Hi all,
I am a hiring manager at Amazon. We are hiring for SDE and Applied Science
roles in my team. Please send a short note about yourself, the role you
wish to apply for and your updated CV.
Thanks,
Kingston
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Hi all,
I've interviews for SDE-1 in amazon bangalore this weekend. Can any one
tell me the recent questions they are focusing on ? or if anyone have the
recent experience with them please share it.
Thanks
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Algorithm
There are openings in my team for SDE-Is, SDE-IIs, SDM and QAE-1.
Please send me resumes if you or your friends are interested in any of
these roles.
Thanks,
Kingston
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Hi Immanuel,
Please find attached resume and process it for mentioned opening.
Regarding,
Abhijeet Srivastva
09871147025
On Sun, Jul 27, 2014 at 8:27 PM, immanuel kingston
kingston.imman...@gmail.com wrote:
There are openings in my team for SDE-Is, SDE-IIs, SDM and QAE-1.
Please send me
If I understood the problem, the following works fine.
if(A[i][j] == 'o' and it is not and edge element) {
if(A[i][j] is surrounded by either 'x' or 'o') {
A[i][j] = 'x';
}
}
On Mon, Jun 10, 2013 at 8:38 PM, Jai Shri Ram del7...@gmail.com wrote:
Given a 2D board containing 'X' and 'O',
sorry.. pls ignore the above post.. that doesn't work..
On Wed, Jun 19, 2013 at 10:5
5 PM, bharat b bagana.bharatku...@gmail.com wrote:
If I understood the problem, the following works fine.
if(A[i][j] == 'o' and it is not and edge element) {
if(A[i][j] is surrounded by either 'x' or 'o') {
Start BFS from a position which is 'o'. the neighbour elements are as
defined in the question. Mark neighbour as 'Z' if it is not a boundary and
either 'x' or 'o', else *failure*. If there is no more for the considered
connected component *and if it is* *success*, then mark all 'Z's to 'x'. Do
Given a 2D board containing 'X' and 'O', capture all regions surrounded by
'X'.
A region is captured by flipping all 'O's into 'X's in that surrounded
region .
For example,
X X X X
X O O X
X X O X
X O X X
After running your function, the board should be:
X X X X
X X X X
X X X X
X O X X
--
Round 1 :
Q2: given BST is not correct.
correct BST passing this condition will be 7
/
5
/
Hi ravi ,
I think he is trying to find the longest possible increasing chain in
matrix .
he needs to start traversing from first row choosing columns one by one and
move in all direction. but he needs to maintain the already visited nodes.
On Thu, Jun 6, 2013 at 12:07 AM, sourabh jain
Round 1:
1.Design a Data Structure supporting 3 queries a)push b)pop c) find
minimum
Ans : Do it using Two Stacks . in first stack use it as normal stack.
second stack use it to find minimun as the value inserted is greater than
the top ignore it else push it. if pop operation happens and the
Can u clear Round3 --- Qstn-3. the language is not cleared
On Wed, Jun 5, 2013 at 1:52 PM, sourabh jain wsour...@gmail.com wrote:
Round 1:
1.Design a Data Structure supporting 3 queries a)push b)pop c) find
minimum
Ans : Do it using Two Stacks . in first stack use it as normal stack.
Can any one give some points on Round 3 : 1st and 3rd question ?
On Thu, May 2, 2013 at 9:27 PM, Guneesh gunees...@gmail.com wrote:
Round 1:
1.Design a Data Structure supporting 3 queries a)push b)pop c) find
minimum
2.Given post order of a BST find whether each node of the tree(except
these are for which position? and experience?
