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Piyush Grover
*Be a traveler, not a tourist*
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The problem is an array sorting problem.
You are given an array of size N containing values 1 to N only. Sort it in
O(N).
Start from floor 1 till ith floor until you find a person on the wrong
floor, take him to his respective floor
take the guy from that floor and take him to his respective
I have a practical problem, need an optimal solution for this
*What is given?*
Given *N* sets, each containing some jobs to be executed, such that no two
sets are subsets of each other and number of jobs in *i-th* set is *ni N*
.
The jobs can have values between *1...k where k N*. Priority of
, Piyush Grover piyush4u.iit...@gmail.comwrote:
I have a practical problem, need an optimal solution for this
*What is given?*
Given *N* sets, each containing some jobs to be executed, such that no
two sets are subsets of each other and number of jobs in *i-th* set is *ni
N*.
The jobs can have
@Don can you give the logic of your rnd() function?
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For more options,
The number which has 5 as last digit can be written as
n = 4*a + b where b = 1 or 3
so if 4*a + 1 == n or 4*a + 3 == n; then n has 5 as a last digit.
So code looks like:
void main()
{
int n, m;
scanf(%d, n);
m = 2;
m = 2;
if(add(m, 1) == n || add(m,3) == n)
puts(yes);
else
puts(no);
}
forgot to assign:
m = n;
On Sun, Oct 14, 2012 at 2:27 AM, Piyush Grover piyush4u.iit...@gmail.comwrote:
The number which has 5 as last digit can be written as
n = 4*a + b where b = 1 or 3
so if 4*a + 1 == n or 4*a + 3 == n; then n has 5 as a last digit.
So code looks like:
void main
@Partha
try with:
A = {2, 2, 9}
B= {1, 6, 6}
On Mon, May 21, 2012 at 7:08 PM, partha sarathi Mohanty
partha.mohanty2...@gmail.com wrote:
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
you can do it in nlogk or n+klogn time.
create a min-heap tree of size k with first k nodes.
Add k+1..n nodes in the tree. Everytime you insert a node in the tree check
if it's greater than the root node. If yes remove
root node and insert the new node (logk operations). So time complexity
will
Given a set S, find all the maximal subsets whose sum = k. For example, if
S = {1, 2, 3, 4, 5} and k = 7
Output is: {1, 2, 3} {1, 2, 4} {1, 5} {2, 5} {3, 4}
Hint:
- Output doesn't contain any set which is a subset of other.
- If X = {1, 2, 3} is one of the solution then all the subsets of X {1}
:
@Piyush :
you are re-posting same problem which you had posted on 5 dec 2011.
check this link :-
http://groups.google.com/group/algogeeks/browse_thread/thread/8a58ea05c96f811b/ee74f8a4d7b68561?lnk=gstq=Maximal+possible+subsets+Algorithm#ee74f8a4d7b68561
On Mon, Jan 9, 2012 at 3:27 AM, Piyush
insertNode(node *head, int value){
node *new;
if(head == null){
new = (node*)malloc(sizeof(node));
new-data = value;
new-left = new-right = new-inoredrsucc = null;
head = new;
}else{
node *root = head;
node *l, *r;
.
Please check it.
Regards,
Aman.
On Sat, Dec 10, 2011 at 6:30 PM, Piyush Grover
piyush4u.iit...@gmail.comwrote:
insertNode(node *head, int value){
node *new;
if(head == null){
new = (node*)malloc(sizeof(node));
new-data = value;
new-left = new-right
not taken that into consideration.
Please correct me if I am wrong.
Regards,
Aman,
On Sat, Dec 10, 2011 at 7:37 PM, Piyush Grover
piyush4u.iit...@gmail.comwrote:
I don't think so.
First you added 15.
15-left = null=15-right=15-succ;
add 13. (13 15) so
13-left = 13-right = null; 13-succ
-Piyush
On Sat, Dec 10, 2011 at 8:28 PM, AMAN AGARWAL mnnit.a...@gmail.com wrote:
Hi.
