hello friends
i NEED UR HUMBLE SUPPORT
I will post a question after an hour, if
anybody can solve it for me as early as possible (probably within 45
minutes), then please solve it and send it ti me. I urgently need it
my id
hello friends
I need your humble support.
after 1 hour i will poet a question if some can solve it as soon as
possible (within 45 minutes after post) , then please solve it for me
.
and send me
languages c++, java
my idkamleshlu2...@gmail.com
--
Kamlesh Kumar Yadav
MCA
plaese be online at 8.30 pm today if possible and please help me.
On Tue, Apr 26, 2011 at 7:19 PM, kamlesh yadav kamleshlu2...@gmail.com wrote:
hello friends
i NEED UR HUMBLE SUPPORT
I will post a question after an hour, if
anybody can
2º Semestre de Análise e Desenvolvimento de Sistemas
FATEC
On Tue, Apr 26, 2011 at 11:08, kamlesh yadav kamleshlu2...@gmail.com
wrote:
plaese be online at 8.30 pm today if possible and please help me.
On Tue, Apr 26, 2011 at 7:19 PM, kamlesh yadav kamleshlu2...@gmail.com
wrote:
hello
can anyone give me list of books like ( cracking the coding
interview )
for placement preparation
thanks in advance.
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To
given an array of elements (all elements are unique ) , given a sum
s find all the subsets having sum s.
for ex array {5,9,1,3,4,2,6,7,11,10}
sum is 10
possible subsets are {10}, {6,4} ,{7,3} {5,3,2}
{6,3,1} etc.there can be many more.
also find the total number of these
.
On Apr 18, 4:16 pm, kamlesh yadav kamleshlu2...@gmail.com wrote:
given an array of elements (all elements are unique ) , given a sum
s find all the subsets having sum s.
for ex array {5,9,1,3,4,2,6,7,11,10}
sum is 10
possible subsets are {10}, {6,4} ,{7,3} {5,3,2}
{6,3,1} etc
[])
{
search(set, setSize, sum);
return 0;
}
On Apr 18, 6:16 am, kamlesh yadav kamleshlu2...@gmail.com wrote:
given an array of elements (all elements are unique ) , given a sum
s find all the subsets having sum s.
for ex array {5,9,1,3,4,2,6,7,11,10}
sum is 10
possible
I didnt understand what you mean. Please clarify.
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|.
This surely can be done in O(n) and code for the same goes like
S1=0; M1=1; M=1; S=N*(N+1)/2.
for(i=0;iN;i++)
{
S1+=arr[i];
M1*=arr[i];
M*=(i+1);
}
t1=abs(S-S1); t2=abs(M1-M);
Repeated_element = M1*t1/t2;
Missing_element = M*t1/t2;
Thanks,
Kamlesh
; temp=t;
swaps++;
}
}
}
Thaks,
Kamlesh
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Btw this is just O(n) Algo as no of swaps are restricted to N-2
overall.
Kamlesh
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