CareerCup

WE can improve the algorithm in the controller which is controlling the elevators.

great...I think this will work Jackie

I think it will not work for this, try it out
{2, -3, 5}
and is 5 but your maxSumArray will give 4,

Hi

this wont work for numbers which are all negative
{-1,-3,-5,-2,-1,-4}

Nice solution Loler. I think it will work

dumbo there can be numbers in the subarrays which are common to both hence they can overlap so I dont think it will work
Can you please explain your algo using the example given above

Can someone please explain what does this means
"Patterns can be mutated"

Nice answer Manohar. Yes it will give the desired result in O(n) time

Yes it will.. there is a loop in the list and if a pointer isnt able to move 3 steps then obviously there is no loop. This means you have found a loop much faster than the method where you move 2 times

I will use a min heap according to priorities in this case.
struct heap
{
int priority
int service
heap *left;
heap *right;
}

or best approach to use a heap is arrays

Do you think aka your approach will lead to minimum lift movements?

A sorting algorithm which takes minimum seeks.
How about bubble sort which will take about 2N seeks in every step(maximum)

Ashish can you please explain how it will not work>
For no right child I am using a blank in the array.

Do a level order traversal. Remember for null leaves enter blank in an array
eg
1
2 3
4 6 7

post order traversal is : 1234_67
Now here we follow the property that child will always be at 2i and 2i+1 index or vise versa parent is always at |i/2| index
therefore parent of 1 will be no one
parent of 2 = 2/2 = 1 i.e 1
parent of 3 = 3/2 = 1 i.e 1
parent of 4 = 4/2 = 2 i.e 2 and rest copy from 2
parent of 6 = 6/2 = 3 i.e 3 and rest copy from 3
parent of 7 = 7/2 = 3 i.e 3 and rest copy from 3

Oh yes I forgot to add that logic.
You can use with a simple check in the beginning to know. Use a trie for the dictionary.

Looks good
Chintamani can you please elaborate your algo with a small example.
you can use a adjacency matrix to represent the graph.

This can be solved using recursion
Please let me know if I have messed up somewhere here.

void FromTo(char *source, char *dest, char **stackofWords, int wordCount)
{
if(strcomp(source, dest) == 0)
{
print(stackofWords, wordcount);
return;
}
else
{
for(int i=0;i<strlen(source);i++)
{
for(int j=0;j<26;j++)
{
if((char)((int)'A' + j) != source[i])
source[i] = (char)((int)'A' + j);
else
continue;
stackofwords[wordcount] = source;
Fromto(source, dest, stackofWords, wordcount+1);
}
}
}

}

Algo

I am changing every letter with all the alphabets in the english dictionary and then comparing with the desitnation.

I think you are right gen-y-s

Here is my question
We can an array of bits. When we convert it to a integer, isn't we are scanning each and every array element? I mean
We divide the array into 2 halfes and take the first half. Now to know whether first half has a 1 we convert the array into integer.
My question is how we are converting the array into integer?Are we not scanning the complete array for this?

We can hash the list 1 in one hash table keeping the count of each word in hash
Now as we scan the next list decrement the count in hash once it is present in the hash

This is an invalid string right ?
11+*6

Can you please give an example

Can you please provide an example. Is it that value in key of one hash table needs to be compared to value in every key in other hash table?

How are hash tables compared. Is it values with the same keys or what?

In the first pass remove all 'b'
Then in the second pass search for a 'c' and once you find a 'c' then go back and look for its immediate 'a' and found delete it.

A matrix with printer and system as rows and columns. i.e a graph representation.
Each cell can contain the distance between printer and system.

Oh yes, I just slipped out on that case. well you have your full solution
1) If there are no right child of a given node its parent is the next highest node
2) Else the left most node of the right node of the given node

Do you want me to code for the solution? I think once an algorithm of a problem is known, coding may no be a problem.
Please do let me know if you are finding anything wrong with the above solution, we can discuss.
Also let me know if you want concrete code, I will write it down for you

why do we need a parent pointer just wondering. The next largest node will be the left most node of the right node of the given node.

