void findPaths(int x, int y, int depth)
{
if (isEnd(x,y)) showSolution(); // One path will be marked by
letters A,B,C...
else
{
maze[x][y] = 'A' + depth;
if (x (maze[x-1][y]=='1')) findPaths(x-1,y,depth+1);
if (y (maze[x][y-1]=='1')) findPaths(x,y-1,depth+1);
BFS
On Sun, Oct 14, 2012 at 4:21 PM, Rahul Kumar Patle
patlerahulku...@gmail.com wrote:
response awaited!!!
anyone??
On Sat, Oct 13, 2012 at 12:31 AM, Rahul Kumar Patle
patlerahulku...@gmail.com wrote:
Pls help to solve this que.. does any one have DP solution for following
que.
@jaspreet : i dont find much difference between using BFS or
backtracking...which is doing similar to DFS.
On Tue, Oct 16, 2012 at 4:28 AM, Jaspreet Singh
jassajjassaj...@gmail.comwrote:
BFS
On Sun, Oct 14, 2012 at 4:21 PM, Rahul Kumar Patle
patlerahulku...@gmail.com wrote:
response
response awaited!!!
anyone??
On Sat, Oct 13, 2012 at 12:31 AM, Rahul Kumar Patle
patlerahulku...@gmail.com wrote:
Pls help to solve this que.. does any one have DP solution for following
que.
http://www.geeksforgeeks.org/archives/24488
section 5/question 2
Write a program to find all the
here are the steps :
1) Construct a temporary array left[] such that left[i] contains product of
all elements on left of A[i] excluding A[i].
2) Construct another temporary array right[] such that right[i] contains
product of all elements on on right of A[i] excluding A[i].
3) To get OUT[],
well we can do with just one array. Overwrite the answer directly on left[]
array.
On Thu, Aug 16, 2012 at 6:47 PM, mohit mohitsingh1...@gmail.com wrote:
here are the steps :
1) Construct a temporary array left[] such that left[i] contains product
of all elements on left of A[i] excluding
yeah we can do it without using an extra array too. first calculating the
product of elements on its left and putting in OUT[] and then multiplying
with the product of elements on its right . no auxiliary space used.
On Thursday, August 16, 2012 2:26:58 PM UTC+5:30, ram wrote:
Hi,
We have to consider cases when an element is zero.
On Thu, Aug 16, 2012 at 7:07 PM, shady sinv...@gmail.com wrote:
well we can do with just one array. Overwrite the answer directly on
left[] array.
On Thu, Aug 16, 2012 at 6:47 PM, mohit mohitsingh1...@gmail.com wrote:
here are the steps
@Navin: Why? No division is used.
Dave
On Thursday, August 16, 2012 9:20:03 AM UTC-5, Navin Kumar wrote:
We have to consider cases when an element is zero.
On Thu, Aug 16, 2012 at 7:07 PM, shady sin...@gmail.com javascript:wrote:
well we can do with just one array. Overwrite the answer
http://www.geeksforgeeks.org/archives/17629
On 2 August 2012 19:50, Daksh Talwar dakshtal...@gmail.com wrote:
When asked , try to make the most balanced one .
otherwise there are many possible BSTs for a given array/linked list.
On Thu, Aug 2, 2012 at 5:05 PM, Umer Farooq
When asked , try to make the most balanced one .
otherwise there are many possible BSTs for a given array/linked list.
On Thu, Aug 2, 2012 at 5:05 PM, Umer Farooq the.um...@gmail.com wrote:
A LinkedList is by itself is a BST such that one child node of each node
is null. Do we need a simple
A LinkedList is by itself is a BST such that one child node of each node is
null. Do we need a simple BST or height balanced BST?
On Tue, Jul 31, 2012 at 2:39 PM, Ashish Goel ashg...@gmail.com wrote:
how would you do convert sorted doubly linked list to bst using same
nodes as in DLL
Best
Ishan,
i am assuming that the list to BST should give a inorder traversal, and the
logic of yours does not seem to give a right solution.
try two different trees with 7 nodes, convert into LL and then back to BST,
the answer is not same as the trees that we start with.