So can you please tell me the modifications required so it works correctly.
Regards,
Aman.
On Sat, Dec 10, 2011 at 8:22 PM, Piyush Grover
piyush4u.iit...@gmail.comwrote:
yup, you are right..
On Sat, Dec
:
The possibility is ruled out by your question itself.There are exponential
subsets of a set,so finding all subset is not possible in polynomail time.
A backtracking approach is what you should think on,
On Mon, Dec 5, 2011 at 12:51 PM, Piyush Grover
piyush4u.iit...@gmail.comwrote:
Given a set S
subsets where
the addition of the weights results in maximum (each weight being
considered only once)...
Say for ex-
The max weight is X=Wmax , then find all subsets which add up to X..
On Dec 5, 1:51 pm, Piyush Grover piyush4u.iit...@gmail.com wrote:
Yeah but there's one more condition
As I mentioned earlier solution with exponential time-complexity is the
obvious one. Is there any way to solve this problem by greedy/Dynamic algo?
On Mon, Dec 5, 2011 at 5:24 PM, WgpShashank shashank7andr...@gmail.comwrote:
@piyuesh , Saurabh isn't 3-Sum Suffics Here ?
Another thought
Given a set S of objects having weights Wi and values Vi, and given a
maximum weight Wmax.
Find *ALL* the maximal subsets of set S such that Sum(Wi) = Wmax.
Maximal subset means if {a, b, c} is a solution (such that Wa+Wb+Wc =
Wmax) it means there doesn't exist any other object x in S such that
Optimal split: [0,0][1,1][0,0][1,1][0,1][1,0]
Expected value of optimal split: 0 + 1 + 0 + 1 + 1/2 + 1/2 = 3
why this is not the optimal split???
On Sun, Nov 27, 2011 at 6:58 PM, Ankur Garg ankurga...@gmail.com wrote:
You have an array with *n* elements. The elements are either 0 or 1. You
, Nov 22, 2011 at 1:59 PM, anshu mishra
anshumishra6...@gmail.comwrote:
first try to understand the sol then comment. it is for binary tree not
for BST.
On Mon, Nov 21, 2011 at 10:25 PM, Piyush Grover
piyush4u.iit...@gmail.com wrote:
For BST it would be rather simpler. find the first node
For BST it would be rather simpler. find the first node which lies in
between the two.
On Wed, Nov 16, 2011 at 1:44 PM, anshu mishra anshumishra6...@gmail.comwrote:
Node *LCA(node *root, node *e1, node *e2, int x)
{
Node *temp =NULL;
Int y = 0;
As you mentioned, the numbers designating the gaps can only be repeated
so in your example 2 20 can be broken into 19 1 but 19 is already there in
the list (the
first couplet) and it's not the one which describes the gap. can you please
clarify?
On Mon, Nov 21, 2011 at 5:23 AM, Ankur Garg
Given array A.
Compute array B such that
B[0] = 1;
for(i = 1; i n; i++)
B[i] = B[i-1]*A[i-1]
now,
mul = 1;
for (i = n-2; i =0; i--){
mul = mul*A[i];
B[i] = B[i]*mul;
}
On Fri, Sep 30, 2011 at 2:18 AM, Hatta tmd...@gmail.com wrote:
are the algorithm instance always a sequence
sorry, one mistake...
mul = mul*A[i];
it should be
mul = mul*A[i+1]
On Fri, Sep 30, 2011 at 2:57 AM, Piyush Grover piyush4u.iit...@gmail.comwrote:
Given array A.
Compute array B such that
B[0] = 1;
for(i = 1; i n; i++)
B[i] = B[i-1]*A[i-1]
now,
mul = 1;
for (i = n-2; i =0; i
as such there is no specific method to generate random number but if you
have to implement it then you can generate a pseudo random generator
using cur_time_value_in_mili_seconds mod n.
