tree *left;
struct tree *right;
}node;
node *createNode(int index)
{
node *temp;
@Sourabh - whats the running time?
On Mon, Nov 28, 2011 at 3:28 AM, Ankur Garg ankurga...@gmail.com wrote:
Cool Solution...I was thinking of DP but wasnt clear on the recurrence...
Nice thinking man and thanks :)
On Mon, Nov 28, 2011 at 2:47 AM, sourabh sourabhd2...@gmail.com wrote:
Hey Sourabh
Could you please explain the solution in a bit detail perhaps using an
example or so..It wud be really helpful ..Just logic not code
On Mon, Nov 28, 2011 at 1:03 AM, sourabh sourabhd2...@gmail.com wrote:
Looks like a dynamic programming problem
Say F(n,k) denotes the maximum
Cool Solution...I was thinking of DP but wasnt clear on the recurrence...
Nice thinking man and thanks :)
On Mon, Nov 28, 2011 at 2:47 AM, sourabh sourabhd2...@gmail.com wrote:
Consider the example that you have given:
[0,0,1,1,0,0,1,1,0,1,1,0] , here n = 12 and k=3
Now we need to
since its a phone number storing problem, you can sort the numbers and store
the differences. That way you can generate the required number on the go
On Thu, Jun 30, 2011 at 4:39 AM, juver++ avpostni...@gmail.com wrote:
@Navneet
Please read problem again - it is about memory efficient storing.
In TRIE , we can store nodes by characters. So it is very easy to search by
names and hence their corresponding numbers :) .
TRIE saves a lot memory coz it stares a character only once.
LIKE :-
if i want to save sagar with phone no. 123456789 then we store it in TRIE
as :
s-NULL
|
a-NULL
|
Sorry for the previous post it got a mistake here take a look again :-
In TRIE , we can store nodes by characters. So it is very easy to search by
names and hence their corresponding numbers :) .
TRIE saves a lot memory coz it stares a character only once.
LIKE :-
if i want to save sagar
I wonder why people have discarded the idea of Hash map here.
Searching is obviously the most important task here and if we are to
assume that names can be uniformly hashed over a table of 1000 rows
with each row containing 1000 names each.
Further optimization can be achieved by having names
@samby
You are wrong anyway. Main problem is to reduce memory while storing phone
numbers.
We have 1 million of phones, they have many common prefixes which can be
addressed by trie.
For storing names, you may use any data structure which is best for the
particular problem.
Key is name, and
@Navneet
Please read problem again - it is about memory efficient storing.
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Please explain why you think TRIE use more space?
To my knowledge TRIE says lot of memory as the common numbers are saved only
once.. If you have any good reason then please explain and don't make any
single line statements.
On Wed, Jun 29, 2011 at 9:21 PM, MONSIEUR monsieur@gmail.com wrote:
@Swathi :We can't use trie data structure to store the phone numbers.
The most sound reason is that the users require phone numbers to be
sorted by name, but by using the trie data structure we can't get the
phone numbers which are sorted by name. Again we can't use trie whose
nodes are numbers,
http://anandtechblog.blogspot.com/2011/06/bitonic-merge.html
On Fri, Jun 24, 2011 at 2:17 AM, sankalp srivastava
richi.sankalp1...@gmail.com wrote:
1,2,43,41,5 , 6
Start at a[3] and a[5] Swap them up .
Reversing it , we get
1,2,43,5,6,41
This does not work
.
On Jun 23, 9:05 pm, Swathi
@Anand : Plz explain ur algo ???
On Fri, Jun 24, 2011 at 10:55 AM, Anand anandut2...@gmail.com wrote:
http://anandtechblog.blogspot.com/2011/06/bitonic-merge.html
On Fri, Jun 24, 2011 at 2:17 AM, sankalp srivastava
richi.sankalp1...@gmail.com wrote:
1,2,43,41,5 , 6
Start at a[3] and
We just need to find the start and end of the decreasing sequence then we
have to reverse the elements in that decreasing sequence by swapping the
elements at both the edges...
On Thu, Jun 23, 2011 at 2:13 PM, sankalp srivastava
richi.sankalp1...@gmail.com wrote:
@piyush Sinha
How can you do
I heard somewhere in some online video lecture that sites like tinyurl
change the address using more than base three {base 6 } and then apply
hash
On 6/4/11, bittu shashank7andr...@gmail.com wrote:
well i can speak much on these question.as these algorithms are part
of web crawler ..but do u
Nishant's soln is incorrect because he assumes, ctrlA and ctrlC are pressed
each time ctrlV is pressed.
As saikat has pointed out, this is incorrect.
According to me:
*buff = 0; //keeps track of last ctrlC*
*for each i*
*{*
* dp(i)=max(dp(i-1)+1, 2*dp(i-3), dp(i-1) + buff)*
*
@ Nikhil sir : I have coded the same solution, but was waiting for its
correctness to be proved... thanx.. :)
On Fri, Jan 28, 2011 at 1:01 PM, Nikhil Jindal fundoon...@yahoo.co.inwrote:
Nishant's soln is incorrect because he assumes, ctrlA and ctrlC are pressed
each time ctrlV is pressed.