Best Regards
Ashish Goel
can you give the link within geeksforgeeks please
Best Regards
Ashish Goel
Think positive and find fuel in failure
+919985813081
+919966006652
On Tue, Jul 31, 2012 at 4:13 PM, a g ag20071...@gmail.com wrote:
check on geeksforgeeks.org
On Tue, Jul 31, 2012 at 3:09 PM, Ashish Goel
how would you do convert sorted doubly linked list to bst using same nodes
as in DLL
Best Regards
Ashish Goel
Think positive and find fuel in failure
+919985813081
+919966006652
On Sun, Jul 29, 2012 at 10:30 PM, Purani Sekar nagapur...@gmail.com wrote:
convert sorted doubly linked list to bst
check on geeksforgeeks.org
On Tue, Jul 31, 2012 at 3:09 PM, Ashish Goel ashg...@gmail.com wrote:
how would you do convert sorted doubly linked list to bst using same
nodes as in DLL
Best Regards
Ashish Goel
Think positive and find fuel in failure
+919985813081
+919966006652
On Sun, Jul
1. get the middle of the linked list and make it root
2. same for left half and right half recursivly
a. get the middle of left half and make it left child.
b. get middle of rite half and make it rite child.
this is must b he logic for the qstn. :)
Thank You,
Ishan Sood.
I think they asked same questions for intern and full time. The second
round questions were:
1. given a string , remove the duplicates and print it in ascending order
eg : accommodate
op: acdemot
2. given a sorted array with a few numbers in between reversed. fix the
array
eg : 1 2 3 4 9 8 7
Also if smeone cn post some questions/experience for microsoft
intern/placementit will be a grt help
Thanks
Tanuj Makkar
Delhi College Of Engineering
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No It will not.
For negative number, it will fail.
On Sun, Jul 15, 2012 at 11:45 PM, Tanuj Makkar
tanujmakkar.de...@gmail.comwrote:
double round(double num)
{ return (int)(num+0.5)
}
will it work all the time?
..
didnt get itcan anyone explain it.thnx in advance.
thnx amit.jst asked a stupid quesiton:)
On Mon, Jul 16, 2012 at 11:55 AM, Amit Jain aj201...@gmail.com wrote:
No It will not.
For negative number, it will fail.
On Sun, Jul 15, 2012 at 11:45 PM, Tanuj Makkar
tanujmakkar.de...@gmail.com wrote:
double round(double num)
{ return
double round(double num)
{ return (int)(num+0.5)
}
will it work all the time?
..
didnt get itcan anyone explain it.thnx in advance.
On Friday, 26 August 2011 21:58:05 UTC+5:30, rahul sharma wrote:
hi guys...microsoft is coming to our campus..plz nyone tell their
@Atul
thanx for the code its working for the example you took...Please check
the same for i/p abcmno,abcmnop
Algo displays:- mno
It should display mno
mnop...
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@All *Here is the working code: test on input {-1,5,3,-8,4,-6,9} and
{1,-1,2}*
Algo:
increment current till first +ve number(p) and decerement end till last
-ve number(n)
now consider only array between [p..n]
If current is negetive, increment current
If current is positive, swap it with end and
@Dave :- a minor change
Initially, decrement the end pointer till it points to positive number,
Now end points to the last negative number.
Now,
If current is negative , increment current
If current is positive , swap it with the element at end and decrement
current and end both
If current =
@Navin: If I am correctly executing your algorithm on the data in the
original posting, {-1,5,3,-8,4,-6,9}, I get {-1,-6,-8,4,3,5,9}, when the
correct answer is {-1,-8,-6,5,3,4,9}. The array contains the correct
numbers, but the order of the positive numbers and the order of the negtive
+1 naveen
On Thursday, June 28, 2012 8:29:26 PM UTC+5:30, Navin Gupta wrote:
Keep two pointers - one at start of the array and other at end of the
array
Now current points to start of the array
If current is negative , increment current
If current is positive , swap it with the element
@Navin: Try this with {1,-1,2}. current points to the 1 and end points to
the 2. Since 1 is positive, the algorithm swaps the 1 and the 2, giving
{2,-1,1}. Then it decrements current to point outside the array and end to
point to the -1. How can this be right?
Dave
On Thursday, June 28, 2012
keep swaping left most -ve and left most positive untill counter reaches at
the end of array, can be done in o(n) no extra space required..
