for a naive algorithm. But a little thought can give a
divide-and-conquer algorithm which is O(n).
Dave
On Jun 24, 8:44 am, sunny agrawal sunny816.i...@gmail.com wrote:
hmm i also doubt that
but it is Strictly O(32n) not O(nlgn) where lgn = 32 depending upon
value
of n
On Fri, Jun
@Adarsh
fails on this
a[] = {2,2,2,6,6,6}
On Sat, Jun 25, 2011 at 10:54 AM, Adarsh s.adars...@gmail.com wrote:
int n = A.size();
for (int i=0; in; i++)
total += A[i];
findMinMax(A[1...n]);
int mean = (max+min)/2;
if ((total - n*mean) == 0)
printf(consecutive\n);
else
@adarsh
no it will Fails for both even and odd no of elemets
a[] = {2,2,2,4,6,6,6}
seems like we need to check for the presence of each no between min to max
using a good hashing approach.
On Sat, Jun 25, 2011 at 11:20 AM, Adarsh s.adars...@gmail.com wrote:
Missed that... also, my method seems
initially compute xor of all the values from 0 to n in a variable Temp
so temp = 0^1^2^n
let result is used to store the missing number
for each ith bit of missing number where i = 0-31 we can find it as
following
ith bit of result = (xor of all ith bits of values of array) xored with
LCS is abcdcba or abcecba
not abc or cba
On Wed, Jun 22, 2011 at 10:15 PM, ankit mehta mehta.bond...@gmail.comwrote:
Consider string: abcdecba
Reverse of above string: abcedcba
Longest common substring: abc and cba : Both not Palindromes!
On Jun 22, 9:29 pm, sanjay ahuja
LCS stands for Longest Common Substring
--
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B-Tech IV year,CSI
Indian Institute Of Technology,Roorkee
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ohh sorry my mistake
LCS stands for Longest Common Subsequence
On Wed, Jun 22, 2011 at 11:21 PM, SVIX saivivekh.swaminat...@gmail.comwrote:
ah... i misunderstood the question... sorry..
On Jun 22, 12:01 pm, ankit mehta mehta.bond...@gmail.com wrote:
You dont have to create longest
:11 AM, oppilas .
jatka.oppimi...@gmail.comwrote:
Sunny,
Can but can we modify this code to get the *node X and node Y*?.
On Wed, Jun 22, 2011 at 8:32 AM, Anantha Krishnan
ananthakrishnan@gmail.com wrote:
@sunny agrawal
*Thanks a lot.*
*Great code. I got the logic
last line is
*in worst case k=1 only 2*n comparisons will be there hence O(n)*
On Thu, Jun 23, 2011 at 11:26 AM, sunny agrawal sunny816.i...@gmail.comwrote:
Lets Consider the case of Naive matching in which at some shift s first k
characters are matched and next character does not match so
but it is O(N^2)...will give TLE
On Tue, Jun 21, 2011 at 2:33 PM, ross jagadish1...@gmail.com wrote:
A small correction, j = 0 to i.
On Jun 21, 2:01 pm, ross jagadish1...@gmail.com wrote:
Hi,
Ya DP with memoization is best..
nCr= n-1Cr-1 + n-1 C r
DP]i,j] = DP[i-1,j-1] +
DP is good when we need to use intermediate values too..
here we only need to compute nCr as a final result
so we cannot use DP here
the Logic Vandana used will be helpful here.
as nCr = n! / r! * n-r!
without loss of generality we can assume that r n-r else we can just
replace r = n-r
then
will increase vry fast and will go out of limit..
-
PRAMENDRA RATHI
NIT ALLAHABAD
On Tue, Jun 21, 2011 at 2:52 PM, sunny agrawal sunny816.i...@gmail.comwrote:
DP is good when we need to use intermediate values too..
here we only need
this problem is same as
*http://www.codechef.com/problems/MARBLES/*
solution to this are public
u can check for more clarification
On Tue, Jun 21, 2011 at 3:51 PM, sunny agrawal sunny816.i...@gmail.comwrote:
if it is given tha final answer fits in 64 bit signed integer then
we can run a loop
i am doing the same thing without using array
long long int res = 1;
int j = 2;
for(int i = n-k+1; i = n; i++){
res *= (long long int)i;
while(j = k res%j == 0){
res/=j;
j++;
}
}
On Tue, Jun 21, 2011 at 4:25 PM, kartik sachan
because j doesn't divide it, it can
overflow.