As
hope this works :
#includestdio.h
#define MAX(A,B) AB?A:B
#defineMIN(A,B) AB?A:B
int FindMaxA(int n)
{
int i,k,factor,max = 0,cur,prev;
int* arr = new int[n+1];
int p = MIN(n,4);
for( int j = 1;j = p;j++)
arr[j] = j;
for(i=5;i=n;i++)
{
k = i-4;
According to me Nishaanth's solution is incorrect, as let for n =10, your
output : m=16
but my output : m =20: For first 5 times hit 'A', then ctrl+A, ctrl+C
resulting in 7 keystrokes. then 3 times ctrl+V, which result in m = 20.
On Thu, Jan 20, 2011 at 9:24 PM, abhijith reddy d
but my output : m =20: For first 5 times hit 'A', then ctrl+A, ctrl+C
resulting in 7 keystrokes. then 3 times ctrl+V, which result in m = 20.
Try this on a notepad. you will only 15A's
On Thu, Jan 20, 2011 at 12:46 PM, Saikat Debnath saikat@gmail.comwrote:
According to me Nishaanth's
Hi,
I think this method will work:
Possible Number of A's = N/2(1+R)
where R=N-(N/2+3)
assuming 11/2 = 5
Thanks
Preetam
On Fri, Jan 21, 2011 at 2:29 AM, Anand anandut2...@gmail.com wrote:
but my output : m =20: For first 5 times hit 'A', then ctrl+A, ctrl+C
resulting in 7 keystrokes. then
How about the following dynamic programming solution.
Let dp[i] be the max no of As with i keystrokes.
dp[i]=max(dp[i-1]+1,2*dp[i-3])
dp[N] is the required solution.
Correct me if i am wrong.
On Wed, Jan 19, 2011 at 9:20 PM, Raj rajmangaltiw...@gmail.com wrote:
this will not work out
a[0]b[0] doesn't mean that a[0]+b[i] is ith largest sum
try
int a[]={10,8,6,4,1};
int b[]={9,6,3,2,1};
Best Regards
Ashish Goel
Think positive and find fuel in failure
+919985813081
+919966006652
On Wed, Oct 6, 2010 at 11:36 PM, ligerdave david.c...@gmail.com wrote:
)x(n-3) would leave n-biggest
out? What is the criterion to chose the Ist quardrant?
manish...
From: topojoy biswas topojoy.bis...@gmail.com
To: algogeeks@googlegroups.com
Sent: Thu, 22 July, 2010 10:10:58 AM
Subject: Re: [algogeeks] a google question
im sry
...@gmail.com
*To:* algogeeks@googlegroups.com
*Sent:* Thu, 22 July, 2010 10:10:58 AM
*Subject:* Re: [algogeeks] a google question
im sry the L should look like this:
109 8 5 321
7 17 16
8 18 17
9 19 18
10 20 19
11 21 2019 16
consider these arrays:
10 9 8 5 3 2 1
and
13 12 11 10 9 8 7
form a grid like this:
109 8 5 321
7 17 1615 1210 98
8 18 1716 1311 10 9
9 19 1817 1412 12 10
10 20 1918 1513 12 11
11 21 20
im sry the L should look like this:
109 8 5 321
7 17 16
8 18 17
9 19 18
10 20 19
11 21 2019 1614 13 12
12 22 2120 1715 14 13
13 23 2221 1816 15 14
I missed a row in the last mail.
On Thu, Jul
this ques was asked by google.. it could find any gud soln than order n*n
On Mon, Jul 19, 2010 at 10:55 AM, 王奇凡 wqfhena...@gmail.com wrote:
@Kushwaha
Your programe is wrong.
Try this input:
a[ ] = {30,25,19,16,14};
b[ ] = {20,18,12,10,8};
the right answer should be 50 48 45 43 42
but
this ques was asked by google.. i* could find any gud soln than order n*n
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i think ,you can use the idea of merg (merg sort)
--- On Mon, 7/19/10, manish bhatia mb_mani...@yahoo.com wrote:
From: manish bhatia mb_mani...@yahoo.com
Subject: Re: [algogeeks] a google question
To: algogeeks@googlegroups.com
Date: Monday, July 19, 2010, 5:03 PM
Given two sorted postive
: [algogeeks] a google question
Here is a solution of O(n) , taking 4 pointers 2 for each array
#include cstdio
#includeiostream
using namespace std;
#define N 10
int main(void)
{
int arr1[N] = {8,7,4,3,2,1,1,1,1,1};
int arr2[N] = {34,23,21,19,15,13,11,8,4,2};
int *p11,*p12
@Kushwaha
Your programe is wrong.
