@ll...thnx a lot
On Sat, Aug 25, 2012 at 8:22 PM, GAURAV CHAWLA wrote:
> see A(n)ie the average case will always be smaller or equal to the
> worst case...
>
> ie something like... A(n)<= c. w(n) for some c as constt ... which the
> definition of big O...
>
>correct me if i'm wrong..
see A(n)ie the average case will always be smaller or equal to the
worst case...
ie something like... A(n)<= c. w(n) for some c as constt ... which the
definition of big O...
correct me if i'm wrong..
On Sat, Aug 25, 2012 at 7:11 PM, rahul sharma wrote:
> *Let w(n) and A(n) denote re
You have to discard option d because , according to definition of small o
notation if f(n) =o(g(n)) then for ALL constants c> 0 you have f(n) <
cg(n). or Lim(n->infinite) f(n)/g(n) = 0.
On Sat, Aug 25, 2012 at 10:24 PM, vishal yadav wrote:
> Because you can always find a positive constant c for w
Because you can always find a positive constant c for which following
inequality hold true.
A(n) <= cW(n) i.e. the avg. case time complexity always upper bounded by
worst case time complexity. Which is the definition of Big O.
On Sat, Aug 25, 2012 at 7:11 PM, rahul sharma wrote:
> *Let w(n) and
*Let w(n) and A(n) denote respectively, the worst case and average case
running time of an algorithm executed on an input of size n. which of the
following is ALWAYS TRUE?*
(A) [image: A(n) = \Omega(W(n))]
(B) [image: A(n) = \Theta(W(n))]
(C) [image: A(n) = O(W(n))]
(D) [image: A(n) = o(W(n))]
ans
Thanku Sasi .
On Tue, Jul 26, 2011 at 10:49 PM, sasi kumar wrote:
> hi
> > void XYZ(int a[],int b[], int c[])
> > {
> > int i,j,k;
> > i=j=k=0;
> > while((i > {
> > if (a[i] > c[k++]=a[i++];
> > else
> > c[k++]=b[j++];
> >0
> > }
>
> In this case either i value or j value is
hi
> void XYZ(int a[],int b[], int c[])
> {
> int i,j,k;
> i=j=k=0;
> while((i {
> if (a[i] c[k++]=a[i++];
> else
> c[k++]=b[j++];
>0
> }
In this case either i value or j value is incremented at a time
in an iteration . So its impossible that both the conditions (i< n)
and (j
> Which of
Consider the following C-function in which a[n] and b[m] are two
sorted integer arrays and c[n+m] be an other integer array.
void XYZ(int a[],int b[], int c[])
{
int i,j,k;
i=j=k=0;
while((ihttp://groups.google.com/group/algogeeks?hl=en.
C )
On Tue, Jul 26, 2011 at 9:02 PM, Vijay Khandar wrote:
> Consider the following C-function in which a[n] and b[m] are two
> sorted integer arrays and c[n+m] be an other integer array.
>
>
> void XYZ(int a[],int b[], int c[])
> {
> int i,j,k;
> i=j=k=0;
> while((i {
> if (a[i] c[k++]=a[i++];
>
Consider the following C-function in which a[n] and b[m] are two
sorted integer arrays and c[n+m] be an other integer array.
void XYZ(int a[],int b[], int c[])
{
int i,j,k;
i=j=k=0;
while((ihttp://groups.google.com/group/algogeeks?hl=en.
ASCII value of 'A' is 65 and Asciivalue of 'E' is 69.
69-65=4
On Sat, Jul 2, 2011 at 7:12 PM, abhijith reddy wrote:
> p[3] = 'E'
> p[1] = 'A'
> p[3]-p[1] = 4
> ?
>
>
> On Sat, Jul 2, 2011 at 7:10 PM, KK wrote:
>
>> 10. What does the following fragment of C-program print?
>>
>> char c[] = "GATE20
p[3] = 'E'
p[1] = 'A'
p[3]-p[1] = 4
?
On Sat, Jul 2, 2011 at 7:10 PM, KK wrote:
> 10. What does the following fragment of C-program print?
>
> char c[] = "GATE2011";
> char *p =c;
> printf("%s", p + p[3] - p[1]);
>
> (A) GATE2011 (B) E2011 (C) 2011 (D) 011
> Answer: - (C)
>
> why is p[3] - p[1]
10. What does the following fragment of C-program print?
char c[] = "GATE2011";
char *p =c;
printf("%s", p + p[3] - p[1]);
(A) GATE2011 (B) E2011 (C) 2011 (D) 011
Answer: - (C)
why is p[3] - p[1] returning 4
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