@Vicky
Piyush's algo is based on a little trick to determine divisibility by 3.
Number of bits set in odd position - Number of bits set in even position 'll
be divisible by 3
For example, take 9. 1001. No. of bits set in odd position - number of bits
set in even position is 0. Hence divisible by
yup...thank u:)
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Thanks :)
On Wed, Jul 20, 2011 at 10:55 AM, SAMM somnath.nit...@gmail.com wrote:
This comes from the below
n-n/2-n/2^2-..-n/2^k
until n/2^k=1
Think from bits prospective
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To
I doubth .
For (d r0 + r1) ignore the point with smaller radius as it will
overshadowed the bigger circle completely
There may be a case where the circle is partially overlapped by the
other circles. Then this algo will fail .
The area will be of like these :-
Suppose 3 circles are there X,YZ
gn array - say a
hav extra array - say b - initialise all values to zero
ques 1:for(i=1;i=n;i++)
{
b[a[i]]++;
}
On Jul 20, 12:07 pm, siva viknesh sivavikne...@gmail.com wrote:
1.Given an array of size n. It contains numbers in the range 1 to n.
Each number is present at least once except
@SkRiPt KiDdIe
kindly explain the working of ur algo with the given example
On Wed, Jul 20, 2011 at 1:09 AM, SkRiPt KiDdIe anuragmsi...@gmail.comwrote:
#includeiostream
#includestring
using namespace std;
void permute(string str,int x,string print)
{
int mask=0;
check the above solution.. urs is O(logn) mine is O(log k).
On Jul 19, 11:35 pm, sagar pareek sagarpar...@gmail.com wrote:
well i dont think so...any better solution will be there compare mine one
On Wed, Jul 20, 2011 at 12:00 AM, Dumanshu duman...@gmail.com wrote:
heres the code
let A:: ((n(n+1)/2) - sum)
let B:: ((n(n+1)(2n+1)/6) - (sum of squares of elements))
then missing number = ((B/A) + A)/2;
complexity O(n).
space complexity O(1).
On Wed, Jul 20, 2011 at 12:47 PM, saurabh singh saurab...@gmail.com wrote:
Q1 can be solved using some simple maths:)
ya nitish above condition will do
On 7/20/11, Nitish Garg nitishgarg1...@gmail.com wrote:
I think:
s[i] = max(s[i-2], s[i-2]+a[i], s[i-1], a[i]) should satisfy all the cases,
even when all the numbers are negative.
Pleas check.
On Wed, Jul 20, 2011 at 12:44 AM, pnandy
int a = 20 ; //whatever value you like
int b = 30 ; // again
int sum = a - (-b) ;
cout sum ;
//eh ?
On Wed, Jul 13, 2011 at 6:17 PM, vaibhav shukla vaibhav200...@gmail.comwrote:
:D :D ;)
On Wed, Jul 13, 2011 at 6:11 PM, Anika Jain anika.jai...@gmail.comwrote:
@ vaibhav: ohh sorry,
Q2 o(1) space o(n) sol.
traverse through the array.
do -1*a[abs(a[i])-1] if a[abs(a[i])-1) +ve else do nothing
traverse again to check for the indexes with +ve values.
On Wed, Jul 20, 2011 at 1:01 PM, Shubham Maheshwari
shubham.veloc...@gmail.com wrote:
let A:: ((n(n+1)/2) - sum)
let B::
Given an infinite length list. u got to find index of an element k.
use this approach-
initially, take length as 2^x where x increases from 1 to ...
while (still not found)
{
now if arr[2^x-1] k,
increment x
else
binarysearch on length 2^(x-1) to 2^(x)
}
Please help me to find the
@SkRiPt KiDdIe
I got your logic.
Nice.
Thanks.
