merge sort both lists: O(nlogn)
Now, for both lists to be identical, just compare the corresponding elements
in the lists i.e. L1(1) == L2(1), L1(2) == L2(2) ...
= O(n)
--Sundeep.
On Wed, Jun 2, 2010 at 10:47 PM, Raj N rajn...@gmail.com wrote:
@Antony: The 2 lists should have the same
construct bst from both list and check if the bst are equal by printing
their inorder traversal.
On 2 June 2010 22:47, Raj N rajn...@gmail.com wrote:
@Antony: The 2 lists should have the same elements as well the number must
be equal
On Wed, Jun 2, 2010 at 5:38 PM, Antony Vincent Pandian.S.
@ Raj,
Just to clarify, same elements refer to copy of identical data or link list
members pointing to same data ?
On Wed, Jun 2, 2010 at 10:47 PM, Raj N rajn...@gmail.com wrote:
@Antony: The 2 lists should have the same elements as well the number must
be equal
On Wed, Jun 2, 2010 at 5:38
for three stacks u can use indices 0 3 6 etc for stack1. 1 4 7 for stack 2
and 2 5 8 etc for stack 3.
now if any of the stack overflows but there is still space in array as other
stacks have few element then now u can grow ur stack in reverse direction as
shown below :
indices of array : 0
three or more stacks can be made by using linked array representation
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u have not sorted the array . first sort the array nd then apply the
approach. u ll get the ryt ans
On 1 June 2010 21:32, Jitendra Kushwaha jitendra.th...@gmail.com wrote:
Using subset sum method we can solve this. It actually DP
check out Paybill question and its solution on topcoder website
same question as wat i asked in partioning of array such that the diff is
min.
On 31 May 2010 22:07, Senthilnathan Maadasamy
senthilnathan.maadas...@gmail.com wrote:
Same question with interesting answers in stackoverflow :
1.calculte the postfix of given expression.
2.now remove a particular parenthesis from expression and check if the
postfix of this expression is equal to the postfix of original expression.
if yes then the parenthesis we have removed were extra. if no then the
parenthesis were not exta.
3 now
@divya: but still haven't you seen Jagadish's example? It is a
counterexample to your greedy approach.
--
Rohit Saraf
Second Year Undergraduate,
Dept. of Computer Science and Engineering
IIT Bombay
http://www.cse.iitb.ac.in/~rohitfeb14
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You
You can restore the infix expression from the postfix expression
calculated, only add parenthesis when necessary in each step.
On 2010-6-3 15:35, divya jain wrote:
1.calculte the postfix of given expression.
2.now remove a particular parenthesis from expression and check if the
postfix of
oh...okay...
good example...
On 3 June 2010 13:23, Rohit Saraf rohit.kumar.sa...@gmail.com wrote:
@divya: but still haven't you seen Jagadish's example? It is a
counterexample to your greedy approach.
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Rohit Saraf
Second Year
@Divya: nope, your Q requires equal count too whereas this doesn't.
On Thu, Jun 3, 2010 at 1:06 PM, divya jain sweetdivya@gmail.com wrote:
same question as wat i asked in partioning of array such that the diff is
min.
On 31 May 2010 22:07, Senthilnathan Maadasamy
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
The question is that you have to print all the valid permutations of
the given number of brackets
for example for input 3 we have the output
1 ((()))
2 (()())
3 (())()
4 ()(())
5 ()()()
total five valid permutation
this can be solved in O( 2^(2n) ) where n is number of brackets .
Algo will be
@Raj N : You are right overflow is top2-(top1+1)==0
@Anand : The oveflow condition is dependent how underflow condition are
implemented. Considering underflow above condition for overflow suffice
For 3 stacks a efficient algo is not easy to think in a single array , yet
we can optimise
@jitendra
#includeiostream
using namespace std;
void parens(int pairs)
{
int i, j, s, n = 2*(pairs - 1);
for (i = 0; i 1 n; i++) {
for (s = 1, j = 0; (j n) (s = 0) (s = pairs); j++)
s += ((i j) 1) ? 1 : -1;
if ((j != n) || (s != 1))
Still the solution will be similar and actually a bit simpler.
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Second Year Undergraduate,
Dept. of Computer Science and Engineering
IIT Bombay
http://www.cse.iitb.ac.in/~rohitfeb14
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A*(B+C*D)
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Rohit Saraf
Second Year Undergraduate,
Dept. of Computer Science and Engineering
IIT Bombay
http://www.cse.iitb.ac.in/~rohitfeb14
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So there is a prob algoose A*(B*C) and a*(b*c+d)
i hope you understood
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Rohit Saraf
Second Year Undergraduate,
Dept. of Computer Science and Engineering
IIT Bombay
http://www.cse.iitb.ac.in/~rohitfeb14
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why don't you make use of the optimal substructure.
You can easily use the recurrence relation as DP
@all : people don't just paste your code. Words are better than code
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simple in place O(n lg n) solution.
Choose a pivot in first array and partition it like in quicksort.
Find the pivot in second array and partition. Now recurse on both
halves. At any point if no of elements in array are not equal...
Abort!
--
--
A recursive way of filling each position is better, so we can cut it down
when its invalid and not proceed further. Something like below,
// 1-based indexing
- global n , A[2*n+1]
- go( ci , cs ) // cur.index , cur.sum
{
if( ci == 2n+1 ) print A with [ 1 = '(', -1 = ')' ]
Discard illegal prefix as soon as possible, and only proceed with legal one.
* For some position, if the number of '(' and ')' are equal, then do not
try ')' on that position.
* If the number of '(' reaches n, fill the rest positions with ')' to
get a valid permutation.
On 2010-6-3 16:15,
yep, constraints are fewer here
but in terms of problem statement 'same' and 'similar' can create
diversions.
On Thu, Jun 3, 2010 at 6:01 PM, Rohit Saraf rohit.kumar.sa...@gmail.comwrote:
Still the solution will be similar and actually a bit simpler.
--
But perhaps the elements in lists may not be in order.
Anurag Sharma
http://anuragsharma-sun.blogspot.com/
On Thu, Jun 3, 2010 at 6:38 PM, Rohit Saraf rohit.kumar.sa...@gmail.comwrote:
simple in place O(n lg n) solution.
Choose a pivot in first array and partition it like in quicksort.
Find
This should be solvable by dp as in knapsack problem
where value = weights
and W = S/2.
where S = sum of all elements in the array.
The claim here is diff in sum of two subsets will be minimum
when one maximizes a sub-set sum (say s) where sum (s) is closest to S/2.
Divya's problem is looking
@Jitendra: Catalan number grows at least exponentially:
http://mathworld.wolfram.com/CatalanNumber.html
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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
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Second Year Undergraduate,
Dept. of Computer
If the base logic follows the below rule, it should work.
If any operator within parenthesis is of less precedence to the
operator before the opening parenthesis, the parenthesis can not be
removed.
For cases of equal precedence of operators before parenthesis and
within parenthesis,
Consider two linked lists l1 and l2. Create 2 hash maps,h1 and h2, one
for each list with the format, element, number of occurrence.
Traverse one element in each list at a time. For each element in list
l1, check whether it is present in h2. If yes, remove the element from
h2 if the number of
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