Find the distance between each of the points and the origin(0,0) and sort
the points based on this distance.
Also, divide the points based on which quadrant they belong. If the
difference between their distance(from origin) between 2 points is less and
they belong to the same quadrant, then they
Yup, it could be O(n^2) in the worst case.
On Wed, Dec 21, 2011 at 1:59 PM, Carol Smith carol.interv...@gmail.comwrote:
@Algoose, in worst case, this is still O(n^2), ain't it?
On Wed, Dec 21, 2011 at 12:50 PM, Algoose chase harishp...@gmail.comwrote:
Find the distance between each
n = x%2 ?
x can be any integer.
On Fri, Dec 2, 2011 at 5:19 PM, Don dondod...@gmail.com wrote:
(!x || !(x^1))
!(x1)
!((x|1)-1)
(x*x)==x
(x==(x==x))||(x==(x!=x))
etc.
On Nov 29, 9:07 pm, Nitin Garg nitin.garg.i...@gmail.com wrote:
*What are the different ways to say, the value of x
Reverse the 2nd part of the Array so that they are sorted in descending
order.
Then apply bitonic sort
On Wed, Jun 22, 2011 at 2:34 PM, ross jagadish1...@gmail.com wrote:
@himanshu: I dont think, the approach works, in present form.
in place merge sort or insertion sort is fine.
Test with,
Will this work ?
Order by choosing the last element of the permutation first.
initially Calculate T = Total time of all tasks
and calculate for all i, (T-Ti)*Ci and choose the task with min among them
as last.
To find the next last element , re-calculate T = T-Ti(i being the element
chosen in
1. Insert the node(order of insertion is irrelevant) in any order according
to the binary search tree properties.
2. Compare the j value of node with its parent recursively (bottom up) and
then perform rotations to restore the Heap property.
On Thu, Jun 9, 2011 at 12:38 AM, mukesh tiwari
, Mar 9, 2011 at 5:10 PM, Algoose chase harishp...@gmail.comwrote:
Hi,
Any solution other than brute force(exponential growth) for this problem ?
On Sun, Mar 6, 2011 at 6:42 PM, UTKARSH SRIVASTAV
usrivastav...@gmail.com wrote:
can anyone please tell me why i am getting wrong answer
Hi,
Any solution other than brute force(exponential growth) for this problem ?
On Sun, Mar 6, 2011 at 6:42 PM, UTKARSH SRIVASTAV
usrivastav...@gmail.comwrote:
can anyone please tell me why i am getting wrong answer for
problem.https://www.spoj.pl/problems/TRT/
.
.
.
MY CODE IS THIS
@Jammy
in case of
1X
Since X is the start of a character, thus the preceding byte couldn't the
first byte of the 2-byte character.. So the MSb of the byte preceding X must
be a dont care. So in that case shouldn't we delete 2 bytes preceding X ?
On Thu, Feb 17, 2011 at 12:20 AM, bittu
I think we can build a n-ary tree from the n directory paths that are
already available in the computer and then for each of the m directory paths
we can traverse the tree up until the directory which is already available
in the tree and then count the remaining directories in the path.
1
of n lesser the error proneness.
On Thu, Feb 3, 2011 at 9:51 PM, Srikar srikar2...@gmail.com wrote:
@algoose I see what you are saying. what do you propose? checking out your
link now...
On Thu, Feb 3, 2011 at 11:44 AM, Algoose chase harishp...@gmail.comwrote:
@Srikar
In your first
@Srikar
In your first approach you cant simply ignore the queries that are not
present in the heap because you have a stream of queries and you never know
if the query that you are about to ignore is going be received frequently or
not in future. Your approach is like find the top 100 queries
The DP solution to this problem is very similar DP solution for counting the
number of Dyck words with some additional conditions.
while calculating DP[i][j] you need to check if i+j equals one from the
list of k values. if yes copy the value from the prev row(i.e DP[i-1][j])
instead of
@ritu
how would you find a successor without extra space if you dont have a parent
pointer ?
for Instance from the right most node of left subtree to the parent of left
subtree(root) ?