On Thu, May 2, 2013 at 9:27 PM, Guneesh gunees...@gmail.com wrote:
Round 1:
1.Design a Data Structure supporting 3 queries a)push b)pop c) find
minimum
2.Given post order of a BST find whether each node of the tree(except
leaf) has only 1 child
kindly explain visualizing a balanced BST as a graph and unable to
underdstand your point
thanks
--mac
On Fri, May 24, 2013 at 6:04 AM, Avi Dullu avi.du...@gmail.com wrote:
On Sat, May 11, 2013 at 10:29 AM, Aman Jain pureama...@gmail.com wrote:
2. from this no*de u*, again apply*
My bad, I wrote out degree where I should have written in degree.
My proposal:
Topological Sort http://en.wikipedia.org/wiki/Topological_sorting the
graph and keep removing the nodes from vertices which have in degree of 0
(given the DAG, there will always be at least one such vertex).
For a BST,
On Sat, May 11, 2013 at 10:29 AM, Aman Jain pureama...@gmail.com wrote:
2. from this no*de u*, again apply* dfs* to find longest distant node
from this node.Let us say that node is v.
Small doubt about the solution. Consider this graph a - b - c - d
You randomly choose vertex 'b' and do a
@atul anand : got it thanks for pointing it out
On Mon, May 13, 2013 at 12:19 AM, atul anand atul.87fri...@gmail.comwrote:
@Karthikeyan : Given graph is directed and does not carry edge
weight.So you cannot use pris/kruskal algo to convert them to
tree.Even if you tweak prism/krukal algo
@karthikeyan : It is an acyclic graph not a binary tree. your solution
will work if given graph is a binary tree.
problem can be solved using dfs as suggested above
On 5/11/13, Karthikeyan V.B kartmu...@gmail.com wrote:
Since it is an acyclic graph, find the appropriate node that can be the
@atul anand : acyclic graph can be converted to a tree using prim/kruskal
or by finding an appropriate node that can act like the root of a tree
On Sun, May 12, 2013 at 5:55 PM, atul anand atul.87fri...@gmail.com wrote:
@karthikeyan : It is an acyclic graph not a binary tree. your solution
@Karthikeyan : Given graph is directed and does not carry edge
weight.So you cannot use pris/kruskal algo to convert them to
tree.Even if you tweak prism/krukal algo to form a tree , you can
never guarantee that each node will have only 2 child , it can have
more than 2 children.So your
Given an acyclic graph. Give an algorithm to find the pair of nodes which
has the maximum distance between them, i.e. the maximum number of edges in
between them
any suggestion ?
thanks
--mac
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In a connected and acyclic graph,that is a tree, we can apply this approach
1. apply *dfs *on any random node and find the longest distant node from
this node.Let us say this node i*s u*.
2. from this no*de u*, again apply* dfs* to find longest distant node from
this node.Let us say that node is
Since it is an acyclic graph, find the appropriate node that can be the
root of this tree.
Now apply the logic to find the diameter of the tree here, which finds the
longest path in the graph as follows:
int diameter(Tree * t) { if (t == 0) return 0; int lheight =
height(tree-left); int rheight
Round 1:
1.Design a Data Structure supporting 3 queries a)push b)pop c) find
minimum
2.Given post order of a BST find whether each node of the tree(except
leaf) has only 1 child or not.
eg5
\
7
/
3
/
2
is correct as
A = {5, 3, 8, 9, 16}
After one iteration A = {3-5,8-3,9-8,16-9}={-2,5,1,7}
After second iteration A = {5-(-2),1-5,7-1} sum =7+(-4)+6=9
Given an array, return sum after n iterations
my sol/
void abc(int arr[],n)
{
for(i=0;in;i++)
arr[i]=arr[i+1]-arr[i];
abc(arr,n-1);
}
I wana ask that the
Your solution fails for a number of reasons:
1. If your array is size 1 or 0.
2. The last digit in the array is found by arr[n-1] - [n-2] not
arr[i+1]-arr[i]
3. Recursion here creates unnecessary overheard
How many times are you calling abc? Draw the recursion tree.