On Jun 21, 4:12 pm, sunny agrawal sunny816.i...@gmail.com wrote:
i am doing the same thing without using array
long long int res = 1;
int j = 2;
for(int i = n-k+1; i = n; i++){
res *= (long long int)i;
while(j = k
@Piyush
good to start with
But i think a recursive O(n) is possible
in downward calls pass sum from root to node
and on return calls pass sum from leafs to root of each subtree and using
this collective information updating a global ans max.
On Tue, Jun 21, 2011 at 5:05 PM, Piyush Sinha
one of the possible
create the min heap O(n)
keep extracting elements and look for missing elements
worst case it will be O(nlgn)
equivalent to heap sort
or
Simply sort using quick sort or merge sort and then traverse through the
array and look for missing ones
On Tue, Jun 21, 2011 at 7:09 PM,
i dont know why r u thinking so but this was my accepted solution for the
problem.
On Tue, Jun 21, 2011 at 7:34 PM, kartik sachan kartik.sac...@gmail.comwrote:
hey sunny suppose u have to calculate 100c50 then there is a lot of
chances of over flow becoz u have to multiply 100 to
both will work fine
in first case in each step what is done is ans = (ans*(n+1-i))/i;
what happens is
first iteration ans = n, i =1 so i divides ans
sec. iteration ans = product of 2 consecutive nos at least one even so
divisible by 2
in third, ans is prod of 3 consecutive no.s so divisible by 3
see this
*
https://ideone.com/1ZtIq*
On Tue, Jun 21, 2011 at 10:23 PM, Anantha Krishnan
ananthakrishnan@gmail.com wrote:
Thanks.
I expect more details in implementation point of view.
Thanks Regards,
Anantha Krishnan
On Tue, Jun 21, 2011 at 6:41 PM, sunny agrawal sunny816.i
after first Roll no of Expected No of Dices will be 2.5
1/2(0)+1/6(4+5+6)
so after second 2.5^2
so after N rounds 2.5^N
On Mon, Jun 20, 2011 at 3:56 PM, Navneet Gupta navneetn...@gmail.comwrote:
We have an unlimited number of dice at our disposal. Let's roll the
die. If the outcome is 1, 2, or
and Expected value should be
2.5^(N-1)*(7/2)
--
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B-Tech IV year,CSI
Indian Institute Of Technology,Roorkee
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i think u r missing the cases when o is winning and and countx counto then
answer should be no
On Fri, Jun 17, 2011 at 12:19 PM, KK kunalkapadi...@gmail.com wrote:
https://www.spoj.pl/problems/TOE1/
For which test case does this program fail
#includeiostream
#includevector
using
no i didn't mean that
in first test u checking if count of X should be either equal of one more
than that of O
and in last u r checking if both are winning or only only one
but what i meant is if O has already won but no of moves of X are greater
than O the answer should be No but your solution
as you can see in this case no of moves of X are 4 and that of O are 3
as X starts first, after both players has played 3 moves each, O would have
already won the game so next move of X is invalid
i got your solution AC after adding this condition :)
On Fri, Jun 17, 2011 at 2:48 PM, KK
you need to try something better as limits of A and B are very large :)
you can not run a loop from A to B
i have not tried it but the logic is there will be many nos which will give
the same value and we dont need to calculate for them all explicitply :)
On Fri, Jun 17, 2011 at 2:52 PM, KK
where n is ??
On Fri, Jun 17, 2011 at 3:23 PM, Arpit Sood soodfi...@gmail.com wrote:
i have got AC with O(n)
On Fri, Jun 17, 2011 at 2:59 PM, sunny agrawal sunny816.i...@gmail.comwrote:
you need to try something better as limits of A and B are very large :)
you can not run a loop from
but limits of A and B are very large
10^15
how is this possible
am i missing something,
like Max(B-A) = 10^6 or 10^7
On Fri, Jun 17, 2011 at 3:30 PM, Arpit Sood soodfi...@gmail.com wrote:
lol, i mean in linear time
On Fri, Jun 17, 2011 at 3:27 PM, sunny agrawal sunny816.i
the final result in the order of no of bits
O(64)
On Fri, Jun 17, 2011 at 3:38 PM, sunny agrawal sunny816.i...@gmail.com
wrote:
but limits of A and B are very large
10^15
how is this possible
am i missing something,
like Max(B-A) = 10^6 or 10^7
On Fri, Jun 17, 2011 at 3:30 PM, Arpit Sood
Try this:
say i is the index of the first occurrence of the first character
say j is the index of the first occurrence of the second character
say n is length of array
int Min = n+1;
while(i n j n){
int Min = min(Min, abs(i-j))
if(i j){
find next occurrence of first character
}
else{
find
n^2
On Thu, Jun 16, 2011 at 12:41 PM, Lavesh Rawat lavesh.ra...@gmail.comwrote:
*Number Trick Riddle - 16 june * *
* * ***
**
*Sonal asked the class to see if they could find the sum of the first 50
odd numbers. As everyone settled down to their addition, Lavesh ran to her
and said, 'The
The place where strict constness is not enforced is with character
array literals. You can say
char* cp = howdy;
and the compiler will accept it without complaint. This is
technically an error because a character array literal (“howdy” in
this case) is created by the compiler as a constant
stored at p , and also increments p .