@Juver++
Internal stack does count as extra space !!
On Wed, Jan 26, 2011 at 3:02 PM, ritu
26, 2011 at 3:27 PM, Algoose chase harishp...@gmail.comwrote:
@ritu
how would you find a successor without extra space if you dont have a
parent pointer ?
for Instance from the right most node of left subtree to the parent of
left subtree(root) ?
@Juver++
Internal stack does count as extra
If we shouldn't use recursion at it uses internal stack, then I hope we can
use Morris tree traversal and use a counter to find the 5th max.
On Fri, Jan 21, 2011 at 11:19 PM, juver++ avpostni...@gmail.com wrote:
@above yes, posted solution needs parent links.
Another solution is to process
To add to what nishaanth mentioned, I think we should also track all the
state transitions so that we can back track and make alternate transitions
if the path that was already taken was later found to be incorrect one which
will not help to reach the given target (with the given set of words).
hope this works :
#includestdio.h
#define MAX(A,B) AB?A:B
#defineMIN(A,B) AB?A:B
int FindMaxA(int n)
{
int i,k,factor,max = 0,cur,prev;
int* arr = new int[n+1];
int p = MIN(n,4);
for( int j = 1;j = p;j++)
arr[j] = j;
for(i=5;i=n;i++)
{
k = i-4;
@avpostnikov
In my reply I meant TCHAR ( maybe it doesnt apply to the example given in
the problem)
when your project is being compiled as Unicode, the TCHAR would translate to
wchar_t. If it is being compiled as ANSI/MBCS, it would translated to char
Not always we explicitly use wchar_t.
On
Pipeline : Choice B : 165 ( this pipeline wastes 20 ns between last 2 stages
for each item )
Scheduling : Choice B
Ram : Choice B
Synchronization : Choice B
On Mon, Jan 17, 2011 at 10:01 AM, Bhavesh agrawal agr.bhav...@gmail.comwrote:
On Mon, Jan 17, 2011 at 10:00 AM, Bhavesh agrawal
Apart from that ,
In Unicode application each char would be 2 bytes in length and its always
advisable to use malloc(sizeof(char) * 25) which seamlessly works fine in
ASCII application as well.
On Wed, Jan 12, 2011 at 7:20 AM, Kiran K kira...@gmail.com wrote:
p= (char*)malloc(25);
KK: p
@Juver++
I am not sure if your input represents a path as asked in the problem.
We typically think of a path within a binary tree as a downward path(from
root to a leaf) thats not spread across different branches.
Of course, if you consider that example as a valid case, then DFS wont work
!
On
Will this work ?
Find the node x, then the node y within the sub tree rooted at x and then z
within the sub tree rooted at y since they must within a unique path
If any of those searches fails there are no such nodes
On Sun, Jan 9, 2011 at 6:02 AM, Gene gene.ress...@gmail.com wrote:
The
Brute force :
Have 2 pointers one pointing to last character and other pointing to the
first occurrence of last character. compare the chars at the corresponding
positions and decrement both pointers. when the latter hits -1 increment the
counter and reset it to its original value. if the
To insert a node into a tree with such a property:
First insert the node into the tree using the rules of Binary Search tree
based on Value i .
Now compare Node-j and Node-Parent-j. Depending upon the result of
comparison perform left rotation or right rotation so that the Heap property
is also
I believe the space complexity should be O(n) since you need to store the
cumulative sum corresponding to each of the elements from the input starting
from the first element.
On Wed, Dec 1, 2010 at 12:04 AM, Prims topcode...@gmail.com wrote:
I got the solution to use a hash table storing
://tech-queries.blogspot.com/
On Thu, Nov 25, 2010 at 11:18 AM, Algoose chase harishp...@gmail.comwrote:
For this specific case since only 2 operators are used : + , * and we
know that * is the operator that maximizes the value(provided both the
operands are not equal to one / none
I am not sure if this what you are looking for.
