Use inplace merge algorithms along with merge sort.
http://www.logiccoder.com/Downloads/krnrd20.pdf
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hi ,
this is not a research kind of problem i expect a simple answer.
On Wed, Dec 29, 2010 at 2:33 PM, juver++ avpostni...@gmail.com wrote:
Use inplace merge algorithms along with merge sort.
http://www.logiccoder.com/Downloads/krnrd20.pdf
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please visit this blog
http://code-forum.blogspot.com/2010/12/in-place-merge-sort.html
to see a simple and straight solution to this problem.however if first
array does not have enough space then also it can be done inplace and
still lots of research is going on for this.
On Dec 29, 2:25 pm,
@vishal thanx
On Wed, Dec 29, 2010 at 3:39 PM, vishal raja vishal.ge...@gmail.com wrote:
Perform simple merging taking the ends of the list.
So You compare the last elements of both the list which ever is larger you
copy that at the end of the first array and so on.
you got to maintain three
you are given a string.It may have repeated characters.each character has
it's own weight.find combination of characters whose sum value is equal to
given value.
ex: given string 'abcbacbabcc'
weight of each character a-5,b-4,c-6.
character combination which gives sum 15 is 'aaa'
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Thank You
Is'nt this just a knapsack problem!
On Dec 29, 3:45 pm, rajeev kumar rajeevprasa...@gmail.com wrote:
you are given a string.It may have repeated characters.each character has
it's own weight.find combination of characters whose sum value is equal to
given value.
ex: given string
String gives no of occurrences of each character.no of times you select a
character should be less than the no of times it appears in the string.
ex: given string 'abcbacbabcc' has 3 a's, 4b's, 4 c's.
resulting string should have a less than 3 times,b less than 4 times and c
less than 4 times.
It's a knapsack problem with bounds. Solve it using DP - for each state keep
number of used characters and preserve to exceed the bounds.
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@juver++
can you elaborate this sentence 'It's a knapsack problem with bounds. Solve
it using DP' or send me any link which has good explanation.
On Wed, Dec 29, 2010 at 4:52 PM, vishal raja vishal.ge...@gmail.com wrote:
i think we don't need to store the total no. of occurance of any
Yes, it's true. But we should process DP in the following order not to take
one element more than once.
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@juver, ofcourse , and that's not a big deal.
On Wed, Dec 29, 2010 at 5:11 PM, juver++ avpostni...@gmail.com wrote:
Yes, it's true. But we should process DP in the following order not to take
one element more than once.
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dp[i][j] - true, if there is a way to have sum j after processing the first
i elements, and false otherwise.
so transitions wil be:
dp[i][j] = dp[i - 1][j] OR dp[i - 1][j - weight[i]], (OR means logical
or),first term means that we don't want to use i-th character, second - we
use it, so the
It's not a big problem when you use 2D matrix instead of 1D for the DP :)
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Given set of words, find an area of the text where these words are as
close to each other as possible.
Distance is measured on number of words.
e.g. for words [rat”, “jat”, “pat”] and text “rat stop the pat blah
blah jat by pat jat tra la la” such area is “cat by mat bat”
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But the same solution I've given above can give you the solution for this
problem .
In the formed table of P[i][j] , you can take another variable attached to
it as count[i][j] for how many items we have selected yet.
So you gotta find , the max. value of j which has count = 50.
count[i][j] =
Thanks everybody for wonderful support and special thanks to Vishal
raja. . But i was bit apprehensive about your last solution . . i will
test it :) and let you know as well . Thanks . . . .
On Thu, Dec 30, 2010 at 11:52 AM, vishal raja vishal.ge...@gmail.com wrote:
But the same solution I've
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