Regards
Anantha Krishnan
On Wed, Jul 20, 2011 at 2:04 PM, SkRiPt KiDdIe anuragmsi...@gmail.comwrote:
While you are on a state do not change ur state on encountering a specific
character if u have already done so earlier.This check in addition to
pls explain...i cant get the idea...:(
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Hi
Given an array how to find a sequence whose sum is maximum
*but condition is that no two adjacent elements* of the given Array.
ex:- Array:-[3,6,7,10,4]
output:- {6,10}=16
Array::[12,3,5,30]
output:-{12,30}= 42
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simple one!! . all the missing nos which r not present b/w 1 to n will
be printed!! TC O(n)
#include stdio.h
#define size 1
int main()
{
int i, j, n;
scanf(%d, n);
int a[n + 1] ;
int b[n + 1] ;
a[0] = 0;
b[0] = 0;
for (i = 1; i =
I would like to redefine my algo with cases clarified...
Create a queue that is made to contain the points...
say points queue [1000];
for i:1 to n
for j:i+1 to n
Calculate d (distance between the two centers)
if (d = r0 + r1) keep them in two separate queues //the circles
don't
space o(n) toomine takes o(1) space
On Wed, Jul 20, 2011 at 3:50 PM, JIƬΣN BAJIYA j.s.baj...@gmail.com wrote:
simple one!! . all the missing nos which r not present b/w 1 to n will
be printed!! TC O(n)
#include stdio.h
#define size 1
int main()
{
int i, j, n;
http://groups.google.com/group/algogeeks/browse_thread/thread/2ada4a0cd27036c2
This has already been discussed follow the above link .
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needs explicit function specialisation. be careful with constant strings.
T Add(T a, T b)
{return a+b ;}
template
char* Add char* a, char* b)
{return strcat((char*)a,b); }
surender
On Tue, Jul 19, 2011 at 10:17 PM, Anika Jain anika.jai...@gmail.com wrote:
here T becomes char *.. u r trying
@Dumanshu..i am not partitioning them into just two queues...
Moreover I just gave a raw idea...and yeah the complexity is in the order of
n^2 only.
There are many chances of improvement in it..
On Wed, Jul 20, 2011 at 5:30 PM, Dumanshu duman...@gmail.com wrote:
@Piyush:
Initially for
http://www.codechef.com/FEB10/problems/M5/
On Wed, Jul 20, 2011 at 5:35 PM, Piyush Sinha ecstasy.piy...@gmail.comwrote:
@Dumanshu..i am not partitioning them into just two queues...
Moreover I just gave a raw idea...and yeah the complexity is in the order
of n^2 only.
There are many
Describe
an
algorithm
that
takes an
unsorted
array
of
axis‐
aligned
rectangles
and returns
any
pair
of
rectangles
that
overlaps,
if
there
is such
a
pair.
Axis‐aligned means
that
all
the
rectangle
sides
are
either
parallel
or
perpendicular
to
the
x
and y‐axis.
You
if a[i] is negative then what u can do is first find the min of the
array(say Min) and then
do map[a[i]-Min]++
On Wed, Jul 20, 2011 at 6:17 PM, anonymous procrastination
opamp1...@gmail.com wrote:
Hello,
Usually whenever we use index as key to count frequency or other
similar algos.
The
Given an array with + and - numbers, including zero, Write an algorithm to
find all the possible sub arrays which sum up to zero.
For example, if given array is
20 , -9 , 3 , 1, 5 , 0, -6 , 9
Then possible sub arrays are:
-9, 3, 1, 5
0
1, 5, 0, -6
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You received this message because you are
but its showing output
On Wed, Jul 20, 2011 at 6:53 PM, mohit verma mohit89m...@gmail.com wrote:
hey guys...
1. char c='a';
while(c=='a')
{
printf(%c,c);
c=getchar();
}
..
2.
char c='a';
while(c!='b')
{
printf(%c,c);
c=getchar();
}
i want to ask one thing...the way some are saying first check with 2
then 4 and then 16to reach at that place we are suppose to
traverse it and also hav eto put a condition say like countn or
something...in this case also we are comparing so whats the
usecorrect me if im wrong.
On
its showing same output...
On Jul 20, 6:40 pm, chetan kapoor chetankapoor...@gmail.com wrote:
but its showing output
On Wed, Jul 20, 2011 at 6:53 PM, mohit verma mohit89m...@gmail.com wrote:
hey guys...