On Thu, Jun 16, 2011 at 1:10 PM, sunny agrawal
sunny816.i...@gmail.comwrote:
The place where strict constness is not enforced is with character
array literals. You can say
char* cp = howdy;
and the compiler will accept it without complaint
in such cases for
still better understanding :)
Be it a book or a website.
On Thu, Jun 16, 2011 at 8:13 PM, sunny agrawal
sunny816.i...@gmail.comwrote:
yes copy pasting the exact thing :)
for better understanding :)
On Thu, Jun 16, 2011 at 8:06 PM, Navneet Gupta navneetn...@gmail.comwrote
@kartik sachan
This function is *not* defined in ANSI-C and is *not* part of C++, but is
supported by some compilers.
and +1 to Shachindra's post...i also think memory allocation will be
from heap...not stack
On Wed, Jun 15, 2011 at 2:34 PM, Shachindra A C sachindr...@gmail.comwrote:
i think u r wrong
what if heap size -1 is 0
i think one should pick atleast one coin else game will draw
On Wed, Jun 15, 2011 at 5:17 PM, immanuel kingston
kingston.imman...@gmail.com wrote:
First Player can always win.
For each heap
Pick heap-size - 1 coins if this is not the n-1th
consider the case.
n = 2;
heap 1 - no of coins 1
heap 2 - no of coins 2
On Wed, Jun 15, 2011 at 5:34 PM, sunny agrawal sunny816.i...@gmail.comwrote:
i think u r wrong
what if heap size -1 is 0
i think one should pick atleast one coin else game will draw
On Wed, Jun 15, 2011 at 5:17 PM
the the heaps are of size 1 the Player 1 can win always.
Thanks,
Immanuel
On Wed, Jun 15, 2011 at 5:36 PM, sunny agrawal sunny816.i...@gmail.comwrote:
consider the case.
n = 2;
heap 1 - no of coins 1
heap 2 - no of coins 2
On Wed, Jun 15, 2011 at 5:34 PM, sunny agrawal
sunny816.i
the other coin from heap1.
Player 1 will take both the coins in heap 2.
Thanks,
Immanuel
On Wed, Jun 15, 2011 at 6:33 PM, sunny agrawal sunny816.i...@gmail.comwrote:
check out this case
n = 2
both heaps having 2 coins
player 2 will win i think
On Wed, Jun 15, 2011 at 6:26 PM, immanuel
@Nitish
n=2
heap 1 = 2
heap 2 = 3
Xor = 1
still player one can win :)
On Wed, Jun 15, 2011 at 6:49 PM, sunny agrawal sunny816.i...@gmail.comwrote:
@immanuel
ohh, i read the Question wrong. :(
i was thinking player1 is starting from least numbered heap and player 2
from highest no heap
i think solution depends on no of heaps having single coin
if there are even number of such heaps player 1 will win
if there are odd number of such heaps player 2 will win
On Wed, Jun 15, 2011 at 6:49 PM, Nitish Garg nitishgarg1...@gmail.comwrote:
Player 1 can still win in this case: Player 1
no that also wont work
n=2
3,1
On Wed, Jun 15, 2011 at 6:59 PM, sunny agrawal sunny816.i...@gmail.comwrote:
i think solution depends on no of heaps having single coin
if there are even number of such heaps player 1 will win
if there are odd number of such heaps player 2 will win
On Wed
@Raghavan
if you want to fing the shortest path from a node u to v then its better to
use BFS
but according to your second post it seems you want to find path from root
node to a leaf, in that case DFS seems to be best
On Wed, Jun 15, 2011 at 9:56 PM, bittu shashank7andr...@gmail.com wrote:
I
0.00 with priority_queue stl for me :)
On Thu, Jun 16, 2011 at 10:55 AM, Kunal Yadav kunalyada...@gmail.comwrote:
Whats ur running time. Mine is 0.05 without using priority queque.