On Tue, Apr 9, 2013 at 11:29
@Rashmi: I did not get your approach. I do not get emails from the group
just in case you have posted a solution :( What do you mean by keeping a
count? Also, are you using a hashmap? If yes, whats ur K,V?
#Pralay
On Tue, Feb 12, 2013 at 10:00 AM, rashmi i rash...@gmail.com wrote:
Hey Pralay,
Guys,
Why can't we simply use a hashset for both the questions. hash the arr[i]
and the frequencies in the hash map in the first pass. Then iterate over
the array to find the first arr[i] whose freq is 1 in the hash map. For
second part, keep a count and find the kth element in the array whose
first problem can be solved using a fixed sized array if we know the
range of numbers or a hash table with an appropriate hash function and
it is O(N) for time and space as some solutions above already
mentioned this.
for the second problem, I don't think heap is the right data structure
which is
One solution for the 2nd question can be LinkedHashMap (linked list +
hashmap) .
Store the integer in linked list in the order of occurrence in stream and
make an entry in hashmap on first occurence. Delete the integer entry from
linked list on 2nd occurence and replace the reference with some
Hi All,
I need ur help in solving few questions.
Would you please help me out *BY GIVING THE ALGORITHM AND THE LOGIC BEHIND
IT and it's DEEP EXPLANATION IF POSSIBLE?*
*
*
*a. Kadane’s Algo.*
*
*
*b. Linked-list intersection point.*
*
[A tree with only parent pointer, how to find LCA?]
*
*
Hey Pralay,
Sorry, if I have missed any point.Why would we need to map the
frequencies when the second problem can be solved by simply keeping a count
and comparing the index values that have been already mapped.
On Fri, Feb 8, 2013 at 11:19 AM, sourabh jain wsour...@gmail.com wrote:
One
was trying to do something clever. Knew it couldnt be that simple. Missed
some cases. I still feel with some hacks and handling some special cases
this approach would work.
But considering it still takes O(n) time I am not thrilled. I am still ok
with my algo taking Space for time. But probably
I guess the part 1 can be solved in O(n) time and space both. Here is my
approach.
Assume: Array given is arr[] = {2,3,1,2,3,2,4,1,5,6}
1. Given an array, scan thru it, element by element and hash it on a
hashmap with key as the arr[i] as the key and i+1 as the index.
2. There would be two cases
Navneet,
For 2nd problem, i need a clarification, whether the Kth number is wrt
mathematical ordering of numbers or
the kth number is wrt to the order in which the number are input ?
On Wed, Feb 6, 2013 at 10:00 AM, navneet singh gaur
navneet.singhg...@gmail.com wrote:
nice algo ankit, so it
what would happen of the input is like this : 5, 5, 66, 66, 7, 1, 1, 77,7
i think in this case the moment window reaches to point (66,7,1) it will
take 7 as unique number but
that too window will not move any futhur , but 7 is not unique .
Please comment if i misunderstood ur explanation .
Also, for the part two of the question, you can simply go in for the *kth
largest positive index *in the same hashmap (described earlier for part 1).
That solves the part two of the problem :)
Hth,
*Pralay Biswas*
*MS,Comp Sci, *
*University of California, Irvine*
On Tue, Feb 5, 2013 at 8:46 PM,
for 2nd question you can make a heap with their index as a factor to
heapify them. whenever a integer in stream gets repeated you just nead to
remove it from heap and heapify it.
On Wed, Feb 6, 2013 at 10:00 AM, navneet singh gaur
navneet.singhg...@gmail.com wrote:
nice algo ankit, so it will
@sourabh : how do u find whether the element in stream gets repeated in
heap.-- O(n) time...totally its O(nk) algo ..
If we maintain max-heap with BST property on index, then it would be
O(nlogk).
On Wed, Feb 6, 2013 at 12:25 PM, sourabh jain wsour...@gmail.com wrote:
for 2nd question you can
*APPROACH 1:*
use a hashtable for both the questions ?