1. char c='a';
while(c=='a')
{
printf(%c,c);
c=getchar();
}
guys just give input a stream of characters as: a a a a a a a a a
a a a ---(press enter)
for second case input as: c d e f g h i j b -(press enter)
and see the difference. (I know that printf() is line buffered. So
why?)
1.printf() shows 'a' one time only.
2.here it shows
I think sunny meant that in IA-64 the pointer is of size 8 but here we are
assigning it to int *p. So, the segfault can occur.
On Wed, Jul 20, 2011 at 6:52 PM, Sandeep Jain sandeep6...@gmail.com wrote:
@Sunny: I couldn't understand how size of int int* matter here??
Regards,
Sandeep Jain
One a not so standard way of doing that is while creating the link list,
keep an extra pointer in the node to point to the jump location.
Just hold the previous node in a temp say address of 2 node and when i/p
reaches 4, point the jumper pointer to 4 and make 4 the next jumper pointer.
On
create another array with sum of the elements from 0 to i.
20 , -9 , 3 , 1, 5 , 0, -6 , 9
20 , 11,14,15,20,0,14,23
above thing can be done in O(n) .
now start hashing them (sum as keys and index as value), and if the sum
exist previously , then that means from that point , from where you are
In first case first character input is 'a' and second is space so loop
breaks
in second loop runs till b is not read hence all characters including spaces
are found in output
On Wed, Jul 20, 2011 at 7:29 PM, mohit mohit89m...@gmail.com wrote:
guys just give input a stream of characters
I think the novelty of this is that we are avoiding the comparison of new
element with all the members of data in LL
On Wed, Jul 20, 2011 at 7:27 PM, Ankur Khurana ankur.kkhur...@gmail.comwrote:
@Popli : Bingo , that is what i was thinking and mentioned in my previous
post..
On Wed, Jul
Question - you are given a1b3c7d8.. an array like this. This is a
compressed array. (Similar to
Run Length) You are supposed to expand the given input IN-PLACE. That is,
your array is large
enough to hold the output.
Example -
Input - a1b4 - Array Length - 5
Output - a
--
With Regards,
You can try this out :---
If the range of the number are in between -N to +N eg: -10^6 to
+10^6 , then take the the Absolute( Lower bound.) i:e N (10^6) .
For each number add it in the Index [i+N] ..
While printing the value of Number at Index X, Print the value in the
Index [X-N]..
You can make the following structure :
#define MAX 100
typedef struct
{
int count_positive;
int count_negative;
}Element;
typedef Element Map[MAX];
Now you can just create a map as :
Map map;
Now for every element read, first check whether it is +ve or -ve. Use
the absolute value of the
This works.
code
#include iostream
#include map
using namespace std;
int main()
{
mapint,int a;
a[-1] = 0;
couta[-1]endl;
}
/code
On Wed, Jul 20, 2011 at 7:50 PM, ankit sambyal ankitsamb...@gmail.comwrote:
You can make the following structure :
#define MAX 100
typedef struct
{
int
I doubt if it would work on the following example : size of array is
10, 5 and 7 are missing numbers
{1,2,3,4,6,6,6,8,9,10};
On Jul 20, 12:31 pm, Shubham Maheshwari shubham.veloc...@gmail.com
wrote:
let A:: ((n(n+1)/2) - sum)
let B:: ((n(n+1)(2n+1)/6) - (sum of squares of elements))
then
I doubt if it would work on the following example : size of array is
10, 5 and 7 are missing numbers
{1,2,3,4,6,6,6,8,9,10};
On Jul 20, 1:04 pm, Shubham Maheshwari shubham.veloc...@gmail.com
wrote:
@saurabh.
kindly use a lil bit of indentation ... ur algo is illegible.
On Wed, Jul
Hey
there are three answers for the given example.
But you solution will give only 2
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ohh ...got it.
On Jul 20, 7:06 pm, sunny agrawal sunny816.i...@gmail.com wrote:
In first case first character input is 'a' and second is space so loop
breaks
in second loop runs till b is not read hence all characters including spaces
are found in output
On Wed, Jul 20, 2011 at
Sorry.
