Given array of integers (0 or +ve or -ve value) find two elements having
minimum difference in their absolute values.
e.g. Input {10 , -2, 4, 9,-20,17,-8,14,12)
output {9,-8}
I have solved it in O(nlogn). can it be solved in O(n).
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Single queue only satisfies the given condition. I think it's the correct
one..
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**~Sathish Babu~**
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From: Sathish babu satbrucei...@gmail.com
Date: Wed, Jul 25, 2012 at 9:52 PM
Subject: general tree into BST
To: algogeeks@googlegroups.com
Hey hi, can anyone provide the code to convert general tree into BST and
BST into general
Can you please tell the kind of algos that will be asked for the interview
and what all subjects to focus for?
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thanks for the code.
On Wednesday, July 25, 2012 7:29:36 PM UTC+5:30, Navin Kumar wrote:
logic is very simple ...just trace the program you will understand.
On Wed, Jul 25, 2012 at 7:28 PM, Navin Kumar algorithm.i...@gmail.comwrote:
#include stdio.h
#include stdlib.h
#include ctype.h
thx alot
How silly my mistake was!! :( ...
On Thu, Jul 26, 2012 at 11:16 PM, payal gupta gpt.pa...@gmail.com wrote:
perhaps its this quoted line...
permu(c,i+1,j); permu(c,i+1,n)
On Thu, Jul 26, 2012 at 9:37 PM, teja bala pawanjalsa.t...@gmail.comwrote:
// plz anyone tell
@ ashish the example which you given here ie aaab and at the end you said
that ba is the beautiful string how com ba can be derived or produced from
aaab becoz as per the definition String s2 is *producible* from string s1,
if we can remove some characters of s1 to obtain s2. and ba sequence
#includestdio.h
int main(){
printf(%d %d\n, 321, 320);
printf(%d %d\n, 32-1, 32-0);
printf(%d %d\n, 321, 320);
printf(%d %d\n, 32-1, 32-0);
return 0;
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Can you please give the algo of solving it in O(nlogn)
On Friday, 27 July 2012 15:35:24 UTC+5:30, Navin Kumar wrote:
Given array of integers (0 or +ve or -ve value) find two elements having
minimum difference in their absolute values.
e.g. Input {10 , -2, 4, 9,-20,17,-8,14,12)
output {9,-8}
and just for confirming.. will the algo for O(n2) be:
// uses kind of selection sort technique
mindiff = abs(a[1]-a[2]) //take a default min value
for i= 1 to n
for j= 1 to n
if i==j // same no.
continue
if abs(a[i]-a[j]) mindiff // if difference of some other
@s_m154: Sort the array in O(n log n) and then compare adjacent numbers to
find the closest pair in O(n). Total: O(n log n).
Dave
On Friday, July 27, 2012 10:41:28 AM UTC-5, s_m154 wrote:
Can you please give the algo of solving it in O(nlogn)
On Friday, 27 July 2012 15:35:24 UTC+5:30,
64 32
16 32
16 32
64 32
On Fri, Jul 27, 2012 at 6:48 PM, Hraday Sharma hradaysha...@gmail.comwrote:
#includestdio.h
int main(){
printf(%d %d\n, 321, 320);
printf(%d %d\n, 32-1, 32-0);
printf(%d %d\n, 321, 320);
printf(%d %d\n, 32-1, 32-0);
return 0;
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You
@Dave Sir : Sir if u sort the array(given above) the array would be:
-20,-8-2,4,9,10,12,14,17, and according to ur suggestion, the only ans is
{9,10}...but one of the ans {9,-8} is also possible...as he is asking
the difference in absolute values.
correct me if i m wrong...
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compiler dependent
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@Arun: I left out that you sort the array by absolute values and compare
absolute values of adjacent elements. The sorted array would be {-2, 4, -8,
9, 10, 12, 14, 17, -20}. Then abs(abs(-8) - abs(9)) is the smallest
difference.
Dave
Dave
On Friday, July 27, 2012 11:21:52 PM UTC-5, Arun
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