Assuming that the arrays are sorted in ascending order.
Choose one of the 2 arrays as a Reference Array.
for each element element in reference array, do a binary search to find all
elements that are less than the current element in reference and
The Solution is pretty straight forward when you long number is represented
in reverse order in linked list.
If the number is not in reverse order, We need an Explicit stack or we must
Use Recursion .
Other way around this is to construct another parallel linked list along
with Sum(linked list)
@ Anand, No , It doesnt
Try with String2 = LH
String1 = HELLO.
I think LCS solves a different problem from the one being asked here.
I think we must generate all possible combination of strings and for each
combination , check if all chars from str2 is present in it.
On Sun, Aug 1, 2010 at
Add Each number from the stream to a Doubly linked list in sorted fashion
i = 1;
j = 1;
while( stream not empty)
{
if( i == 1 j == 1)
{
Median = Cur ; /*Create New list with ist Number*/
}
Add_Node_In_Sorted_Order(Cur);
If( If new node is added
@jalaj
TRY
A:16, 12, 10, 6 ,2
B:11, 10,7, 2, 1
num: 26
On Sat, Jul 24, 2010 at 5:13 AM, jalaj jaiswal jalaj.jaiswa...@gmail.comwrote:
Take two pointers both at the start of each array...
i=0,j=0
let the size of sorted arrays be m and n
int func(int num,int m,int n){
int i=0,j=0;
can we sort the boxes based on their volume in descending order
and start highest volume box to lowest volume box (outer loop)
-Inner loop start from lowest volume box to highest volume box upto
the box considered in outer loop.
Running time : O(n^2)
On Tue, Jul 20, 2010 at 8:28 PM,
Given the collection of nodes that constitute the diameter of the tree, The
node with Max height among them i.e the node thats closest to the root(top
level node) need not necessarily have children with equal height.
For Instance:
If the the top_level_node-left-height
@ Amir:
I think you cannot find two numbers in B and C such that their sum is -A[k]
in O(n) in all cases with the logic that you mentioned.
for instance you can try with :
A : 2 7 8 10
B :-2, 0, 1, 9
C: -7 -2 1 3
On Wed, Jun 16, 2010 at 5:56 AM, Amir hossein Shahriari
The Definition of isomorphic trees given in the first post is incomplete
We say two rooted unordered trees are isomorphic if
1. a tree with a single node (the root) is only isomorphic to a tree with a
single node;
2. two trees with roots A and B, none of which is a single-node tree, are
isomorphic
For multiple ordered lists you can build a single Max heap out of elements
from all this list and Process as its done in heapsort
On Wed, Jun 9, 2010 at 9:14 PM, Rohit Saraf rohit.kumar.sa...@gmail.comwrote:
I had answered this question(of multiple lists) 2 or three days back.
Go into the
Hi ,
To add to your logic, I hope we must also be checking at the precedence of
the first operator that appears after the closing parenthesis ')' before we
can decided if the parenthesis can be removed or not .
On Thu, Jun 3, 2010 at 11:37 PM, Antony Vincent Pandian.S.
sant...@gmail.com wrote:
Will this work ?
consider A+(B*C)
have an operator stack to hold the operators. As we scan elements from left
to right,push the operators in operator stack.
when you encounter a '(' , then scan to find the first operator that comes
after '(' (in this case *).
If this operator has a higher
Thats right !!!
On Thu, Jun 3, 2010 at 6:08 PM, Rohit Saraf rohit.kumar.sa...@gmail.comwrote:
So there is a prob algoose A*(B*C) and a*(b*c+d)
i hope you understood
--
--
Rohit Saraf
Second Year Undergraduate,
Dept. of Computer
Hi ,
We can do Breadth first search but without any additional Memory like Queue.
Since we connect the siblings we can traverse through siblings.
Going from Top to bottom, Each Internal node(non-leaf) must connect its
children. If that internal node has a right sibling then connect the right
most
...@gmail.com wrote:
@Algoose Chase
point taken
Thanks
Mohit Ranjan
Samsung India Software Operations.