On Mon, Sep 19, 2011 at 9:24 PM, Don dondod...@gmail.com wrote:
The most common way that it works would be
for(i = 0; i n; i++)
for(j = 0; j n; j++){
setColor(i, j) = black;
if(A[i][j] == str[0]){
setColor(i, j) = blue;
a = findWord(A, i, j, str, 1)
if(!a) setColor(i, j) = black;
else break;
}
}
findWord(A, i, j, str, k){
if(k
sry, in the findWord function all a's are different e.g
a0, a1a7
and if(!a) is actually if(a0||a1||...||a7)
thnx
piyush
On Mon, Sep 19, 2011 at 1:10 AM, Piyush Grover piyush4u.iit...@gmail.comwrote:
for(i = 0; i n; i++)
for(j = 0; j n; j++){
setColor(i, j) = black
=Take a function rand() which returns value between [0, 1) uniformly or use
function rand(n) = n*rand() which return value between 0- (n-1) using
uniform probability distribution.
=Now create a array A[0..n-1] = [0..n-1]
now rake an array R.
k = n;
for(i = 0; i n; i++){
a = rand(k);
Don is right
if R = 0, P = 1 and Q = -1 then the given expression is UNDEFINED!!!
On Thu, Sep 15, 2011 at 10:16 PM, abhinav gupta
guptaabhinav...@gmail.comwrote:
Shut up...its 3,,
On Thu, Sep 15, 2011 at 9:43 AM, Don dondod...@gmail.com wrote:
It might be 3, but it doesn't have to
to 3.
Don
On Sep 15, 11:57 am, abhinav gupta guptaabhinav...@gmail.com wrote:
u cnt divide a number by 0..that thing is self undrstod
On Thu, Sep 15, 2011 at 9:49 AM, Piyush Grover
piyush4u.iit...@gmail.comwrote:
Don is right
if R = 0, P = 1 and Q = -1 then the given
it should be:
(1/n)^n * (1 + 2 + 2^2 + 2^3 +(n/2)+1 terms) = {2^(1 + n/2) - 1}*(1/n)^n
when n even
(1/n)^n * (1 + 2 + 2^2 + 2^3 +(n-1/2)+1 terms) = {2^(1 + (n-1)/2) -
1}*(1/n)^n when n is odd
-Piyush
On Mon, Sep 12, 2011 at 8:01 PM, sandeep nagamalli nsandee...@gmail.comwrote:
Sry i was little wrong:
nC0*(1/n)^n + nC2 *2*(1/n)^n + nC4*2^2*(1/n)^n++nCn*2^(n/2)*(1/n)^n
when n is even
nC0*(1/n)^n + nC2 *2*(1/n)^n +
nC4*2^2*(1/n)^n++nCn-1*2^(n-1/2)*(1/n)^n when n is odd
On Mon, Sep 12, 2011 at 10:08 PM, Piyush Grover
piyush4u.iit...@gmail.comwrote:
it should
O(n^4)
Just think...
for(i = 0; i n; i++)
for(j = 0; j i; j++)
has Time Complexity O(n^2)
then how come for(i = 0; i n; i++)
for(j=0; ji*i; j++)
it's O(n^2)..
it's...1 + 2^2 + 3^2 + ...n^2 which is O(n^3)
but adding one for(k=0; k j; k++) will not make
I got..
1.) 2/pi
2.) 3.) 0.5 - (1/pi)
On Sat, Sep 10, 2011 at 2:47 PM, Ishan Aggarwal
ishan.aggarwal.1...@gmail.com wrote:
three points are randomly chosen on a circle.what the probability that
1.triangle formed is right angled triangle.
2.triangle formed is acute angled triangle.
