Hashmapping O(n)
On 25-Dec-2011 2:15 AM, sravanreddy001 sravanreddy...@gmail.com wrote:
any better approach than O(N log N) time?
maintain a heap of nodes value, count
for each element, if already present increase the count. Else add the
elements.
Max-Heap -- fetch the node, print it
What about the case of the numbers having same frequency?
Which one should come first?
On Dec 24, 11:18 pm, atul anand atul.87fri...@gmail.com wrote:
first sort the given array , you will get
1,1,1,1,2,2,2,3,3,3,3,3,4,4,5
now count frequency of each number and insert it into min heap.
each
@Atul..your solution is correct and would do the job but its complexity wud
be nlogn .
Any better way of solving it ?
Regards
Ankur
On Sun, Dec 25, 2011 at 2:10 AM, sravanreddy001 sravanreddy...@gmail.comwrote:
any better approach than O(N log N) time?
maintain a heap of nodes value, count
AND, for two points to be close, they need not be on the same quadrant
On Dec 21, 10:55 pm, atul anand atul.87fri...@gmail.com wrote:
to find all points which lies on the same quadrant for a specific node(say
1) , we have to check all nodes...rite??
we have to find difference b/w 2 node(frome
fpr the nos with same frequency , the one having lower value shud come first
For ex
3,1,1,3,1,3,2
Here 1 should come first
so
2,1,1,1,3,3,3
On Sun, Dec 25, 2011 at 11:18 AM, top coder topcode...@gmail.com wrote:
What about the case of the numbers having same frequency?
Which one should
When two persons have same name and we use one hashtable(HT), what happens
inside HT is entirely depends on HT implementation. In this case, HT may
chain the values when keys match. When querying the HT for this key, HT
returns all the values chained under this key and developer has to iterate
and
@Gene,
The algorithm which you have given below is O(n^2). I tried too and could
not figure out anything better than this unless we go for data structures
like kd-trees, as you already suggested.
Is it really the case that w/o using kd-tree, there is no better algorithm?
On Fri, Dec 23, 2011 at
refer stanford page
1,2,3,4,5,6
will become
5,3,1
6,4,2
void MoveNode(struct node** destRef, struct node** sourceRef) {
struct node* newNode = *sourceRef; // the front source node
assert(newNode != NULL);
*sourceRef = newNode-next; // Advance the source pointer
newNode-next = *destRef; // Link
@ankur : if i am able to figure out a better solution , i will post.but i
guess nlogn is the best we can get.
On Sun, Dec 25, 2011 at 4:50 PM, Ankur Garg ankurga...@gmail.com wrote:
@Atul..your solution is correct and would do the job but its complexity
wud be nlogn .
Any better way of
@atul:
I dont think you should make this assumption entering at should list down
all the phone numbers starting with at. It should be rather taken as at
can occur anywhere in the name (beginning, mid or end). This is how it
works in Android(I have android based phone and I checked it personally).
@ashish : please provide link for that page
On Sun, Dec 25, 2011 at 10:36 PM, Ashish Goel ashg...@gmail.com wrote:
refer stanford page
1,2,3,4,5,6
will become
5,3,1
6,4,2
void MoveNode(struct node** destRef, struct node** sourceRef) {
struct node* newNode = *sourceRef; // the front
@ashish : acc to Karthikeyan question ...output should not be in reverse
order.
there is not much difference b/w both implementation.
it will become little interesting if you try to solve using recursion .
On Sun, Dec 25, 2011 at 10:36 PM, Ashish Goel ashg...@gmail.com wrote:
refer stanford
LSD Radix sort can be used:
- First, it will sort the no. This will take O(k.n) (k: max no of digits in
a number; n: no of integers)
- Second, since the no are now sorted, scan all the integers and get their
frequency count. O(n)
- Use radix sort again to sort the frequency count in decreasing
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