This will work
Best Regards,
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You can do it using stack concept:--
Pop the element from the end , taking two variable index1, index2 and
Ch(character to Iterate)
Eg:- a1b4 here index1=4 , Ch='b', index2=1;
Start filling the element of Ch from the extreme end of the array ..
From right hand side .
The array will look
Yaa even if it is 8 bytes long . Compiler will treat the value 10 as 8
bytes only . It should be able to assign it to the pointer of the same
size (type) ..
Try to free the dynamically allocated memory just before return 0
and tell me the result after compilation . Try this :-
int main()
{
http://groups.google.com/group/programming-puzzles/browse_thread/thread/4fecd0d904624a0d
this will clarify all doubts :)
On Wed, Jul 20, 2011 at 8:52 PM, SAMMM somnath.nit...@gmail.com wrote:
Yaa even if it is 8 bytes long . Compiler will treat the value 10 as 8
bytes only . It should be able
I would like if any here have attended informatica interview
and also about the interview procedure.
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One question wht is the data structure used for the
Dictionary ... The algo is dependent on the Implementation of the
dictionary .
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oh common mine is log(k) too... i m taking only k elements
On Wed, Jul 20, 2011 at 1:01 PM, Dumanshu duman...@gmail.com wrote:
check the above solution.. urs is O(logn) mine is O(log k).
On Jul 19, 11:35 pm, sagar pareek sagarpar...@gmail.com wrote:
well i dont think so...any better
Nice solution dude . Like that one
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try from back end
surender
On Wed, Jul 20, 2011 at 9:54 PM, Soumya Prasad Ukil
ukil.sou...@gmail.comwrote:
Two passes over the original array is required.
On 20 July 2011 08:10, SAMMM somnath.nit...@gmail.com wrote:
You can do it using stack concept:--
Pop the element from the end ,
when n is not defined , you want complexity in terms of ?
On Wed, Jul 20, 2011 at 3:16 PM, Dumanshu duman...@gmail.com wrote:
Given an infinite length list. u got to find index of an element k.
use this approach-
initially, take length as 2^x where x increases from 1 to ...
while (still
There is a clock at the bottom of the hill and a clock at the top of
the hill. The clock at the bottom of the hill works fine but the clock
at the top doesn't. How will you synchronize the two clocks.
Obviously, you can,t carry either of the clocks up or down the hill!
And you have a horse to help
that is why m confused. how would u rate this algo? in what order?
On Jul 20, 10:44 pm, Ankur Khurana ankur.kkhur...@gmail.com wrote:
when n is not defined , you want complexity in terms of ?
On Wed, Jul 20, 2011 at 3:16 PM, Dumanshu duman...@gmail.com wrote:
Given an infinite length
first count the no of elements to be printed say it is 'n'.
now start from n-1 and start expanding the array from right to left..
On Wed, Jul 20, 2011 at 10:56 PM, surender sanke surend...@gmail.comwrote:
try from back end
surender
On Wed, Jul 20, 2011 at 9:54 PM, Soumya Prasad Ukil
@Kamakshi...
answer must be sub sequence of the array
anyone pls explain ANKUR's approach
On Thu, Jul 21, 2011 at 12:12 AM, Kamakshii Aggarwal
kamakshi...@gmail.comwrote:
for the above example,there are more possible sub arrays like
-9,9
-9,0,9
On Thu, Jul 21, 2011 at 12:06 AM, sagar
*Contiguous elements*
On Thu, Jul 21, 2011 at 12:12 AM, Kamakshii Aggarwal
kamakshi...@gmail.comwrote:
for the above example,there are more possible sub arrays like
-9,9
-9,0,9
On Thu, Jul 21, 2011 at 12:06 AM, sagar pareek sagarpar...@gmail.comwrote:
@ankur
pls explain through an
how to deal with it??
surender
On Wed, Jul 20, 2011 at 9:02 PM, sunny agrawal sunny816.i...@gmail.comwrote:
http://groups.google.com/group/programming-puzzles/browse_thread/thread/4fecd0d904624a0d
this will clarify all doubts :)
On Wed, Jul 20, 2011 at 8:52 PM, SAMMM
@Pankaj
did u get ankur's approach?
