naved momin

Ok i got it anyways thanks for the help

ok i got it thanks for the help

Thanks for the reply sir,

My question is Set doesn't store duplicate then how 2 stack with similar content & similar hashcode is stored in Set lists?

My Main Set
Set<Stack<Integer>> lists = new HashSet<Stack<Integer>>();

My Temp Set
Set<Stack<Integer>> tmp = new HashSet<Stack<Integer>>();

Suppose main Set contains
{{0,3}{1,2,4,5,6}}

Suppose tmp Set contains
{{5,6,8},{1,2,4,5,6}}

When I am doing

lists.addAll(tmp);

what I get in Set lists is
{{0,3},{1,2,4,5,6},{5,6,8},{1,2,4,5,6}}

So i thought why not to check their hashcode so I did
int max = getMaxSize(); for (Stack<Integer> s: lists) { if(s.size() == max){ System.out.println(s.hashCode()); for (Integer integer : s) { System.out.print(integer + " "); } System.out.println(); } }

since the max size value is 5
Output is 1,2,4,5,6
1,2,4,5,6
twice which I don't want in fact I m using Set because i don't want duplicate Stack in Set lists
Where my Understanding is going wrong?

Fist I m very sorry i didnt knew that editing my orignal post will have that impact, i will keep this thing in mind

Second thing is this question is asked to me in one of mine interview

Third ya still i dont know how to solve this problem by hands (because then only I can code it)

I have already spent hours trying to figure it out how this problem might me solved but no luck, so asked here, I guess there is some math involved in this type of questions so unless and until i don't have that particular topic of maths knowledge I wont be able to solve it.

Can you help please ?

I have corrected the question please have a look again , thanks for providing your hands to help me

Multiple “intervals” are to be given as tasks. Calculate the maximum work time (minutes) to assign to one worker

Implement the method “int Problem2#getMaxWorkingTime(List intervals)”.

Return the maximum time to work on a task when one worker takes it.

Time assignment unit shall be based on tasks. (i.e. Task assignment shall be either to complete the whole task, or to do nothing for it.)

The length of the interval, time required to complete the task, is calculated by subtracting "end time" from "start time". (e.g. If [12:00, 13:00] is given, the length of the task is 60 min.)

Return 0 if the argument is null or an empty list.

The argument (a list of “intervals”) must not contain null.

The number of “intervals” must not exceed 10,000

In Figure 3 , work time is maximized when three tasks [“06:00”, “08:30”], [“09:00”, “11:30”], and [“12:30”, “14:00”], are assigned. Therefore, the answer is 390 (minutes).

Figure 3: An example of input and answer 6 7 8 9 10 11 12 13 14 ---------------- --------------- --------------- ------- -------------------------------- -------------------

Intervals = [06:00-08:30], [08:00-09:00], [09:00-11:00], [09:00-11-30], [12:30-14:00], [10:30-14:00] (which is shown in figure 3 also)
Total Intervals = 6

What I have tried so far

To me it appears to as a problem whose answer I can find simply by subtracting the Interval with max end-time i.e 14:00 with Interval which has smallest start-time, say 14:00 minus 06:00 answer is 480 minutes (since 6 to 14 has 8 hours in it so 8*60 minutes = 480 minutes) but the correct answer is 390 minutes


Can any one please help me explaining how to solve this problem...?

The thing which I want to understand is out of 6 Intervals why only those 3 Intervals (mentioned above) is selected ?

According to me its nlogn, what you think ?

Hi all,

Just a simple question, suppose i have given a List<Intervals> list; I am iterating over list and inserting each element from list into BST, if time require to insert into BST is logn then what is the total time require to insert all the elements into tree ?

logn or nlogn ?

Is this algo is an in-place, according to me yes, because tmp[] is consider as auxiliary space, so space complexity still remains O(1)
Do you agree with my answer if no please explain me too
thanks for your time and effort..!!

public String transformAlgo(String k, int len) { int med = len/2; int init = 0; char[] tmp = new char[k.length()]; for (int i = 0; i < k.length(); i++) { if( (i%2)==1 ){ tmp[med++] = k.charAt(i); }else tmp[init++] = k.charAt(i); } k = String.valueOf(tmp); return k; }

thanks for the answers...
i asked because when i was reading a placement crack book ...they used pointer size as well to determine the optimality of the code for a given problem .....

hi all

what is the size of reference variable in Java ?

eg

MyClass foo = new MyClass();
what is the size of foo in my case ?
JVM is 32bit.

what i understand is ...

if JVM is 32bit , reference variable is of 4bytes since (8*4=32bit)
if JVM is 64bit , reference variable is of 8bytes since (8*8=64bit)

is my answer correct ??

thanks ..appreciate your input....


so conclusion is ..its O(n^2) ....

hi All...

Suppose i have written a method like this
public void myWork(){ mergeSortMyData(); O(nlgn) .... someProcessing(); O(n^2) someprocessing2(); O(n^2) }

then what will be time complexity of my code...according to me it will be O(nlgn + n^2) ...is it correct ?

Ulf Dittmer wrote:A few comments in no particular order.

1) In the client code, the ByteArrayOutputStream is unused and unnecessary.

2) If the file size exactly 1024 bytes? If it is larger, the code is sending partial content that will be corrupted. If it is smaller then the code sends too many bytes, again possibly corrupting the contents - the number of bytes in the packet should be what fis.read returns, not the size of the byte[].

3) Have you gotten this client/server pair to work? If not, start with that.

silly mistake ..filesize in kbs * 1024 = buffersize ...lolz...thanks mate!