Sorry guys.. I was wrong, radius also matters here...
still investigating
On Sat, Sep 10, 2011 at 3:56 PM, sarath prasath prasathsar...@gmail.comwrote:
@Piyush Grover:
please explain ur answer
On Sat, Sep 10, 2011 at 3:30 PM, Piyush Grover
piyush4u.iit...@gmail.comwrote:
I got
be 1/3 for each of the three options...
On Sat, Sep 10, 2011 at 4:48 PM, Piyush Grover
piyush4u.iit...@gmail.comwrote:
Sorry guys.. I was wrong, radius also matters here...
still investigating
On Sat, Sep 10, 2011 at 3:56 PM, sarath prasath
prasathsar...@gmail.comwrote:
@Piyush
it would be close to zero but not exactly zero.
On Sat, Sep 10, 2011 at 6:28 PM, Ishan Aggarwal
ishan.aggarwal.1...@gmail.com wrote:
How u calculated this??
On Sat, Sep 10, 2011 at 6:19 PM, Neha Singh neha.ndelhi.1...@gmail.comwrote:
The correct ans is :
for right angled triangle : 0
pseudo algo:
=array idx[0...k-1] indicates the current pointer position in the ith
stream(initialized to 0).
=heap tree of size k where each node stores value of the data and value of
stream which the node belongs to.
do{
for all i = 0:k-1
=insert idx[i] value of ith stream to the
It depends on the use cases.
If you have less elements of order of( say 100) then even insertion sort can
be a better choice. It's in-place sorting algo and can
perform sorting as it receives the elements like a stream.
If you have large number of elements and all the elements can't be
Option a) i didn't understand properly, please elaborate!!
Option b) certainly yes.
Option c) No (remember, the size of the table is subjective)
Option d) yes (use index on the column used in the where clause)
On Wed, Sep 7, 2011 at 8:58 PM, Mani Bharathi manibharat...@gmail.comwrote:
when
i don't think it's of O(n)
for even it's actually easy. Take the xor of all the elements you will left
with the element which appears odd number of times. O(n)
but for the odd case, we need to use extra space. Use hash map to keep the
count of each number and
take the odd one out, i mean the even
1.)a
2.)b
3.)b
4)b
On Wed, Sep 7, 2011 at 11:08 PM, Mani Bharathi manibharat...@gmail.comwrote:
“Kya-Kya” is an island inhabitants of which always answer any question with
two
answers , one of which is true and the other is false .
1.You are walking on a road and came to a park . You ask
4) a and b
On Thu, Sep 8, 2011 at 12:08 AM, Piyush Grover piyush4u.iit...@gmail.comwrote:
1.)a
2.)b
3.)b
4)b
On Wed, Sep 7, 2011 at 11:08 PM, Mani Bharathi manibharat...@gmail.comwrote:
“Kya-Kya” is an island inhabitants of which always answer any question
with two
answers , one
it's a subjective question.
If you have 100GB of main memory and 2 GB of virtual memory and you are
running very light weight programs then
nothing will happen. The system memory has enough space to take care of all
the applications but if your system is heavily loaded with
processes then it
One thing i would like to know is
01/02/2011 needs to be considered or 1/02/2011.
thnx
On Mon, Sep 5, 2011 at 11:48 PM, Neha Singh neha.ndelhi.1...@gmail.comwrote:
@yogesh: its stiil not correct. There r many test cases where ur solution
will fail
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use hashing u'll get O(1) space..
@rahul...why not??
On Sat, Sep 3, 2011 at 7:13 PM, priya ramesh love.for.programm...@gmail.com
wrote:
use hashing
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:09 pm, Piyush Grover piyush4u.iit...@gmail.com wrote:
if I have understood the question correctly then:
a[n-1] + b[i] a[j] + b[i] for all 0 = j n-1
and a[j] + b[n-1] a[j] + b[i] for all 0 = i n-1
therefore,
i = j =n-1;
count = 1;
S[0] -- (a[n-1], b[n-1])
p = a[n-1] + b[n-2
if I have understood the question correctly then:
a[n-1] + b[i] a[j] + b[i] for all 0 = j n-1
and a[j] + b[n-1] a[j] + b[i] for all 0 = i n-1
therefore,
i = j =n-1;
count = 1;
S[0] -- (a[n-1], b[n-1])
p = a[n-1] + b[n-2];
q = a[n-2] + b[n-1]
while(count n){
if(p q){
j--;
Since A(n) and B(n) are sorted so in the pair (a[i], b[j]) either i = n-1 or
j = n-1 or both.