3rd year
manit bhopal
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Keep two pointers - one at start of the array and other at end of the array
Now current points to start of the array
If current is negative , increment current
If current is positive , swap it with the element at end and decrement
current and end both
If current = end , then break.
Navin
@wgp the ques is to maintain the order intact..
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refer to this link http://www.ics.uci.edu/~eppstein/161/960130.html.
or Using quicksort , it can be done in O(n).
One more way to do this is to make min heap of size-k. Then insert elements
from the original array.If element is greater than root of the heap,swap
them.In the Last, root of the heap
@akshat ur code doesn't give intact output, so i modified ur code and
here is the code :
int j=0,k=0;
for (i = 0; i n; ++i)
{
if(a[i]0)
{
a[j] = a[i];
j++;
}
else
{
temp[k] = a[i];
k++;
}
}
k=0;
for (i = j; i n; ++i)
{
guys this is my solution to the problem, it's a bit sloppy but as far as I
checked it was working please have a go at it?
#include stdio.h
#include stdlib.h
int main() {
int arr1[] = {0,-5,3,0,4,-6,-9};
int arr2[7];
int j = 0;
for ( int i = 0 ; i 7 ; i++ ) {
//loop for
can this be done in single pass in O(n) .
On Thu, Jun 21, 2012 at 8:10 PM, rusty dafc1...@gmail.com wrote:
guys this is my solution to the problem, it's a bit sloppy but as far as I
checked it was working please have a go at it?
#include stdio.h
#include stdlib.h
int main() {
Can't we use k iterations of bubble sort ?
On Jun 18, 2012 2:11 PM, Ramindar Singh ramin...@gmail.com wrote:
We can use Median of medians
http://en.wikipedia.org/wiki/Selection_algorithm#Linear_general_selection_algorithm_-_Median_of_Medians_algorithm
On Sunday, 17 June 2012 08:13:18
single traversal n O(n) are 2 diff things...plz specify???
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Well they are the same you're going over an array once. As long as they are
not nested they are still counted as O(n) because leading constants are
dropped, at least that's what my acumen says. Need inputs on this guys!
On Friday, June 22, 2012 12:53:02 AM UTC+5:30, suzi wrote:
single
i mean o(n) in single traversal .
On Fri, Jun 22, 2012 at 12:53 AM, sanjay pandey
sanjaypandey...@gmail.comwrote:
single traversal n O(n) are 2 diff things...plz specify???
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Hi,
It looks like, The interviewer is expecting MinHeap from you,
If modification to array is permitted just build the heap in place (from
end bubbleUp n-1 time) and extract K elements in sorted order
Otherwise you need minimum O(K) memory
Again if you want to use the quick-sort, I
This can be done using the concept of median of medians, wherein we can
achieve linear time complexity in the worst case.
This is a concept used in parallel algorithms too, check it out.
On Mon, Jun 18, 2012 at 5:38 PM, Prem Nagarajan prem.cmna...@gmail.comwrote:
@enchantress : This problem can
We can use Median of medians
http://en.wikipedia.org/wiki/Selection_algorithm#Linear_general_selection_algorithm_-_Median_of_Medians_algorithm
On Sunday, 17 June 2012 08:13:18 UTC+5:30, Prem Nagarajan wrote:
Give an array of unsorted elements, find the kth smallest element in the
array.
Hi all,
A variation of selection sort can be used which takes O(nk) time.
for i 1 to k
min_index = i
for j i+1 to n
if a[j] a[min_index]
min_index = j
swap(a[i],a[min_index])
required element : a[k]
On Sunday, 17 June 2012 08:13:18 UTC+5:30, Prem Nagarajan wrote:
Give
@enchantress : This problem can be solved using quicksort in O(nlogn). No
need to go for selection sort.
But O(n) solution is needed.
On Mon, Jun 18, 2012 at 2:50 PM, enchantress elaenjoy...@gmail.com wrote:
Hi all,
A variation of selection sort can be used which takes O(nk) time.
for i 1
@ayush goel couldnt really understand your algo , can you please explain
little bit more .
On Wednesday, 13 June 2012 21:49:49 UTC+5:30, Krishna Kishore wrote:
Given a array of integers both positive and negative and you need to shift
positive numbers to one side
and negative numbers to
Queue can be defined as a priority queue where priority decreases with
time. Hence an element inserted later has lower priority and goes to back
of queue. (FIFO)
Stack can be defined as a priority queue where priority increases with
time.Hence an element inserted later has higher priority and
Root of a graph can be any node whose in-degree is zero. i.e. there are no
nodes pointing to that node.
