@Dave solution is perfect. I prefer it over mind.
However, here is an alternate solution.
We know that this is an equation in 'x' with a degree of 1. It means it is
a straight line and root is guaranteed unless of course the equation is of
the form y = c. That is not the case here as it would m
I like that solution better than the one I suggested.
Don
On Apr 4, 4:45 pm, Dave wrote:
> @Kumar0746: Technically, you can't solve an _expression_; you can solve an
> _equation_, which is a statement of the form expression = expression, which
> is what you have.
>
> Don's suggestion is a good on
@Kumar0746: Technically, you can't solve an _expression_; you can solve an
_equation_, which is a statement of the form expression = expression, which
is what you have.
Don's suggestion is a good one. Another way is to call the expression on
the left side of the equation f(x) and the expressi
Simplify the expression by evaluating expressions inside parenthesis
first. Follow the order of evaluation, doing multiplications first and
then addition and subtraction. It should be possible to reduce any
expression to the form
ax+b=0. Then x=-b/a.
Don
On Apr 4, 11:18 am, arun kumar wrote:
> Gi
@Ankur: Since the list is of infinite length, equal probability of
selecting any given node is impossible. The probability distribution
must be such that
inf
sum p(i) = 1.
i = 0
I.e., the individual probabilities must form a convergent series, and
thus p(i) --> 0. But a uniform distribution has
The thing is ..will it ascertain that the probability is equal
I am not sure how ur method guarantees that...
May be if you and Dave can explain the algo a bit better that wud be great .
regards
Ankur
On Sat, Dec 24, 2011 at 5:48 AM, Piyush Kansal wrote:
> Hey Ankur,
>
> What is the order of ti
@piyush : O(n/2) is O(n) .
your approach is good , it will efficient than linear search.
On Sat, Dec 24, 2011 at 5:48 AM, Piyush Kansal wrote:
> Hey Ankur,
>
> What is the order of time complexity we are looking for in this case. The
> option which Dave suggested can give us random node by
Hey Ankur,
What is the order of time complexity we are looking for in this case. The
option which Dave suggested can give us random node by traversing that many
number of nodes from the head. That will be O(n).
This can be further reduced to n/2 if we use two pointers, both of which
will trave
@Ankur: The linked list is isomorphic to the non-negative integers, so
selecting a random integer is equivalent to selecting a random node.
There is no uniform distribution on the integers, so we can't find a
uniform distribution on the nodes. One way to find a non-uniform
distribution is by taking
Correct me if I am wrong,
If the given number is lets say 'p'. We have to find N such that
N=(10^m-1)/9 for least m,that is divisible by 'p'.
(10^m-1)/9 = 0 (mod p)
(10^m-1) = 0 (mod p)
10^m = 1 (mod p)
Since the given number 'p' always ends with a 3 in its units place, 10^m and
p are always c
I should have noted that this can handle inputs up to about 2^32 / (10
* x). Run time is proportional to the number of 1's. You can also add
a bit of code to discover the digits of the multiplicand.
I was able to verify with lisp bignums that: 25,514 1's is equal to
76543 * ( a 25,509 digit numb
First note:
0 * 3 = 0
7 * 3 = 21
4 * 3 = 12
1 * 3 = 3
8 * 3 = 24
5 * 3 = 15
2 * 3 = 6
9 * 3 = 27
6 * 3 = 18
3 * 3 = 9
Important is that the final digits of each line count 0 to 9.
Now you can build an answer right to left.
Example: 123.
Check the table to get the rightmost 1.
7 * 123 = 861
N
well , I tried to solve it from Maths
though half way through only :(
N = number required i.e. (10^k -1)/9
given n = 10x + 3
by eq
(10x + 3) * m = N= (10^k - 1)/9
implies
k = 2log 3 + log m + log(10x + 3)
i.e.
k > 1 + log n
This gives the lowerbound on minimum number of 1 to be start with, b
Does "multiple 1" means that number with 3 in units place has all 1's or
there can be other digits in the multiple of number like 34111232?...
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@Prasad: The number with 22 1s, 1,111,111,111,111,111,111,111, is
divisible by 23. The quotient is 48,309,178,743,961,352,657.
Dave
On Oct 11, 5:00 pm, prasad jondhale wrote:
> 23 has 3 in its unit place but it is not multiple of 111 or 11 or .
> will u pls elaborate on the problem statement
I meant *Txy < Txz + Tzy*
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one more detail:
*Txy* is optimal distance so that *Txy > Txz + Tzy*, for any x, y and z
between 1 and N.
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i think steiner tree can be used
On Sep 26, 10:05 pm, drealecs wrote:
> I have a software problem and I'm searching for a solution but tried
> different algorithm approach and nothing came out.
> I'm not very familiar with all the graph algorithms and I hope there
> is already a way to solve this
@Ankur: The least significant digit of a[i], base n, is a[i]%n. The
middle digit is (a[i]/n)%n, and the most signficant digit is a[i]/
(n*n). So the code looks something like
int b[n], c[n], i, j;
for( i = 0 ; i < n ; ++i ) // sort by least significant digit
c[i] = 0;
for( i = 0 ; i < n ; ++i
lets assume numbers are in range 0 - 4 ,and array is aarr . then range
would be 0- 4^3 .. (0 - 64)..
now allocate an arraay .. int *x = calloc(n^3,sizeof(int)) // can use even
bits aalso .. but for simplicity of algo , using array
for(int i=0;i<4;i++)
{
x[arr[i]] = 1;
}
for(int i=0;i<64;i++)
{
if(
@Dave
Dude can u provide a sample code...What do u mean by radix n ..also radix
sort requires some other sorting algo to sort digits
Regards
Ankur
On Sun, Aug 14, 2011 at 1:33 PM, Dave wrote:
> @Ankur: Use a radix sort with radix n. It will take 3 passes to sort
> the 3 base-n digits, each of
Yes..i agree with Dave..Here is what i think.