Go to the middle and reverse the second part of the linked list
Now we have 2 pointers one pointing to the first element and the other to the middle element. Compare and return true if pallindrome

there is a runtime error here
Please add a check for n i.e if (n<0) then return false;

Yes it works aka only thing is as alex has pointed out count%2 == 1 instead of 0

Find longest increasing subsequence where you have the index as well for the elements which contributes to the subsequence.
Now the numbers which do not belong to this subsequnce are the one which needs to be removed

aka what is sum in the array that you have given

Please let me know if something is wrong here

bool IsSumForOdd(int *arr, int n, int count, int sum)
{
if(sum == 0 && count%2 == 0)
return true;
if(sum < 0)
return false;
if(sum < arr[n-1])
return IsSumForOdd(arr, n-1, count, sum);
else
{
return (IsSumForOdd(arr, n-1, count+1, sum-arr[n-1])) || (IsSumForOdd(arr, n-1, count, sum));

}

}

Probably he is asking for the longest increasing subsequence so whichever is longest we have to take that

Starting from any node we can calculate the diff between distance and petrol at each node.
(assuming we can only travel in one direction).
now I will start from a node which will have max positive diff and then work out going through every node. I cannot take nodes as start nodes which have negative diff.

I will have 2 types of parking here. For big cars and small cars. Small cars can be accommodated in big cars parking but not vice versa. .
Will have 2 queues for this.
Once the queue for small cars is full I will assign big cars pool or queue to small cars.

Here can be the classes

Class Garage
{
int id
}

Class SmallGarage : Garage
{
Car objId;
}

Class BigGarage : Garage
{
Car objId;
}

Will have multiple objects of smallGarage and BigGarage each inserted in a queue
As an when we have a car object we dequeue a queue and then enqueu in a queue with the car object assigned to that garage object.

Is it observer design pattern wherein the program register for particular exceptions and when an exception occur the program gets notified.
Please let me know your views

"You have to step on every spot exactly once", is It that once we have stepped on a path we cannot step on it again?

Take a 256 integer array. Now once a charater is scanned increase the frequency of the character in the integer array according to its ASCII value

Why do we need a max heap. Build a min heap here of 10 number here. REad file in chunks and then replace the min element

Your method is same as calculating the distance between centre and the point and checking if it is < R

Here is the modified solution. Hope this helps

bool SubArray(int n, int *arr, int N, Sum, int *val, int index) // N is the number of elements in the array
{
if(n < 0)
return false;
if(Sum \ N == 0)
{
//PrintSubArray
PrintSubArray(val, index);
}
else
{
val[index] = arr[n-1];
SubArray(n-1, arr, N, Sum + arr[n-1], val, index+1);
IsSubArray(n-1, arr, N, Sum, val, index);
}
}

JSDUDE, your structure doesnt care about the timestamp. Let say a customer hits a particular video 6 times in a span of 4 hrs, and the threshhold is 4 then it means customer is having issue but actually he is not facing any issues

bool IsSubArray(int n, int *arr, int N, Sum) // N is the number of elements in the array
{
if(n < 0)
return false;
if(Sum \ N == 0)
return true;
else
{
return (IsSubArray(n-1, arr, N, Sum + arr[n-1]) | IsSubArray(n-1, arr, N, Sum);
}
}

This fn will find whether there is any subarray like this.
We can easily modify this fn to get the elements of the subarray which has the sum divisible by N

Lets take customerId and VideoId as a key. Therefore for this key we can create a hash and where chaining is done for the same key and the data in the same key are sorted according to the time stamp. In this way we can get the difference between 2 time stamp and can predict whether a customer is facing buffering issues

Please do let me know your suggestions

First calculate the distance between all the places with the centre i.e given longitude and lattitude.
Now whichever is equal to and less that the radius are the places needed

Some questions to follow
1) "Knowing a value between 0 and 1 for every one of them" What does this mean? Can you please give example.
2) How do you rank people i.e top 5 ?

We can have a model view controller type architecture in a calculator
where
view can be the display
A controller and a model i.e a small memory storage to store last 10 commands.