On Wed, Jun 15, 2011 at 8:24 PM, kartik sachan kartik.sac...@gmail.comwrote:
ya dude finally i applied that
Did you mean Shortest path from one node to another
Use BFS i think we can find the shortest path
On Tue, Jun 14, 2011 at 5:37 PM, Raghavan its...@gmail.com wrote:
Hi,
How to find the shortest path in a 4-ary tree in an optimal way?
Node tree{
Node *left,*right,*top,*bottom;
int val;
same as left right or top
u can consider as each node as a cell of grid type of thing for easy
understanding
On Tue, Jun 14, 2011 at 5:59 PM, vaibhav agarwal vibhu.bitspil...@gmail.com
wrote:
What actually is meant by bottom pointer
On Tue, Jun 14, 2011 at 5:57 PM, sunny agrawal sunny816.i
it must be any general string
Suffix tree approach seems best both in time and space complexity
On Tue, Jun 14, 2011 at 7:36 PM, Navneet Gupta navneetn...@gmail.comwrote:
This looks like a good problem. Could you confirm if string contains only
two characters as you mentioned in both
, Arpit Sood soodfi...@gmail.com wrote:
i meant if N = { 1, 1, 1, 2, 12}
and M = { 1, 1, 3, 12}
then answer should be = {1, 1, 12}
On Mon, Jun 13, 2011 at 8:06 PM, sunny agrawal sunny816.i...@gmail.comwrote:
no we can take care of duplicates without any extra memory
modify 2nd step of my
Pull the Trigger Again ??
On Mon, Jun 13, 2011 at 1:01 PM, Lavesh Rawat lavesh.ra...@gmail.comwrote:
*Probability Riddle Loaded Revolver - 13* June
*
* * ***
**
*Henry has been caught stealing cattle, and is brought into town for
justice. The judge is his ex-wife Gretchen, who wants to
and
the space requirement of the algorithm proposed does not approach
infinity.
here, even if n-infinity, input size would still be 8kb.. hence O(1)
space.
hope tat helps.
On Jun 13, 12:38 pm, sunny agrawal sunny816.i...@gmail.com wrote:
hello all,
what if character are not ASCII
why do we need 2 bits at all ??
i think single bit per table entry will do.
say table is T[1025], and array is A[M]
1. Go through the N sized array and set bit 0 of the hash table entry to 1
if it is present in the first array.
2. Go through the M sized array and if T[a[i]] is set then this
no we can take care of duplicates without any extra memory
modify 2nd step of my previous solution as follows
if T[a[i]] is set then this element is there in second array so report this
element and Reset T[a[i]].
now no duplicates will be reported. and only 1025 bits will be required.
any
because in all expression is evaluated to true.
On Mon, Jun 13, 2011 at 10:02 PM, sahil sharma sahil18...@gmail.com wrote:
ok i got the values of i ,j ,k ,.in all the 3 codes but can u
please explain me how we get the value of m =1 in all the 3
program.?
--
You received
m= ++i || ++j ++k;
++i will evauate to 3 and which is true so as next is ||
so next part will not be evaluated and m = true
On Mon, Jun 13, 2011 at 11:34 PM, udit sharma sharmaudit...@gmail.comwrote:
Bt y k=0, in 2nd code???
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@tech rascal
10 = 01010
20 = 10100
--
xor = 0
= 30
why r u getting 1 ??
On Tue, Jun 14, 2011 at 2:28 AM, Supraja Jayakumar suprajasank...@gmail.com
wrote:
@Wladimir:
Can you kindly explain the overflow and underflow you mentioned.
Thanks
Supraja J
On Fri, Jun 10,
Piyush Sinha *ecstasy.piy...@gmail.com
*
--
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B-Tech IV year,CSI
Indian Institute Of Technology,Roorkee
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@terence
which game ??
On Fri, Jun 10, 2011 at 3:51 PM, Terence technic@gmail.com wrote:
42, 47
A popular game.:)
On 2011-6-10 18:11, ankur aggarwal wrote:
42, 49
2011/6/10 • » νιρυℓ « • vipulmehta.1...@gmail.com
42, 47
just guessing according to the pattern.
