i am Considering given heap as min heap
Sol - because heap has property that root will has lower value than all
the elements in its left sub tree and right sub tree
so in main we will call a function passing root and value k and x
if at any time root is greater than x and k 0 that means rest of
@sunny : this will work :)
On Wed, Dec 14, 2011 at 4:08 PM, sunny agrawal sunny816.i...@gmail.comwrote:
i am Considering given heap as min heap
Sol - because heap has property that root will has lower value than all
the elements in its left sub tree and right sub tree
so in main we will
Just solved it.. The approach is same as Don's but starting from the
front... Hence, penning it down..
Let F(i) be defined as the no. of integers that are possible to be
formed using the first 'i' integral differences..
Then F(i+1) = No. of integers from set F(i) where the units place = 'i
+1'
Longest path won't work so well because it will return infinity if a
path contains a positive cycle, which doesn't apply here because once
you pick up an apple, it's gone.
On Dec 13, 7:17 am, vikas vikas.rastogi2...@gmail.com wrote:
Graph
take up, right and bottom as nodes connected to current
@atul anand
I am guessing that you referring to the the n^2 solution.. if yes,
then no i am not considering all permutations...
Basically L(i) is the reordering of A[i] which depicts the largest no.
that can be formed when read from left to right i.e MSD to LSD...
Hence, L(i+1) can be found by
We need not necessarily pick the apple. We need to suggest a way that could
have maximum apples and path need not be longest. Another person will
follow our path and pick the apples.
we can start from specified (i,j) and have to reach specified (i, n)
traversing either top, right, down(not left).
@dipit
Its not really a strict comparison..
To sort all we are doing is, keeping in mind that all the elements are
integers, is as follows:
Say, we need to sort two elements.. A and B
i = no. of digits in A, i.e Log10 (A)
j = no. of digits in B, i.e Log10 (B)
k = min (i, j);
x1 = A / pow(10,
I need to study both space and time complexities. What is the best source?
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Given a word, convert it into a palindrome with minimum addition of
letters to it. letters can be added anywhere in the word.
. for eg if hello is given, result should be hellolleh
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From (a,b) to (c,n) = What about traversing all columns n = j = a
as we can move up-down-right anytime we want.
On Dec 14, 5:57 pm, Azhar Hussain azhar...@gmail.com wrote:
We need not necessarily pick the apple. We need to suggest a way that could
have maximum apples and path need not be
@ Lucifer :
so i guess this is what you are trying to say:-
input : 2 3 5 78
L(0)= {2)
L(1)= {2,3} , {3,2} {3,2} is greatest
L(2)= {3,2,5} , { 3,5,2 } , {5,3,2 } {5,3,2} is the greatest
L(3)= {5,3,2,78} , {5,3,78,2} , {5,78,3,2 } , {78,5,3,2}
{78,5,3,2} found
nice :)
On
@atul... yup u r right...
There is a faster app. that i have mentioned i.e. O(nlogn)...
On Dec 14, 10:02 pm, atul anand atul.87fri...@gmail.com wrote:
@ Lucifer :
so i guess this is what you are trying to say:-
input : 2 3 5 78
L(0)= {2)
L(1)= {2,3} , {3,2} {3,2} is greatest
@saurabh-as by the above example LCS of HELLO and its inverse would be
LL and how can we form the word HELLOLLEH with it ...
and is your ans for the word NITAN is NITATIN ...?
On Wed, Dec 14, 2011 at 8:39 PM, saurabh singh saurab...@gmail.com wrote:
Find the LCS of string with its reverse
@Mohit
I think what he meant is 2* strlen(Input String) - 2* (Length of
LCS)
On Dec 15, 3:44 am, Mohit kumar lal kumarmohit...@gmail.com wrote:
@saurabh-as by the above example LCS of HELLO and its inverse would be
LL and how can we form the word HELLOLLEH with it ...
and is your ans for the
Introduction to Algorithms Coreman
On Wed, Dec 14, 2011 at 8:21 PM, Shashank Jain shashan...@gmail.com wrote:
I need to study both space and time complexities. What is the best source?
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hey guys ,i am trying to solve TSUM ..on spoj...
http://www.spoj.pl/problems/TSUM/.in which we have to find the sum of any
triplets in n numbers.can any one suggests me any approach other
than brute-force of (n^3)..
..thanks in advance
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@Mohit
Suppose for example
String: NITAN
LCS(Longest Common Subsequence) : NIN
How do you get the palindrome with it?
On Dec 15, 3:47 am, Lucifer sourabhd2...@gmail.com wrote:
@Mohit
I think what he meant is 2* strlen(Input String) - 2* (Length of
LCS)
On Dec 15, 3:44 am, Mohit kumar
@All,
I would like to correct a mistake in my previous post..
The min no. of additions = strlen(Input String) - (Length of LCS)
On Dec 15, 11:24 am, top coder topcode...@gmail.com wrote:
@Mohit
Suppose for example
String: NITAN
LCS(Longest Common Subsequence) : NIN
How do you get the
Correction:
for NAN :
N(IT)A + TI + N = NITATIN
On Dec 15, 11:33 am, Lucifer sourabhd2...@gmail.com wrote:
@topcoder..
String: NITAN
RevStr: NATIN
LCS ( NITAN, NATIN) = { NIN , NAN }
Here all we care about the count which is 2. Hence, 2 additions would
be required to convert it into a
@all , a Naive Approach with Quadratic Time will be for each i=1 to n ,
check if i and a[i]-i makes given sum , so for each each number we will do
the thus can achieve the solution ...i am just thinking about if we can do
it linear time ..will post if able to do it :)
Thanks
Shashank
CSe BIT
@sunny it will 3k not 2k ? u forgot to count the root element so over all
time complexity will O(3K)=O(K)
correct me if am wrong ?
Thanks
Shashank Mani
CSE, BIT Mesra
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more clarification , why number of comparisons are 3k , because we are
looking for only those nodes which are less then x and each nodea can max 2
childs , so tottal 3k comparisons . so it will O(3K) .
Thanks
Shashank
CSE , BIT Mesra
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@anubhaw ..check wiki for O(N^2) algorithm ..u will get the logic
Thanks
Shashank Mani
CSE , BIT Mesra
http://shashank7s.blogspot.com/
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@shashank
nope only 2k comparisions
k numbers which are greater than x and k which are less than x
dont think in terms of root and child
On Thu, Dec 15, 2011 at 12:57 PM, WgpShashank shashank7andr...@gmail.comwrote:
more clarification , why number of comparisons are 3k , because we are
looking
@Azhar , 1st Try to formulate the recursive relation , then draw recursive
tree , then u willl know why it need DP to solve efficiently , then u can
memozation to solve the same .
Hope it help.
Thanks
Shashank Mani
CSE ,BIT Mesra
http://shashank7s.blogspot.com/
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@Azhar , also for 1st question u r trying Array DS will suffices. Its
Standard Coin Change Problem , u need to solve subproblem 1st , store it in
extra array to avoid recalculation , if u are able to produce the sum less
given sum then u can use same approach to reach desired sum . it uses the
HEY DUDE TNX FOR THE REPLY .CAN YOU PASTE ME THE LINK FOR THAT
WIKI PAGE.
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