It can be found by using O( |V| ) space and O( |E| ) time .
Now we can choose any node whose in-degree is zero if present.
or choose any random node
and itf DFS-tree is the desired tree.
@Gaurav: you are taking ia and ib as int so they will have 32 bits in
Java. So you can not set the bits for numbers in the array greater
than 32.
e.g if you have a[i]=59876 so you would want to set the 59876th bit in
ia : ia=ia | (159876) but that is not possible. How do you handle
this?
Also how
@Piyush: Did you even try this on any examples? If not, try a = {0,1,2,3}
and b = {0,2,2,2}.
Dave
On Sunday, May 20, 2012 1:39:25 AM UTC-5, Kalyan wrote:
Piyush. I think we can use your logic. But You should check the product
also.
Have 4 variables, sum_a,sum_b , prod_a, prod_b
No way u can do it in O(1) space and O(n) time.sols above are not
gonna work..yeah, it is possible in O(n) space and O(n) time.
On May 20, 12:29 am, HARSHIT PAHUJA hpahuja.mn...@gmail.com wrote:
given 2 unsorted integer arrays a and b of equal size. Determine if b is a
permutation of a.
Dave,
Cant we have a hash table with the item as key and its count as value (walk
over array A and build HT).
For permutation check, walk over second array and start reducing the count
and remove when count becomes zero for that particular key. If some char
not there in HT, return false, else
^not an O(n)
On May 21, 6:53 pm, Ashish Goel ashg...@gmail.com wrote:
Dave,
Cant we have a hash table with the item as key and its count as value (walk
over array A and build HT).
For permutation check, walk over second array and start reducing the count
and remove when count becomes zero
in space
On May 21, 6:53 pm, Ashish Goel ashg...@gmail.com wrote:
Dave,
Cant we have a hash table with the item as key and its count as value (walk
over array A and build HT).
For permutation check, walk over second array and start reducing the count
and remove when count becomes zero for
@Ashish: Using a hash table violates the O(1) space requirement given in
the original problem.
Dave
On Monday, May 21, 2012 8:53:44 AM UTC-5, ashgoel wrote:
Dave,
Cant we have a hash table with the item as key and its count as value
(walk over array A and build HT).
For permutation
constant space vs no additional space and then O(n) time complexity not
possible..
Best Regards
Ashish Goel
Think positive and find fuel in failure
+919985813081
+919966006652
On Mon, May 21, 2012 at 8:01 PM, Dave dave_and_da...@juno.com wrote:
@Ashish: Using a hash table violates the O(1)
@ashish.. it wont be constant space then.. surely it will be o(n) though
On Mon, May 21, 2012 at 7:23 PM, Ashish Goel ashg...@gmail.com wrote:
Dave,
Cant we have a hash table with the item as key and its count as value
(walk over array A and build HT).
For permutation check, walk over
a[] = [-1,-3,4,0,7,0,36,2,-3]
b[] = [0,0,6,2,-1,9,28,1,6]
b1[] = [0,7,0,36,4,-6,3,0,0]
b2[] =[-1,-3,11,0,0,0,35,0,0]
suma = 42 proda = -84*72*3
sumb = 51 prodb = -84*72*3
sumb1 = 44 prodb1 = -84*72*3
sumb2 = 42 prodb2 = 33*35
do the sum and prod operation w/o 0s and compare the values..
@Partha
try with:
A = {2, 2, 9}
B= {1, 6, 6}
On Mon, May 21, 2012 at 7:08 PM, partha sarathi Mohanty
partha.mohanty2...@gmail.com wrote:
a[] = [-1,-3,4,0,7,0,36,2,-3]
b[] = [0,0,6,2,-1,9,28,1,6]
b1[] = [0,7,0,36,4,-6,3,0,0]
b2[] =[-1,-3,11,0,0,0,35,0,0]
suma = 42 proda = -84*72*3
Hiii!! I have some idea about the solution. Please notify me if i am
wrong
a= [ 4,3,5 ] and b= [ 3,5,4 ]
diff=0;
for (i=0; in;i++)
{ diff= diff+a[i]-b[i];
}
if diff == 0
print: permutation
else
print: not permutation
On 20 May 2012 07:2 0, Dave dave_and_da...@juno.com wrote:
U are checking if the sum is same or not.. which can be same even if the
elements are different.