On Fri, Jun 10,
may be a map can help that maps string to their integer values ans then
using some set of rule to convert to a number
On Sun, Jun 12, 2011 at 12:28 AM, Abhishek Goswami
zeal.gosw...@gmail.comwrote:
@I got one of my interview . I tried to solve this issue but could not
it...I did googling but
find common Ancestor of both. see if y lies on path from x or z to this
ancestor
O(lgn)
On Fri, Jun 10, 2011 at 1:20 PM, aanchal goyal goyal.aanch...@gmail.comwrote:
There is a binary tree(Not a BST) in which you are given three nodes
x,y,z .Write a function which finds whether y lies in the
but that will be O(n) i think
On Fri, Jun 10, 2011 at 2:38 PM, Harshal hc4...@gmail.com wrote:
can be solved using dfs.
On Fri, Jun 10, 2011 at 2:27 PM, sunny agrawal sunny816.i...@gmail.comwrote:
find common Ancestor of both. see if y lies on path from x or z to this
ancestor
O(lgn
y lies on path from lca to z, but it doesn't lie on path from x to z.
Am I missing something?
On Fri, Jun 10, 2011 at 2:50 PM, sunny agrawal sunny816.i...@gmail.comwrote:
but that will be O(n) i think
On Fri, Jun 10, 2011 at 2:38 PM, Harshal hc4...@gmail.com wrote:
can be solved using
@ross if a parent pointer is there lca can be found in lgn
and path can be traversed in lgn too
why it cannot be lgn
what is the problem ??
On Fri, Jun 10, 2011 at 3:06 PM, sunny agrawal sunny816.i...@gmail.comwrote:
thats why i mentioned x OR z and i m considering parent pointer :)
without
First algorithm taht comes in mind
deque a element
if +ve enque again
if(-ve) do nothing
now question is terminating condition
On Fri, Jun 10, 2011 at 3:44 PM, Sangeeta sangeeta15...@gmail.com wrote:
write an algo that deletes all negative integers without changing the
order of remaining
initialy cal size of queue then apply a for loop
On Fri, Jun 10, 2011 at 4:00 PM, sunny agrawal sunny816.i...@gmail.comwrote:
First algorithm taht comes in mind
deque a element
if +ve enque again
if(-ve) do nothing
now question is terminating condition
On Fri, Jun 10, 2011 at 3:44 PM
using parent pointers untill we reach to lca.
On Fri, Jun 10, 2011 at 3:59 PM, aanchal goyal goyal.aanch...@gmail.comwrote:
@sunny finding lca in logn is fine, but how can we traverse the path in
logn.. ?
On Fri, Jun 10, 2011 at 3:50 PM, sunny agrawal sunny816.i...@gmail.comwrote:
@ross
in this question i assumed that we need to write a function which takes an
input a queue and function implements the required.
we don't know how the queue is implemented, means we don't know about
structure of the node.
only 2 standard functions of queue enqueue and dequeue are known
On Fri, Jun
yup, it works, i tried it!!
On Thu, Jun 9, 2011 at 12:15 PM, Abdul Rahman Shariff ears7...@gmail.comwrote:
did u try it ??
nd did u get the msg??
On Thu, Jun 9, 2011 at 1:06 AM, kartik sachan kartik.sac...@gmail.comwrote:
Frnz govt of india has put a condition dat lokpal bill will b
6
On Wed, Jun 8, 2011 at 10:31 PM, tech rascal techrascal...@gmail.comwrote:
20 ? 150 18 11
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series of some random numbers generated ussing some RNG
no logic :P :P :P
On Thu, Jun 9, 2011 at 12:29 PM, Arpit Sood soodfi...@gmail.com wrote:
hey,
what's the logic ?
On Thu, Jun 9, 2011 at 12:22 PM, sunny agrawal sunny816.i...@gmail.comwrote:
6
On Wed, Jun 8, 2011 at 10:31
sum can overflow
Xor method can also be applied to Q1. no need of numbers to be sorted.
2011/6/9 • » νιρυℓ « • vipulmehta.1...@gmail.com
For 1.
sum the numbers in the file, subtract it from sum of first 4 billion
numbers.
On Thu, Jun 9, 2011 at 3:44 PM, Navneet Gupta
yes, but using xor no need of ULL :)
2011/6/9 • » νιρυℓ « • vipulmehta.1...@gmail.com
Sum wont overflow, ULL range will include sum.