1.) so the first element is (a[n-1], b[n-1])
2.) now, we have two numbers to compare p = a[n-1]+ b[n-2] and q =
a[n-2]+b[n-1]
3.) if pq then p = a[n-1]+b[n-3]
else q = a[n-3] + b[n-1]
and repeat in
@Raj
From my personal experience I can just say, being an Indian you will get
more respect in Japan rather in the US.
The problem you may face is the language and the food. If you are non veg,
well n good but if you are veg
you will have to cook it by yourself, there are indian curry restaurants
@Raj
I did two months summer internship there in tokyo. I did internship in
scotland also and what I can say comparing the two is
I got royale treatment in Japan. People are more keen about indians and
Indian culture but in UK no one bother for
you and the same I heard from friends who worked in
This is fine but adding a twist to it.
Find number of nodes which are not in the loop.
On Wed, Aug 31, 2011 at 6:27 PM, rahul sharma rahul23111...@gmail.comwrote:
int loop(void *head1,void *head2)
{
//head1 and head2 initialized to start;
while(head1!=NULL head2!=NULL)
{
head1=head1-next;
loop within a loop, what exactly that means??
Nee to consider the planarity or two loops sharing common edge/s??
On Wed, Aug 31, 2011 at 8:46 PM, MAC macatad...@gmail.com wrote:
Q The question asked in interview of a startup : suppose you have a graph
as shown bellow : please see the
What's wrong with this??
for( i = 0 ; i n ; ++i )
for( j = 0 ; j m ; ++j )
if( a[i][j] != 0 )
a[i][0] = a[0][j] = 1;
for( i = 0 ; i n ; ++i )
for( j = 0 ; j m ; ++j )
if( a[i][0] + a[0][j] != 0 )
a[i][j] = 1;
On Thu, Sep 1, 2011 at 5:45 AM, Dave
and no looping.
int custRand(int n)
{
int a = rand(n*n+n);
int b = a / n;
a %= n;
return (b a) ? n+a-b+1 : n-a+b;
}
On Aug 29, 10:48 am, Piyush Grover piyush4u.iit...@gmail.com wrote:
Given a function rand(n) which returns a random value between 1...n
assuming
equal
I think, it should be C D;
a = i++ = a=i; i = i+1;
a = ++i; = i = i+1; a = i;
in similar fashion, D gets affected.
@ Raj...how come??
On Tue, Aug 30, 2011 at 8:11 PM, raj kumar megamonste...@gmail.com wrote:
code becomes more efficient
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You received this message because you are
#include stdio.h
#include string.h
typedef struct node{
struct node *next;
char *str;
}node;
node *M;
void swap(char *a, char *b){
char t = *a;
*a = *b;
*b = t;
}
void add(char *str){
if(M == NULL){
M = (node *)malloc(sizeof(node));
M-str = (char
the result to be the orthogonal distance from the diagonal.
That is what the formula computes.
Don
On Aug 29, 11:28 am, Piyush Grover piyush4u.iit...@gmail.com wrote:
I understand what you are doing in the loop but return statement is not
clear to me. Can you explain.
On Mon, Aug 29, 2011
to be the orthogonal distance from the diagonal.
That is what the formula computes.
Don
On Aug 29, 11:28 am, Piyush Grover piyush4u.iit...@gmail.com wrote:
I understand what you are doing in the loop but return statement is
not
clear to me. Can you explain.