On Thu, Jul 21, 2011 at 12:16 AM, Pankaj jatka.oppimi...@gmail.com wrote:
*Contiguous elements*
On Thu, Jul 21, 2011 at 12:12 AM, Kamakshii Aggarwal
kamakshi...@gmail.com wrote:
for the above example,there are more possible sub arrays like
-9,9
-9,0,9
nice solution by ankit. but can anyone tell if we were to find sequence that
may or may not be contiguous.
On Thu, Jul 21, 2011 at 12:22 AM, Kamakshii Aggarwal
kamakshi...@gmail.comwrote:
a little mistake in ankur's solution
20 , -9 , 3 , 1, 5 , 0, -6 , 9
20 , 11,14,15,20,20(instead of
@sagar: i read contiguous afterords therefore i deleted my post...
and for ankur's approach
:
original array:20 , -9 , 3 , 1, 5 , 0, -6 , 9
array after addition:20 , 11,14,15,20,20,14,23
now when u hash second array elements
then u fill find collision at value 14 and 20
this means terms b/w 1st
pls anyone explain it
Please please please
On Thu, Jul 21, 2011 at 12:37 AM, varun pahwa varunpahwa2...@gmail.comwrote:
nice solution by ankit. but can anyone tell if we were to find sequence
that may or may not be contiguous.
On Thu, Jul 21, 2011 at 12:22 AM, Kamakshii Aggarwal
@sagar: i read contiguous afterwards therefore i deleted my post...
and for ankur's approach
:
original array:20 , -9 , 3 , 1, 5 , 0, -6 , 9
array after addition:20 , 11,14,15,20,20,14,23
now when u hash second array elements
then u fill find collision at value 14 and 20
this means terms b/w 1st
how collision occur at 14 and 20
actually i m not getting hash function
Thanks in advance
On Thu, Jul 21, 2011 at 12:39 AM, Kamakshii Aggarwal
kamakshi...@gmail.comwrote:
@sagar: i read contiguous afterwards therefore i deleted my post...
and for ankur's approach
:
original array:20 , -9 , 3
Use trie for dictionary.Use permutaion to generate all anagrams and check
finally.
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use line sweep method
On 7/20/11, Dumanshu duman...@gmail.com wrote:
Describe
an
algorithm
that
takes an
unsorted
array
of
axis‐
aligned
rectangles
and returns
any
pair
of
rectangles
that
overlaps,
if
there
is such
a
pair.
Axis‐aligned means
that
all
the
rectangle
sort each string according to their alphabetical order then hash it as key,
for hashing use preferably linked list as value for key
surender
On Thu, Jul 21, 2011 at 12:58 AM, SkRiPt KiDdIe anuragmsi...@gmail.comwrote:
Use trie for dictionary.Use permutaion to generate all anagrams and check
anika can u pls explain ur solution.I am not getting it.
On Wed, Jul 20, 2011 at 1:27 PM, Anurag atri anu.anurag@gmail.comwrote:
int a = 20 ; //whatever value you like
int b = 30 ; // again
int sum = a - (-b) ;
cout sum ;
//eh ?
On Wed, Jul 13, 2011 at 6:17 PM, vaibhav shukla
I think it is related to Joshepus problm... ckeck wikipedia for more info.
On 7/20/11, Vivek Srivastava srivastava.vivek1...@gmail.com wrote:
If n people are standing in a circle ,they start shooting the person
standing next to their neighbour.If they start firing in this way ,
determine
@anika: Now I am able to understand.thanks anyways.:)
On Thu, Jul 21, 2011 at 1:18 AM, Kamakshii Aggarwal
kamakshi...@gmail.comwrote:
anika can u pls explain ur solution.I am not getting it.
On Wed, Jul 20, 2011 at 1:27 PM, Anurag atri anu.anurag@gmail.comwrote:
int a = 20 ; //whatever
O(N^2) naiive algorithm is obvious - check every pair of rectangles.
This problem can be done in O(N log N) using a data structure called an R
tree or a QuadTree.