On Sun, May 20, 2012 at 11:54 AM, Piyush Khandelwal
piyushkhandelwal...@gmail.com wrote:
Hiii!! I have some idea about the solution. Please notify me if i am
wrong
a= [ 4,3,5 ] and b= [
@piyush :
your solution will fail for the case
a={5,1,1}
b={3,3,1}
On Sun, May 20, 2012 at 11:54 AM, Piyush Khandelwal
piyushkhandelwal...@gmail.com wrote:
Hiii!! I have some idea about the solution. Please notify me if i am
wrong
a= [ 4,3,5 ] and b= [ 3,5,4 ]
diff=0;
for (i=0;
Piyush. I think we can use your logic. But You should check the product also.
Have 4 variables, sum_a,sum_b , prod_a, prod_b
Calculate Sum and product of array 'a' and store it in sum_a,prod_a
Calculate Sum and product of array 'b' and store it in sum_b,prod_b
if sum_a=sum_b prod_a==prod_b then
@Piyush: Try this i/p 8,0,0 ; 2,6,0-- Ur algo aint adequate..
On Sat, May 19, 2012 at 11:24 PM, Piyush Khandelwal
piyushkhandelwal...@gmail.com wrote:
Hiii!! I have some idea about the solution. Please notify me if i am
wrong
a= [ 4,3,5 ] and b= [ 3,5,4 ]
diff=0;
for (i=0; in;i++)
@Harshit: These are a few unanswered questions that came to mind when I
read your solution attempt: What do you do with negative elements? What is
the -12th prime number? How do you deal with overflow in the cases where
you have a lot of large prime numbers and the product exceeds your native
@umer : did interviewer told any one of the solution provided in the given
link below or different?
http://www.geeksforgeeks.org/archives/1155
On Tue, Mar 13, 2012 at 11:31 AM, Umer Farooq the.um...@gmail.com wrote:
Yes that is exactly what they wanted. I proposed BFS for this solution.
Well, the interviewer gave a hint to use hash-table.
The key of hash-table will be memory address of original node and value
will be the memory address of the new node.
On Wed, Mar 14, 2012 at 9:43 PM, atul anand atul.87fri...@gmail.com wrote:
@umer : did interviewer told any one of the
Yes that is exactly what they wanted. I proposed BFS for this solution.
Anyway, there was another problem that I was able to solve; but the
interviewer brought up a much more efficient approach.
The problem was:
Given a linked l
On Mon, Mar 12, 2012 at 11:31 PM, Gene gene.ress...@gmail.com
Yes that is exactly what they wanted. I proposed BFS for this solution.
Anyway, there was another problem that I was able to solve; but the
interviewer brought up a much more efficient approach.
The problem was:
- Given a linked a linked list with one pointer pointing to next,
another
http://www.geeksforgeeks.org/archives/1155
On Tue, Mar 13, 2012 at 11:31 AM, Umer Farooq the.um...@gmail.com wrote:
Yes that is exactly what they wanted. I proposed BFS for this solution.
Anyway, there was another problem that I was able to solve; but the
interviewer brought up a much more
This problem is close to copying a graph. For that as you said, just
do DFS or BFS and maintain a map from original nodes to copies. Use
the copy to set pointers whenever it exists. When it doesn't exist,
make a new node and add it to the map.
You can implement the map in various ways. A hash
Copying a full graph requires a BFS or DFS. But here we have a big
advantage. We know the nodes are linked in a single chain. So we
don't need to search to find all nodes. We can just use .next
pointers.
No matter how we do things, we will need Map(A) that returns the copy
of node A if it
Sorry, a small test showed a loop quitting one iteration too soon.
Here is the fix.
import java.util.Random;
public class ListCopy {
class Node {
int val;
Node next, other;
Node() { }
Node(int val, Node next, Node other) {
this.val = val;
@Ashish
We can sort a list in another way as follows:
1) Recursively divide the list into two halves..
2) Call merge while joining the sorted lists..