Insert the integers as they are given in the array into a hashtable. The
structure I am thinking is key (the integer) - [index]. Once all elements
are inserted. Run through the hashtable and select the one which has
len(value) == 1 and is
@srikar :
approach2 is wrong.
ex: [1, 5, 7, 66, 7, 1, 77]
first window [1,5,7] all are unique.oops
On Wed, Feb 6, 2013 at 11:31 PM, Srikar srikar2...@gmail.com wrote:
*APPROACH 1:*
use a hashtable for both the questions ?
Insert the integers as they are given in the array into a
in above algo primary index is no of times that value is repeated till now
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For 1:
i think you can use sorting, sort the array and keep the indices of the
numbers in the sorted list.
Now traverse the sorted list and in the sorted list you need to find the
unique number with the
minimum index which is easy to find.
Eg: Array:5 3 1 2 4 1 4
Indices: 0 1 2 3 4 5 6
1. Given a array,find a first unique integer.
2. Integers are coming as online stream,have to find a kth unique integer
till now.
For 1.
Even we cannot use sorting for solving this as if we sort it than our first
number which is non-repetitive changes.
The best I am able to do is nlogn using
I can think of an o(n^2) algo for 2n question
Make a heap formed acctoring to two indexes
1.Primary -value
2.Secondary - index
Now for each new incoming value first search in head if exist inc its index
else make a new node
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can any one give me anexample which takes worst case of this problem
On Mon, Mar 26, 2012 at 1:56 PM, Arpit Sood soodfi...@gmail.com wrote:
@ankur +1
correct algo, can be done in just one pass.
On Mon, Oct 24, 2011 at 11:03 PM, Ankur Garg ankurga...@gmail.com wrote:
I think this can be
Hi all,
There are multiple openings for SDE1 and SDE2 for Amazon hyderabad and
Bangalore location. Interested candidates please send your resume to
sahilmadaann...@gmail.com
Thanks
sahil
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consider there are N balls in a basket. 2 players play the turns
alternatively ..AT each turn,the player
can take 1 or 2 balls from the basket. the first player starts the game..
Both the players play optimally.
i) Given N,tell whether the 1st player win or loss ?
ii) If player 1
how is winning going to decided
On Sat, Jan 12, 2013 at 6:33 PM, siva sivavikne...@gmail.com wrote:
consider there are N balls in a basket. 2 players play the turns
alternatively ..AT each turn,the player
can take 1 or 2 balls from the basket. the first player starts the game..
Both the
Are we supposed to assume that every ball played is a hit? Or should we
consider a hit or a fail case??
On Sat, Jan 12, 2013 at 7:12 PM, Raunak Gupta raunak.gupt...@gmail.comwrote:
how is winning going to decided
On Sat, Jan 12, 2013 at 6:33 PM, siva sivavikne...@gmail.com wrote:
consider
The player who plays the last turn and finishes the game wins ...
I think the approach would be similar to this ..
http://www.spoj.com/problems/TWENDS/
.. @vamshi .. Can't get your question? .. what you refer to hit or fail
case in picking up a ball?..
At the end any of the 2 players can
For a given number, find the next greatest number which is just greater
than previous one and made up of same digits.
--
loop from the end of given number till you get a digit less than the
previously scanned digit.Let the index of that number be 'i' .
if index = -1,then the given number is the largest one
else do the following
1) swap the digit at the index i with the digit just greater than it in the
scanned
We have a long chain of cuboids in all the six directions (six faces). One
start node is given and one end node is given. Give a data structure to
represent this also search for the given node from start node.
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we can represent in 3-D array ..
what type of elements are those .. is there any special kind of formation
among elements for searching? we have to think about searching based on the
criteria ..
On Tue, Oct 23, 2012 at 3:34 PM, saket narayan.shiv...@gmail.com wrote:
We have a long chain of
If the requirement is only searching in 3-D .. there is a famous data
structure K-D tree.