If you want to use and R Tree - read: http://en.wikipedia.org/wiki/R-tree
The QuadTree approach is very simple.
Maintain something
use sum calculated in 2nd array as index for hashing..and index for second
array as value of hashing;
for example
original array:20 , -9 , 3 , 1, 5 , 0, -6 , 9
array after addition:20 , 11,14,15,20,20,14,23
according to above example hashing will be like this
k[20]=0;
k{11]=1;
k[14]=2;
k[15]=3;
it should involve the exponential increment! somebody plz help
On Jul 20, 10:56 pm, Ankur Khurana ankur.kkhur...@gmail.com wrote:
in my opinion , it is log(indexof(k))
On Wed, Jul 20, 2011 at 11:23 PM, Dumanshu duman...@gmail.com wrote:
that is why m confused. how would u rate this algo?
is it given that string alphabets wud be in alphabetical order...??
On Thu, Jul 21, 2011 at 12:04 AM, Kamakshii Aggarwal
kamakshi...@gmail.com wrote:
first count the no of elements to be printed say it is 'n'.
now start from n-1 and start expanding the array from right to left..
On Wed, Jul
please some1 tell how to do part 2.
On Wed, Jul 20, 2011 at 6:20 PM, kavitha nk kavithan...@gmail.com wrote:
for 3rd v can replace decrement by an increment operator!
//BE COOL// kavi
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Use a modification of the Cristian Algorithm.
Start from the faulty clock.
Go to the accurate clock, get a time sample T, come back.
Set the new time as T + RTT* K/(1 + K), just choose a ratio K which
represents the ratio of the time taken between going downhill and uphill
It's a famous
How to reverse the order of bits of a number in minimum space
complexity?
if the no is 11-1011
the output should b 1101.
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To
Given a linked list with a n*n matrix elements write an algo to
compute product of two such linked list.
say n = 3then
1 2 3
4 5 6
7 8 9
will be given as 1-2-3-4-5-6-7-8-9
now given two matrices , given a algo to perform matrix
multiplication giving result in thrid linked list.
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You
calculate hash value of word then compare with hash value of source string
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write a c code that takes a 32 bit ip address n prints it in dotted notation
as a.b.c.d.
The code shall work for big endian as well as for little endian..
In this how can i make it common for big endian and little endian?? if in
little endian my lower order 8 bits shall be d but for big endian
@Anika..can u give one example??
On 7/21/11, Anika Jain anika.jai...@gmail.com wrote:
write a c code that takes a 32 bit ip address n prints it in dotted notation
as a.b.c.d.
The code shall work for big endian as well as for little endian..
In this how can i make it common for big endian
unsigned int reverse_bits (unsigned int n)
{
if(n==1) return n;
int j,p;
for(j = 31;j0;j--)
if(n(1j)) break; //to find the first set bit from left
p =0;
while(jp)
{
int x1 = (n (1j))j;
int x2 = (n(1p))p;
if(x1!=x2)
{
take an assumption,if an one ppl s there he never die
eg :
no.of ppl person who will not die
11
2 1
3 2
4 4
5 1
6
On 7/21/11, SAMM somnath.nit...@gmail.com wrote:
I think it is related to
Hi
@Somnath my question is some different
if given array :3,2,7,10
So according to last discussion Only Out put is *13 *not the show elements
Output :: {3,10}
So basically my question is that how to pick all elements that return max
sum
Thanks..
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unsigned int reverse_bits (unsigned int n)
{
unsigned int m;
m=0;
while(n) {
m*=2;
if( n1 ) {
m+=1;
}
n=1;
}
return m;
}
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@Nicks : Kindly share the source from where you are practicing these c
output questions ..
Thank You :)
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do both s[i-2] and s[i-1] need to be checked? isn't by construction
s[i-1]=s[i-2] ? what about s[i] = max( s[i-2]+a[i], s[i-1], a[i]) ?
On Wed, Jul 20, 2011 at 3:48 AM, Piyush Sinha ecstasy.piy...@gmail.comwrote:
ya nitish above condition will do
On 7/20/11, Nitish Garg
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