As you have integers upto n^3 in your input, it would need [3*lg(n) + 1]
bits to represent each integer.
So take each group of r = ceil(lg(n)) bits at a time.
So this becomes number of bits needed to represent single digit.
Each digit thus can take 2^r
@Gopi: Explain how a counting sort can be done without zeroing out an
array of size n^3, and then scanning it, or explain how to do these
operations in O(n).
Dave
On Aug 14, 10:52 am, "*$*" wrote:
> if extra space is allowed .. can use counting sort
>
>
>
>
>
> On Sun, Aug 14, 2011 at 8:38 PM, A
@Ankur: Use a radix sort with radix n. It will take 3 passes to sort
the 3 base-n digits, each of O(n), so the overall order will be O(n).
On Aug 14, 10:08 am, Ankur Garg wrote:
> This is one question from Coreman
>
> 3rd Edition -
>
> 8-3-4 -- Sort n integers in the range 0 to n^3 -1 in O(n) ti
I guess it is binary search tree. Try this and you will get what you
want.
On Sep 26, 5:12 am, "mukesh agrawal" <[EMAIL PROTECTED]> wrote:
> What is binary index tree ? I googled it finding irrelevent information ...
> help needed ..
>
> On 9/25/07, Mohammad Naser Zandy <[EMAIL PROTECTED]> wrote:
What is binary index tree ? I googled it finding irrelevent information ...
help needed ..
On 9/25/07, Mohammad Naser Zandy <[EMAIL PROTECTED]> wrote:
>
> I am agree with Sticker,
> I comes to write exactly what Sticker wrote. ;)
> It's O(N^2 / 2) that is 1+2+3+4+5+...+N.
>
> On 9/25/07, Sticker <
To store N we can make a binary tree with extra data store is the no
of left nodes in tree.
now search cost is O(logN) and O(1) is the cost of finding no less
than that particular no.
also , we can update tree in O(logN) tIme.
On Sep 25, 6:16 am, Sticker <[EMAIL PROTECTED]> wrote:
> I saw your id
I am agree with Sticker,
I comes to write exactly what Sticker wrote. ;)
It's O(N^2 / 2) that is 1+2+3+4+5+...+N.
On 9/25/07, Sticker <[EMAIL PROTECTED]> wrote:
>
>
> I saw your ideas. Some of them are correct some are not. Here is what
> I am thinking.
>
> >From the question we know that 1 ≤ M ≤
I saw your ideas. Some of them are correct some are not. Here is what
I am thinking.
>From the question we know that 1 ≤ M ≤ 100 and 1 ≤ N ≤ 4000. So
notice that M >> N. Using a linked list to store M, you have to travel
it from either end, so at least M/2 travels each time, still larger
than
I already know how to solve it.
Binary Index Tree is good approach to solve this problem. Thanks very
body involved in this discussion!
On Sep 22, 6:31 am, KK <[EMAIL PROTECTED]> wrote:
> On Sep 5, 11:07 am, jliang <[EMAIL PROTECTED]> wrote:
>
> > how about a doubled linked list where you can re
On Sep 5, 11:07 am, jliang <[EMAIL PROTECTED]> wrote:
> how about a doubled linked list where you can remove item at any
> index. the removing is O(1). you can also keep track of the current
removing from a doubled linked list is O(1)? How?
> size S so if you are borrowing book at index greate
I can think of an o(n lg n) implementation that uses linked lists.
Maintain LL of original book array: 26-> 1-> 42-> 15 ->3 this shrinks
as books are removed.
maintan LL of borrowed books: 3->4 ..
Original LL now becomes 26->1->15
Sort the borrowed LL --(borrwored order may be 3->4->1 == 1->3->4)
Of course, the index numbers lower than the book being checked out,
don't need to be changed, just the higher numbers.
I like using an array of structs with structure members which can
group all the relevant data together that I want for that book, or
student, etc.
So Books[] might have: title,
how about a doubled linked list where you can remove item at any
index. the removing is O(1). you can also keep track of the current
size S so if you are borrowing book at index greater than S/2, you
traverse the list from the end, if index < S/2, you traverse from the
beginning.
every time a boo
I know the naive method. It's very straight. Just maintain an index
array which records the
index of the books. whenever one book is removed then update all
elements in the index array after its index.
It runs O (M N). I'd like to know is there any efficent data structure
to speedup it?
Thanks.
Nevermind. I found the solution. Thanks anyway. I think the problem is
called "post office location problem" instead of "post office problem"
which is quite complex.
Hi Arunachalam,
Can you please explain that algorithm on how to find the weighted
median? and what is the proof that weighted median is the answer?
Search the net for the post office problem.
The answer is find the weighted median.
On 11/22/05, fangvv <[EMAIL PROTECTED]> wrote:
Given n point coordinate value {x1,y1}.{xn,yn} andweight w1.wn, How to find point P(X,Y) meet :
X >= min{x1,x2,xn} , X <= max{x1, x2,.xn}Y >= min{y1,
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