On Tue, Oct 23, 2012 at 5:54 PM, bharat b bagana.bharatku...@gmail.comwrote:
we can represent in 3-D array ..
what type of elements are those .. is there any special kind of formation
among elements for
@atul-I think This Should work for n dimension-
complexity O(n^no.of dimesions)
:-
have N-dimension check for which Tile can contain which Tile.i.e (3,3,4)
can contain (2,3,4) .
Now
1.Take the titles which cannot contain any-other tile set no-of tile if it
is base =1;
2.now take tiles which can
You are given many slabs each with a length and a breadth. A slab i can be
put on slab j if both dimensions of i are less than that of j. In this
similar manner, you can keep on putting slabs on each other. Find the
maximum stack possible which you can create out of the given slabs
and for
Q1 Solution: . we can use doubly linked list with hash to implement all the
operation in O(1).
By keeping track of head and tail pointer we can do enqueue and dequeue in
O(1) time.
In hash we will keep track of each element present in linked list(queue).
With node value as a hash key and address
its a LIS problem.
need to think for n-dimension...
On 8/26/12, Ravi Ranjan ravi.cool2...@gmail.com wrote:
You are given many slabs each with a length and a breadth. A slab i can be
put on slab j if both dimensions of i are less than that of j. In this
similar manner, you can keep on putting
Yeah Atul is right.
Here is my solution:--
1) first rearrange the dimensions of slabs i.e. put bigger dimension in y
and smaller dimenson in x (rotate the slab)
2) then arrange all slabs in increasing order of x dimension
3) and then find the longest increasing sub-sequence based on y
@kailash : you can simply find area of each slab area=x*y ,,,store it;
then just run LIS on this area.
On 8/26/12, Kailash Bagaria kbkailashbaga...@gmail.com wrote:
Yeah Atul is right.
Here is my solution:--
1) first rearrange the dimensions of slabs i.e. put bigger dimension in y
and
@Atul007: No, because a (1,4) tile will not fit on a (2,3) tile.
Dave
On Sunday, August 26, 2012 7:45:27 AM UTC-5, atul007 wrote:
@kailash : you can simply find area of each slab area=x*y ,,,store it;
then just run LIS on this area.
On 8/26/12, Kailash Bagaria kbkailas...@gmail.com
@dave : correct..
i guess this will work :-
sort in decreasing area.
then run LIS such that for i j , length( i ) length( j ) width(
i ) width( j )
On 8/26/12, Dave dave_and_da...@juno.com wrote:
@Atul007: No, because a (1,4) tile will not fit on a (2,3) tile.
Dave
On Sunday, August
Q2) get to n/2 nodes
Reverse link list after n/2 nodes
Now check from 1st node and n/2 node for equality
Can make the checking for equality function recursive
Sent from my iPhone 4
On 26-Aug-2012, at 12:41 AM, Ashish Goel ashg...@gmail.com wrote:
Q1. Design a data structure for the following
but how to solve this problem for with n-dimension ??
On Sun, Aug 26, 2012 at 6:47 PM, atul anand atul.87fri...@gmail.com wrote:
@dave : correct..
i guess this will work :-
sort in decreasing area.
then run LIS such that for i j , length( i ) length( j ) width(
i ) width( j )
On
Q1. Design a data structure for the following operations:
I.Enqueue
II. Dequeue
III. Delete a given number(if it is present in the queue, else
do nothing)
IV. isNumberPresent
All these operations should take O(1) time.
Q2. Check if a
Hi All,
Has anyone appeared for the online test of amazon recently??
if(yes){
Please share with us :)
}
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Imagine a sequence like this: a, b, c...z, aa, ab, ac...zz, aaa, aab,
aac aax, aaz, aba, abc... (Its same as excel column names). Given an
integer (n), generate n-th string from the above sequence.