On Sat, May 1, 2010 at 8:43 PM, Algoose Chase harishp...@gmail.comwrote:
@mohit
The idea of DP is fine.
When you find the Max i dont think you need to include A[i+1]+B[j+1]
because
OOPs.. sorry
this doesnt work !
On Sun, May 2, 2010 at 6:11 PM, Algoose Chase harishp...@gmail.com wrote:
Hi
will this work ?
since we need Set S with n pairs of largest values , any such pair within
the set would (always) contain A[0] or B[0] because they maximize the value
of the pair
@mohit
The idea of DP is fine.
When you find the Max i dont think you need to include A[i+1]+B[j+1] because
it can never be greater than both A[i+1]+B[j] and A[i]+B[j+1] since both the
lists are sorted in decreasing order.
On Fri, Apr 30, 2010 at 8:47 PM, mohit ranjan
Hi,
How do you define without extra space ?
Doing any order traversal either using recursion or using iteration is going
to take extra space .
If you are given a binary tree represented by pointers that points to
children nodes is it possible to do a heap sort without an array ?
On Thu, Apr 29,
;
root-data = temp;
BinarytoBST(NodeR);
}
On Thu, Apr 29, 2010 at 11:23 AM, Algoose Chase harishp...@gmail.comwrote:
Hi,
How do you define without extra space ?
Doing any order traversal either using recursion or using
I have a logical question about the solution pasted.
The Problem says given N*N numbers filled in a matrix , print the numbers in
spiral .
What the code does is it fills the N*N numbers in spiral in decreasing order
and then prints the matrix contents left to right , top to bottom. Will the
two
Hi all,
How about this ?
set K to the first element i.e Array[0] and set j to the last element i.e
Array[size-1].
Decrement the index j until you find Array[j] Array[j+1] and increment the
index k until you find Array[k] Array[k-1] and do this until you find the
conditions met.
Does it solve
Hi,
@Asish meena and Arun :
I dont think you can simply append a whole tree( BST2) at some
position just because the root of the BST2 is at its correct position.For
instance , Lets say you append BST2's Root anywhere within the left subtree
of BST1's Root. But if the right most leaf node
:03 AM, Algoose Chase harishp...@gmail.comwrote:
Hope you meant a pattern is sub-array containing 2 or more UNIQUE chars.
hope based on dfn, abcd is also a pattern in the input you have given.
On Tue, Feb 2, 2010 at 1:11 AM, ankit mahendru
ankit.mahend...@gmail.com wrote:
Q. Find all
Hope you meant a pattern is sub-array containing 2 or more UNIQUE chars.
hope based on dfn, abcd is also a pattern in the input you have given.
On Tue, Feb 2, 2010 at 1:11 AM, ankit mahendru ankit.mahend...@gmail.comwrote:
Q. Find all the patterns once which are present in the character array
PM, Algoose Chase
harishp...@gmail.com wrote:
Condition:
Can we do it keeping the original lists intact ? ie without
reversing it.
I mean , No recursion no Reversing ... is it possible ?
@kumar :
15234 is represented as 1-5-2-3-4
On Wed, Jan 27, 2010 at 4:09 PM, saurabh gupta
conditions:
NO extra memory (@ stack or Heap) at all. No recursion.
Any body has got any hint about how to get this done ?
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To
void PrintFamilyTree(const short generation)
{
printf(Name : %s Generation = %d\n, m_Name.c_str(),generation);
vectorHuman*::iterator it;
for (it = this.begin(); it this.end() ;it++)
{
it-PrintFamilyTree(generation+1);
}
}
On Tue, Nov 24, 2009 at 9:20 PM, Aditya Shankar
I posted this long b4 but dint see this error : Delivery to the following
recipient failed permanently: algogeeks@googlegroups.com
re-posting:
BST_Spiral(struct node* root)
{
ht = Height(root);
for( i = 0; i= ht; i++)
{
PrintSpiral(root, i, i%2 /*flip 1 and 0 alternately on each
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