On Thu, Jun 9, 2011 at 3:52 PM, sunny agrawal sunny816.i...@gmail.comwrote:
sum can overflow
Xor method can also be applied to Q1. no need of numbers
Discussed many times,
1) add some lines to merge sort
2) use Binary indexed tree for a faster version (i have not tried but get to
know it can be done using binary indexed tree)
On Thu, Jun 9, 2011 at 4:05 PM, D.N.Vishwakarma@IITR deok...@gmail.comwrote:
Inversion means pair(i,j) such that ij
check your this if condition
if(tmap.end()-second=4)
{
ans=tmap.end()-first;
}
it should be == , not =
On Thu, Jun 9, 2011 at 5:09 PM, John Hayes agressiveha...@gmail.com wrote:
Hello every one i tried to code the same problem using STL mapsbut
getting the RUN TIME
hehe
that was also random. :D
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For more
the actual answer ?
On Thu, Jun 9, 2011 at 5:46 PM, sunny agrawal sunny816.i...@gmail.comwrote:
hehe
that was also random. :D
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seems like got it..
there is 150 days difference between 20/6(20 june) and 18/11(18 November)
On Thu, Jun 9, 2011 at 6:14 PM, sunny agrawal sunny816.i...@gmail.comwrote:
ha ha .
i answer randomly...:)
i don't like series questions, but this thread was sleeping so i posted for
fun
Sort in decreasing order of Ci ?
On Tue, Jun 7, 2011 at 8:22 AM, aanchal goyal goyal.aanch...@gmail.comwrote:
Given n jobs, each ith job has a cost Ci associated with it. The waiting
time for a job i is defined as Ci*Delay, where Delay is the number of days
it takes from the first day to
uninterested can filter messeges from sayanna...@gmail.com. :P
On Tue, Jun 7, 2011 at 11:14 PM, sayan nayak sayanna...@gmail.com wrote:
Multiple openings in Samsung Bangalore in following domains:
1. Wireless protocols
2. Android
3. Multimedia
4. Browser
5. mobile application.
Replce all 0's with -1
-1 1 -1 -1 1 1 1 1 -1
perform the following operation
a[i] = a[i-1] + a[i] for i = 1 to n-1
-1 0 -1 -2 -1 0 1 2 1
consider first element as 0
0 -1 0 -1 -2 -1 0 1 2 1
hash all these values to an array of size 2n+1 as sum will lie between -n to
n
where for 2 ij value hash to
@Harshal
Your Solution is optimal but i think in this question whenever there is some
space available in array we should be able to use that space for either of
the stack.
in your solution this thing is missing. I don't know the solution but i
think it can be improved in a way to get the required
@ Maksym Melnychok
Fails i think
some explanation
say we have numbers 1 , 101
divide all by something to get 0.1, 0.101
simple sort will give you 0.101, 0.1
multiply all numbers to get original ones 101,1
join 1011
but correct answer should be 1101, isn't it ??
correct me if i m wrong ??
--
won't this simple algo work ??
start from root node, say it has value 0
at any time if a node has a value v
pass v-1 to left subtree and v+1 to right subtree
keep track of max and min
final answer will be max -min = Diameter of tree.
correct me if i m wrong.
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nope, it will not work :(
got a case
On Mon, May 30, 2011 at 11:57 PM, sunny agrawal sunny816.i...@gmail.comwrote:
won't this simple algo work ??
start from root node, say it has value 0
at any time if a node has a value v
pass v-1 to left subtree and v+1 to right subtree
keep track of max
@sravanreddy001
i don't find any cases for which my algo fails and its O(nlgn)
i may be missing something
can you tell any case where it fails
On Sun, May 29, 2011 at 10:15 PM, sravanreddy001
sravanreddy...@gmail.comwrote:
@vishal,Sanjeev..
for the inputs 18,187.. apply ur method..
18 --
@Logic King
No, My algo will take only O(nlgn) where n is no of elements.
what i mean by editing the comparision function cmp function of sort of
algorithm.h
sort(a,a+n, cmp);
where cmp is the comparision function defined in my prev. post
it will take equal no. of comparision as in sorting.
two strings can be mixed up anywhere .. and yes the ordering of the
characters in the original strings must be preserved while constructing the
third string ??