MergeSort(node * p)
{
if ( p contains only one element)
return p;
p1 = MergeSort(first half of list pointed by p);
@Ashish..
I have something in mind.. but would require verification by u..
Lets say, the structure of the data node is as follows:
struct node
{
int data;
struct node *next;
};
Now, given 2 sorted linked lists we can right a O(N) time and in-place
merge process, to build a sorted merged
Priority Queue:
when popped ... returns the max priority element and if the priorities
of two or more elements are same...then they will popped as they are
inserted ..
when pushed the element : puts the element in the list according to the
priority...
For making priority queue into
On 10/29/11, praveen raj praveen0...@gmail.com wrote:
Priority Queue:
when popped ... returns the max priority element and if the priorities
of two or more elements are same...then they will popped as they are
inserted ..
when pushed the element : puts the element in the list
How about this answer:
b?z:y
int main() {
int a=0,b,y=4,z=5,k;
cinb;
k=(((b+~a+1)7)1);//k will either be 0 or 1
cout (z-int((bool)k(z-y)));
return 0;
}
On Mon, Sep 12, 2011 at 5:32 PM, beginner shubh2...@gmail.com wrote:
although multiplication operator is not allowed..
return (((-1+!x)y) | ((-1+!!x)z)) ;
On Sun, Oct 2, 2011 at 3:18 PM, ravi maggon maggonr...@gmail.com wrote:
How about this answer:
b?z:y
int main() {
int a=0,b,y=4,z=5,k;
cinb;
k=(((b+~a+1)7)1);//k will either be 0 or 1
cout (z-int((bool)k(z-y)));
return 0;
}
On
@Teja Bala
U dont need the last line for a[0][0]
else code will be wrong
conside
0 0 0 0 1
0 0 0 0 0
0 1 0 0 0
0 0 0 1 0
Regards
On Sun, Sep 11, 2011 at 11:56 PM, teja bala pawanjalsa.t...@gmail.comwrote:
//pseudo code dis will work for sure.
for(i=0;irow_count;i++)
congrates dude
On Thu, Sep 22, 2011 at 7:26 PM, Sanjay Rajpal srn...@gmail.com wrote:
Saurabh : Thank u very much :)
Sanju
:)
On Thu, Sep 22, 2011 at 6:15 AM, saurabh sah.saurab...@gmail.com wrote:
thanx to all
@sanjay I have shared my interview experience at
Congrats Saurabh.
On Fri, Sep 23, 2011 at 7:18 PM, sagar pareek sagarpar...@gmail.com wrote:
congrates dude
On Thu, Sep 22, 2011 at 7:26 PM, Sanjay Rajpal srn...@gmail.com wrote:
Saurabh : Thank u very much :)
Sanju
:)
On Thu, Sep 22, 2011 at 6:15 AM, saurabh sah.saurab...@gmail.com
@kartik:
i thnk u r searching for string...that may have characters..in the 2d matrix
..NO MATTER WHERE THEY ARE..
On Wed, Sep 21, 2011 at 7:10 PM, kartik sachan kartik.sac...@gmail.comwrote:
i think i can solve this in O(n^2)
here is code http://ideone.com/Gk69A
# includestdio.h#
thanx to all
@sanjay I have shared my interview experience at
http://msidcinterview.blogspot.com/
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@dheeraj i didn't get what u want to say explain me with the help of example
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Saurabh : Thank u very much :)
Sanju
:)
On Thu, Sep 22, 2011 at 6:15 AM, saurabh sah.saurab...@gmail.com wrote:
thanx to all
@sanjay I have shared my interview experience at
http://msidcinterview.blogspot.com/
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@kartik.,...where are u searching..that ..the next character should be
present..only around the 8 possible locations around that character
On Thu, Sep 22, 2011 at 6:34 AM, kartik sachan kartik.sac...@gmail.comwrote:
@dheeraj i didn't get what u want to say explain me with the help of
example
i think i can solve this in O(n^2)
here is code http://ideone.com/Gk69A
# includestdio.h# includestring.hchar a[100][100];int findword(int
*b,int n,int m){
int i,j,flag=0;
char s[1];
for(i=0;in;i++)
for(j=0;jm;j++)
s[a[i][j]]++;
@piyush: what is time and space complexity of u'r sol..
