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@siddarth
can u explain ur algo for
box1 LBH - 7 8 9
box2 LBH - 6 7 8
box3 LBH - 3 7 2
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Devendra Pratap Singh
B.Tech (IT)
IIIT - Allahabad
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@gene
thanx for the working code
but can u explain its working more clearly
On Jul 13, 11:21 pm, Gene wrote:
> On Jul 10, 5:18 pm, Gene wrote:
>
> > On Jul 9, 3:55 pm, Devendra Pratap Singh
> > wrote:
>
> > > plz write a code to
>
> > > So
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@jalaj
no constraint
but can u tell me how u handle this O(n^2) space when n is around 10^5
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@Jalaj
how using dijikshtra??
can u explain lilbit ... ???
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plz write a code to
Sort n integer in the range 1 to n^2 in O(n)
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@anand
its not a iterative approach
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For m
u can access arr[0],arr[1],arr[2] in following way ...
#include
int main()
{
char name[3][10] = {"jan","feb","march"};
char (*p)[10] = name;
int i;
for(i=0;i<=3;i++){
printf("%s\n",p);
p = p+1;
}
return 0;
}
--
@ashish
how using trasistivity matrix .. will u prepare matrix for each value
of k or something else ... ??
On Jul 5, 8:52 am, Ashish Goel wrote:
> i would prepare the transitivity matrix while inserting the edge into the
> matrix
>
> the search then would be a O(1)
>
> Best Regards
> Ashish Go
reason for that o/p is ...
coz range of short int is -32768 to 32767
and value of i start with i=0
and each time it will increment by 1
so corresponding value of i will be
1
2
3
.
.
.
32766
32767
-32768
-32767
.
.
.
.
-2
-1
but
coz in printf u r using %u so it will print...
1
2
3
.
.
.
3
@Jalaj
if u consider only Space complexity then it can be done in O(n^2) time
and constant space
here i assume string u taken only hv a-z character nothing else
try this
for(i=0;i='a' && st[i]<='z') )
continue;
count=1;
for(j=i+1;j wrote:
> Hi Friends
>
> Ha
Try this .. i think it will work.. correct me if wrong
sort(&x[0],&x[n]);
ysum=y[0]=x[n-1];
ysum=z[0]=x[n-2];
j=0;
k=n-3;
l=n/2;
for(i=1;i wrote:
> my algo on the array 1 200 500 2000
> sort the array therefore we have now 2000 500 200 1
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