On Fri, May 27, 2011 at 1:04 PM, Senthil S senthil2...@gmail.com wrote:
@ sunny agrawal : I misinterpreted the question .. but im
@senthil
nopes, it will not work
eg.
S3 = cba
S1 = a
S2 = bc
count matches but not interleaved
On Thu, May 26, 2011 at 2:43 PM, Senthil S senthil2...@gmail.com wrote:
Can it be done this way ..Take count of each alphabet for all the three
strings and store separately.. Then for each alphabet
answer given by abhishek is the minimum no of flowers that need to be
plucked
in general
no of flowers = 15*x/16
where x is no of flowers to be offered in the temple
so
15,16
30,32
45,48
..
..
all are possible
infinite solutions
On Wed, Mar 16, 2011 at 10:36 PM, Kunal Patil kp101...@gmail.com
i think this will do, using DP
pre compute the sum[i,j] that contains sum of all the values from i to j
let M[i,j] is amount of money that can be made by selling the treats
then M[i,j] = max(a[i] + sum[i+1,j]+M[i+1,j], a[j]+sum[i,j-1],M[i,j-1])
i haven't checked your solution and don't know are
which is nothing but C(2n-2,n-1) Catalan Number
i don't think so??
for a n*n grid answer is 2n!/n!n!
but catalan number gives ans as 2n!/(n+1)!n!
On Thu, Mar 3, 2011 at 8:12 PM, bittu shashank7andr...@gmail.com wrote:
both DP Catalan Number solve this
1st Approach Catalan Number
@ashish
its not that problem, read the question carefully
On Wed, Mar 2, 2011 at 2:32 PM, Ashish Goel ashg...@gmail.com wrote:
refer catalan numbers
Best Regards
Ashish Goel
Think positive and find fuel in failure
+919985813081
+919966006652
On Wed, Mar 2, 2011 at 2:00 AM, bittu
understand why the
problem is different from catalan number sunny...
Best Regards
Ashish Goel
Think positive and find fuel in failure
+919985813081
+919966006652
On Wed, Mar 2, 2011 at 3:03 PM, sunny agrawal sunny816.i...@gmail.comwrote:
@ashish
its not that problem, read the question
@bittu
Question is about to print entire array in sorted order, not searching an
element
On Wed, Mar 2, 2011 at 4:13 AM, bittu shashank7andr...@gmail.com wrote:
@all after 32 Message Discussion I know Everyone is looking for O(N)
solution well it seems odd how we can search an element in a
DP
initialize ways[][] m*atrix to zero*
*
*
*if 0,0 is blocked no solution else*
*ways[0][0] = 1*
*for i in range 1 to N-1 if ways[0][i] is not blocked ways[0][i] =
ways[0][i-1]*
*for i in range 1 to N-1 if ways[i][0] is not blocked ways[i][0] =
ways[i-1][0]*
*
*
*for i in range 1 to N-1*
*for j
step 1: start at Bottom-left corner
step 2: if equal then return else if element is less than number to be
searched for then it cannot be in that row go up
step 3: if element is greater than number to be searched for then it cannot
be in that column go ahead
Repeat untill you got the
for general case to print i to n
create a class with one static member i initialized to 0
in constructor of the class increment i and print i
create n objects of the class in main
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You are given a array with rows sorted and column sorted. You have to print
entire array in sorted order.
--
Sunny Aggrawal
B-Tech IV year,CSI
Indian Institute Of Technology,Roorkee
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int i=1;
#define PRINT1 couti++endl;
#define PRINT2 PRINT1 PRINT1
#define PRINT4 PRINT2 PRINT2
#define PRINT8 PRINT4 PRINT4
#define PRINT16 PRINT8 PRINT8
#define PRINT32 PRINT16 PRINT16
#define PRINT64 PRINT32 PRINT32
int main()
{
//as 100 = (1100100); we need to use PRINT64, PRINT32, and
as there are only 6 possibilities, i think this will do
BCG
BGC
GCB
GBC
CBG
CGB
for each find no of moves, and print min
On Tue, Feb 22, 2011 at 2:51 AM, rgap rgap...@gmail.com wrote:
Hi, I need help with this problem
and also
void change()
{
// Code here
int x = 1;
int*p= x;
while(*p != 5) p++;
*p = 10;
}
On Wed, Feb 16, 2011 at 8:02 AM, Balaji S balaji.ceg...@gmail.com wrote:
The solution is..
_AX = 10;
can anyone explain??
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