On Mon, Sep 19, 2011 at 11:03 AM, Piyush Grover
piyush4u.iit...@gmail.comwrote:
sry, in the findWord function all a's are different e.g
a0, a1a7
and if(!a) is actually if(a0||a1||...||a7)
thnx
piyush
On Mon, Sep 19, 2011 at
hmm, nice questions, can we create an efficient DS to query the
strings ??
tried using trie but memory prints are very large ( O(n^2) )? :-((
On Sep 18, 12:59 pm, Anup Ghatage ghat...@gmail.com wrote:
For the mentioned scenario, it seems to be the only feasible solution.
On Sun, Sep 18,
for(i = 0; i n; i++)
for(j = 0; j n; j++){
setColor(i, j) = black;
if(A[i][j] == str[0]){
setColor(i, j) = blue;
a = findWord(A, i, j, str, 1)
if(!a) setColor(i, j) = black;
else break;
}
}
findWord(A, i, j, str, k){
if(k
sry, in the findWord function all a's are different e.g
a0, a1a7
and if(!a) is actually if(a0||a1||...||a7)
thnx
piyush
On Mon, Sep 19, 2011 at 1:10 AM, Piyush Grover piyush4u.iit...@gmail.comwrote:
for(i = 0; i n; i++)
for(j = 0; j n; j++){
setColor(i, j) = black;
For Stack:
just make a structure:
struct stack_with_priorityqueue
{
int num;
int priority;
struct stack_with_priorityqueue *ptr;
}
now when we add another number just increase the priority... priority++
For Queue:
do same...just decrease priority...priority--
...
char str[10];
int length,count;
void fun(int x)
{
if(x==length)
printf(%d %s\n,++count,str);
else
{
fun(x+1);
str[x]-=32;
fun(x+1);
str[x]+=32;
}
}
int main()
{
scanf(%s,str);
length=strlen(str);
fun(0);
getch();
}
@Vivek Microsoft is open for CSE and IT department (B.Tech/M.Tech)..
The eligibility is 7 CGPA
@Rahul we don't have any information about the profile being offered,
currently just the written test is taken on 19th which would be for 1
hr, no pre placement talks nothing..
@abhinav I already own a
hey ankur..
thnaks for ur concern..
*mca *was not eligible kya..
On Fri, Sep 16, 2011 at 12:04 AM, techankur ankurmitta...@gmail.com wrote:
@Vivek Microsoft is open for CSE and IT department (B.Tech/M.Tech)..
The eligibility is 7 CGPA
@Rahul we don't have any
PEC main mca hai hi nahin :P khaali B.E and M.E hai and from this year
they have started M.Sc
Ankur
On Sep 15, 11:41 pm, vivek goel vivek.thapar2...@gmail.com wrote:
hey ankur..
thnaks for ur concern..
*mca *was not eligible kya..
On Fri, Sep 16, 2011 at 12:04
Shorter. It is assumed that the input string consists of upper and
lower case letters only.
void allCase(string r)
{
int i, n = s.sise();
for( i = 0 ; i (1n) ; ++i )
{
string[i^(i1)] ^= 'a' ^ 'A';
cout s endl;
}
}
The expression i^(i1) is a Gray-code (see
The well known examples of priority queue is minheap and maxheap..
i guess the question is how do we implement one of these(at least) using
queue?
On Wed, Sep 14, 2011 at 9:08 AM, Ankuj Gupta ankuj2...@gmail.com wrote:
I guess the functionality of priority should be maintained
On Sep 13,
@DAVE
dis was the o/p for ur prog.
aBC
abC
abC
abc
abc
abc
abc
abc
#includeiostream.h
main()
{
int i, n = 3;
char *s=ABC;
for( i = 0 ; i (1n) ; ++i )
{
s[i^(i1)] ^= 'a' ^ 'A';
cout s endl;
}
}
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wahat is the logic why 1n?
On Wed, Sep 14, 2011 at 6:44 PM, teja bala pawanjalsa.t...@gmail.comwrote:
@DAVE
dis was the o/p for ur prog.
aBC
abC
abC
abc
abc
abc
abc
abc
#includeiostream.h
main()
{
int i, n = 3;
char *s=ABC;
for( i = 0 ; i (1n) ; ++i )
{
s[i^(i1)] ^= 'a' ^
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