Try this input:
a[ ] = {30,25,19,16,14};
b[ ] = {20,18,12,10,8};
the right answer should be 50 48 45 43 42
but the program output is 50 48 45 43 37。
2010/5/2 Jitendra Kushwaha jitendra.th...@gmail.com
Here is a solution of O(n) , taking 4 pointers 2 for
...@gmail.com
To: algogeeks@googlegroups.com
Sent: Sun, 2 May, 2010 9:13:14 PM
Subject: Re: [algogeeks] a google question
Here is a solution of O(n) , taking 4 pointers 2 for each array
#include cstdio
#includeiostream
using namespace std;
#define N 10
int main(void)
{
int arr1[N
The nature of the problem involves inserting some elements in heap and
retriving back ..It could be solved in worst case O(n * lg(n)).
Average case O(n) solution is not there I believe.
-Arun prasath N
On Fri, Apr 30, 2010 at 5:35 PM, divya sweetdivya@gmail.com wrote:
Given two sorted
@Algoose Chase
point taken
Thanks
Mohit Ranjan
Samsung India Software Operations.
On Sat, May 1, 2010 at 8:43 PM, Algoose Chase harishp...@gmail.com wrote:
@mohit
The idea of DP is fine.
When you find the Max i dont think you need to include A[i+1]+B[j+1]
because it can never be greater
@divya You're rite. Post a solution if you have one.
--
Regards,
Vignesh
On 2 May 2010 13:14, divya jain sweetdivya@gmail.com wrote:
@Mohit
according to ur algo if a[1], b[0] has sum greater than a[0],b[1]
then i is incremented i is now 2 so for next iteration u ll compare a[2]
b[0]
This problem has been discussed before @
http://groups.google.co.in/group/algogeeks/browse_thread/thread/9c1e1aa8cf1ed437/d451cd9468d985f7
I believe, the problem can't be solved in O(n) but only in O(nlog n).
@Divya: Are you sure the interviewer explicitly asked for an O(n) time
algorithm?
On
Hi
will this work ?
since we need Set S with n pairs of largest values , any such pair within
the set would (always) contain A[0] or B[0] because they maximize the value
of the pair.
so
// code snippet
typedef std::vectorint,int Pairs;
Pairs.push(A[0],B[0])
int i = 1; // index for ListA
int j
i found this question on some site and it was mentioned there todo in o(n).
i dont have the solution
@ above
ur solution doesnt include the case of ncluding a[i] b[j] it takes only a[i]
b[0] or b[j] a[0]
On 2 May 2010 18:11, Algoose Chase harishp...@gmail.com wrote:
Hi
will this work ?
OOPs.. sorry
this doesnt work !
On Sun, May 2, 2010 at 6:11 PM, Algoose Chase harishp...@gmail.com wrote:
Hi
will this work ?
since we need Set S with n pairs of largest values , any such pair within
the set would (always) contain A[0] or B[0] because they maximize the value
of the pair.
Here is a solution of O(n) , taking 4 pointers 2 for each array
#include cstdio
#includeiostream
using namespace std;
#define N 10
int main(void)
{
int arr1[N] = {8,7,4,3,2,1,1,1,1,1};
int arr2[N] = {34,23,21,19,15,13,11,8,4,2};
int *p11,*p12,*p21,*p22;
p11 = p12 = arr1;
@mohit
The idea of DP is fine.
When you find the Max i dont think you need to include A[i+1]+B[j+1] because
it can never be greater than both A[i+1]+B[j] and A[i]+B[j+1] since both the
lists are sorted in decreasing order.
On Fri, Apr 30, 2010 at 8:47 PM, mohit ranjan
On Fri, Apr 30, 2010 at 4:35 PM, divya sweetdivya@gmail.com wrote:
Given two sorted postive integer arrays A(n) and B(n) (W.L.O.G, let's
say they are decreasingly sorted), we define a set S = {(a,b) | a \in
A
and b \in B}. Obviously there are n^2 elements in S. The value of such
a pair
Basically the index of ( a + b) in the final array will be ceil(( index of a
+ index of b )/2).
Also if there is a clash of indices you just have to compare the values at
those indices and adjust them.
I have run my concept with below two arrays and it works !!!
Arary A: 1 2 3 4 5 6
Array B: 2 3
oops
Sorry didn't read properly
last algo was for array sorted in ascending order
for this case, just reverse the process
A[n] and B[n] are two array
loop=n, i=0, j=0;
while(loop0) // for n largest pairs
{
print A[i]+B[j]; // sum of first index from both array will be
max
foo =
ignore the previous mail it wrongly send.
On Fri, Apr 30, 2010 at 11:12 PM, Rajesh Patidar patidarc...@gmail.com wrote:
let consider the list in two different part one traversing list B with
respect to A and A with B
(a.len,b.len) is always solution
a1=a2=a.len
On Fri, Apr 30, 2010 at 5:35
Hi Divya,
A[n] and B[n] are two array
loop=n, i=n-1, j=n-1;
while(loop0) // for n largest pairs
{
print A[i]+B[j]; // sum of last index from both array will be max
foo = MAX ( A[i-1]+B[j], A[i-1]+B[j-1], A[i]+B[j-1] ) // using DP
moving backward
if foo=A[i-1]+B[j]; i--
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