Best Regards
Ashish Goel
Think positive and find fuel in failure
+919985813081
+919966006652
Given a singly linked list with a random pointer pointing to any node(can
be even null).
Create a clone of the list.
node structure can be :
struct node {
struct node *next;
struct node *random;
Wow.. You got this question... Lucky Fellow so easy.. I remember SISO
asking such question long back.. its way below Amazon Standard...
Basically they want to make Sure U understand the question.. Implementation
is jst a copy of the content of the current SLL Node and rewiring the rest.
On Sun,
yeah... i got that... just shared with algogeeks ;)
questions were somewhat easy
On Sun, Aug 5, 2012 at 7:25 PM, Prem Krishna Chettri hprem...@gmail.comwrote:
Wow.. You got this question... Lucky Fellow so easy.. I remember SISO
asking such question long back.. its way below Amazon Standard...
Given the array of digits, you have to calculate the least positive
integer value of the expression that could *NOT *have been received by you.
The binary operators possible are *, +, -, / and brackets possible are (
and ). Note that / is an integer division i.e. 9/5 = 1.
For ex- 6 6 4 4 the
Well, I donoo the TO cost Impact on this question. If we need to get the
solution within the given interview time , we can go for the permutation
approach.. i.e permute all the operators and calculate.
But the T(O) and S(O) is seriously jeopardize.
Anyone got better ones??
On Sat, Aug 4, 2012
@Shobhit: Can you give me a few hints on implementing a BS on the 2D?
@neelpulse: That's what I said. A 2D array *might* be a probable candidate.
In your example, the first 2d satisfies the criteria...so we check it --
Not found -- Reject -- Move on to next probable candidate.
On Sat, Jul 21,
Sorry , I've tried but BS will not work here .
On Tue, Jul 24, 2012 at 9:17 PM, algo bard algo.b...@gmail.com wrote:
@Shobhit: Can you give me a few hints on implementing a BS on the 2D?
@neelpulse: That's what I said. A 2D array *might* be a probable
candidate. In your example, the first 2d
Guys i am having amazon support engg. test tonyt...90 min 27 questions
mcq...plz tell how to prepare and wats dis profyl???reply asap..and sory
for posting it in algogeeks as i need quick response.waiting for +ve
response soon...
thnx
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May be I am missing a few details. But Consider this 3D array:
{
{
{1,2},
{7,8} // First 2D array
},
{
{3,4},
{9,10}
}
}
If you search for 3 then your search in first step will give first 2D which
actually does not contain 3. As per my interpretation of
Given a Circuit (with resistors), we need to calculate the total
resistance. Input will be like AB-5ohm, BC-6ohm, BC-10ohm, BC-20 ohm, CD-5
ohm. BC has been repeated twice implying they are in parallel. Write a
program by implementing efficient data structure for storing and
calculating the
How will you search an element in sorted 3D Array ? ( Sorted in all the 3
directions )
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Compare the element with the first([0][0]) and the last element([n-1][n-1])
of each 2D array to pin down the 2D array it *might* be present in.
After that you can follow this approach :
http://www.geeksforgeeks.org/archives/11337
If it's not present in that 2D, move on and search for the next
@algo bard : Why dont you use binary search in your first step (while
comparing first and last element of 2d array)
On Fri, Jul 20, 2012 at 4:25 PM, algo bard algo.b...@gmail.com wrote:
Compare the element with the first([0][0]) and the last
element([n-1][n-1]) of each 2D array to pin down the
Given Preorder and postorder traversals of a tree. Device an algorithm to
constuct a fully binary tree from these traversals.
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Given an array of integers where some numbers repeat once, some numbers
repeat twice and only one number repeats thrice, how do you find the number
that gets repeated 3 times?
Does this problem have an O(n) time and O(1) space solution?
No hashmaps please!
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@ashgoel - Could you please explain what exactly are you doing here ?