On Mon, Aug 29, 2011
http://mathoverflow.net/questions/57371/find-all-maximal-subsets-of-a-set-of-integers-whose-sum-does-not-exceed-a-number
actual question is this.
On Sat, Aug 27, 2011 at 9:56 AM, rahul sharma rahul23111...@gmail.comwrote:
yeah i wen first post answer i said it is for finding sum=k;
n just add
it's a quine problem.
char*f=char*f=%c%s%c;
main(){
printf(f,34,f,34,10);
}%c;
main()
{
printf(f,34,f,34,10);
}
I have used whitespaces to make it understand.
On Sun, Aug 28, 2011 at 11:39 AM, rahul sharma rahul23111...@gmail.comwrote:
program whose output is the program itself???
on paper. You would be able to
understand it.
-Piyush
On Sun, Aug 28, 2011 at 11:46 AM, rahul sharma rahul23111...@gmail.comwrote:
plz expalin char*f=
On Aug 28, 11:12 am, Piyush Grover piyush4u.iit...@gmail.com wrote:
it's a quine problem.
char*f=char*f=%c%s%c;
main
one or two more statements then it will require corresponding
change in char *f...can i open the same file with f open n prin t it
out?
On Aug 28, 11:31 am, Piyush Grover piyush4u.iit...@gmail.com wrote:
char*f=char*f=%c%s%c;main(){printf(f,34,f,34,10);}%c;
f is a global pointer
3x+4y = 60
it's a straight line equation whose x intercept is 20 and y intercept is 15.
Draw it in first quadrant
(as x, y are positive integers)
now x = (60 - 4y)/3 = 4(15-y)/3
now for y = 1, 2...15 you need to check whether (15-y) is divisible by 3 or
not. It's simple y = 3, 6, 9, 12
-Piyush
I don't understand your point if y[i] y[i+1]
how come the slope is positive?
It's just that the i'th line will lie under line i+1th line.
On Sat, Aug 27, 2011 at 5:11 PM, monish001 monish.gup...@gmail.com wrote:
If I correctly understood the question, It says:
1. 0= x[i] =1 for all i and for
, y0)) )
return(i, i+1)
else return(i-1, i);
}
}
On Sat, Aug 27, 2011 at 5:59 PM, Piyush Grover piyush4u.iit...@gmail.comwrote:
I don't understand your point if y[i] y[i+1]
how come the slope is positive?
It's just that the i'th line will lie under line i+1th line.
On Sat, Aug 27
pardon 1 small mistake
instead of
if(i == 0 || i == n)
return (lower, upper);
it should be
if(i == lower || i == upper)
return (lower, upper);
On Sat, Aug 27, 2011 at 8:31 PM, Piyush Grover piyush4u.iit...@gmail.comwrote:
satisfy the point on lines, for point(x,y) to lie in between
Here is a problem:
Given an array of size n. Find all the MAXIMAL subsets whose sum is = K.
The solution is not a concern, the optimization is required.
-piyush
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...@gmail.com wrote:
This is the knapsack problem.
Find a polynomial-time solution and you will be a hero.
Don
On Aug 26, 12:43 pm, Piyush Grover piyush4u.iit...@gmail.com wrote:
Here is a problem:
Given an array of size n. Find all the MAXIMAL subsets whose sum is =
K
it's basic unitary method problem:
x men do work in 10 days
1 man will do-in 10*x days
x-10 men do it in 10*x/(x-10) = (10+10)
Solve it and x = 20
it can't be 130
On Fri, Aug 26, 2011 at 11:53 PM, gmagog...@gmail.com
gmagog...@gmail.comwrote:
@Rahul
Assume the productivity of each man
Cut the rope in 50mtrs and 100mtrs length.
Make a small loop(of negligible length at one end of the 50 mtrs rope)
Tie the other end of the rope at the top and from the loop end side pass the
100mtrs rope
such that you have both the ends of 100mtrs rope in your end.
now get down at 100mtrs peg
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