On Wednesday, 4 July 2012 16:16:38 UTC+5:30, ashgoel wrote:
Q4
vectorString prefix;
prefix[0]=NULL;
prefixCount =1;
for (int i=0;in;i++)
for (int j=0;jn;j++)
for (int k=0; kprefixCount;k++)
{
if
Q1) Given a newspaper and a set of ‘l’ words, give an efficient algorithm
to find the ‘l’ words in the newspaper.
Q2) Find the next higher number in set of permutations of a given number.
Q3) Given preorder of a BST, find if each non-leaf node has just one child
or not. To be done in linear
1. inverted hasp map
2. not clear
3. VLR, how do you identify end of L and start of R, question incomplete
4. One problem: consider
...
a b...
c d...
...
if ab is a prefix, can aba be another prefix, i would assume so. But if
that is true, i am not sure if this program will come to an
Q5 is sorting problem
Best Regards
Ashish Goel
Think positive and find fuel in failure
+919985813081
+919966006652
On Wed, Jul 4, 2012 at 4:13 PM, Ashish Goel ashg...@gmail.com wrote:
1. inverted hasp map
2. not clear
3. VLR, how do you identify end of L and start of R, question incomplete
Q4
vectorString prefix;
prefix[0]=NULL;
prefixCount =1;
for (int i=0;in;i++)
for (int j=0;jn;j++)
for (int k=0; kprefixCount;k++)
{
if (visited[i][j]) continue;
visited[i][j] = true;
String s=prefix[k]+a[i][j];
if (isWord(s) { printWord(s);
Consider trees:
1 3
2 3 2
1
pre order traversal is same still (1 , 2 ,3) even though ist tree doesn't
satisfy the criteria . So I don't think it can be
1. For first question trie of given word will be best option.
Space complexity O(total length of words) (worst case)
Time complexity O(T) . T length of input text (Newspaper)
2. consider it to be a 4 digit number ABCD . Find maximum Most
significant digit say it is C , and out of these
For question 5.
Even this doesn't seems right
Consider this scenario
Match b/w Winner
a vs b a
a vs c c
b vs c b
What will be order ?? acb or bca
On Wed, Jul 4, 2012 at 5:38 PM, Bhupendra Dubey
1
2 3 is not a BST and its pre-order traversal is 1 2 3, pre
order of other is 3 2 1 .
On Wed, Jul 4, 2012 at 5:17 PM, Bhupendra Dubey bhupendra@gmail.comwrote:
Consider trees:
1 3
2 3
Sorry i typed wrongly
tree is
2
1 3
preorder traversal is 123 and same for other tree as well. Please check !
On Wed, Jul 4, 2012 at 5:24 PM, a g ag20071...@gmail.com wrote:
1
2 3 is not a BST and its pre-order traversal is 1 2 3, pre
order of
please provide some good data structure to solve this problem:
http://www.careercup.com/question?id=14062676
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On 6/25/12, Navin Kumar algorithm.i...@gmail.com wrote:
please provide some good data structure to solve this problem:
http://www.careercup.com/question?id=14062676
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Hi,
*There are two arrays of length 100 each. Each of these has initially n
(n=100)
elements. First array contains names and the second array contains numbers
such that ith name in array1 corresponds to ith number in array2.
Write a program which asks the user to enter a name, finds it in
example pls...
On Thu, Jun 14, 2012 at 1:01 PM, Mohit Rathi mohit08...@iiitd.ac.in wrote:
Hi,
*There are two arrays of length 100 each. Each of these has initially n
(n=100)
elements. First array contains names and the second array contains numbers
such that ith name in array1 corresponds
arr1 = [abc,xyz,lmn,def]
arr2 = [3,6,2,8]
if user enters xyz then 6 will be printed
else
if xyz doesn't exist in arr1 then ask for a number and add them in
respective arrays(name in arr1 and number in arr2).
Hope it helps
On Thu, Jun 14, 2012 at 3:58 PM, utsav sharma
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