U can always have a trackback array as we do in LIS.
Regards,
Akash Agrawal
http://tech-queries.blogspot.com/
On Fri, Mar 4, 2011 at 5:48 PM, Vipin Agrawal vipin.iitr@gmail.comwrote:
Solution is good, but its says only Max time that he can spend.
How would he know that which program he should watch for Maximum
utilization ?
On Mar 4, 4:38 pm, Akash Agrawal akash.agrawa...@gmail.com wrote:
http://tech-queries.blogspot.com/2011/03/avid-tv-watcher.html
Regards,
Akash Agrawalhttp://tech-queries.blogspot.com/
On Fri, Mar 4, 2011 at 2:25 PM, Akash Agrawal akash.agrawa...@gmail.com
wrote:
Example:
Channel 1:
Program id
P1
P2
P3
Start time
8:00
9:00
10:30
End time
8:30
10:00
11:30
Channel 2:
Program id
P4
P5
P6
Start time
8:15
9:30
10:45
End time
9:15
10:15
11:15
Sort all programs based on their end time:
Cnt
0
0
0
0
0
0
Pr id
P1
P4
P2
P5
P6
P3
St time
8:00
8:15
9:00
9:30
10:45
10:30
End time
8:30
9:15
10:00
10:15
11:30
11:30
1st Iteration:
Cnt
00:30
0
0
0
0
0
Pr id
P1
P4
P2
P5
P6
P3
St time
8:00
8:15
9:00
9:30
10:45
10:30
End time
8:30
9:15
10:00
10:15
11:30
11:30
2nd Iteration:
Cnt
00:30
01:00
0
0
0
0
Pr id
P1
P4
P2
P5
P6
P3
St time
8:00
8:15
9:00
9:30
10:45
10:30
End time
8:30
9:15
10:00
10:15
11:30
11:30
3rd Iteration:
Cnt
00:30
01:00
01:30
0
0
0
Pr id
P1
P4
P2
P5
P6
P3
St time
8:00
8:15
9:00
9:30
10:45
10:30
End time
8:30
9:15
10:00
10:15
11:30
11:30
4th Iteration:
Cnt
00:30
01:00
01:30
01:45
0
0
Pr id
P1
P4
P2
P5
P6
P3
St time
8:00
8:15
9:00
9:30
10:45
10:30
End time
8:30
9:15
10:00
10:15
11:30
11:30
5th Iteration:
Cnt
00:30
01:00
01:30
01:45
02:30
0
Pr id
P1
P4
P2
P5
P6
P3
St time
8:00
8:15
9:00
9:30
10:45
10:30
End time
8:30
9:15
10:00
10:15
11:30
11:30
6th Iteration:
Cnt
00:30
01:00
01:30
01:45
02:30
02:45
Pr id
P1
P4
P2
P5
P6
P3
St time
8:00
8:15
9:00
9:30
10:45
10:30
End time
8:30
9:15
10:00
10:15
11:30
11:30
Answer will be 2:45 hrs as this is the biggest value in Cnt.
Regards,
Akash Agrawal
http://tech-queries.blogspot.com/
On Fri, Mar 4, 2011 at 2:17 PM, Akash Agrawal
akash.agrawa...@gmail.comwrote:
Seems, it didn't preserve indentation. attached file.
Regards,
Akash Agrawal
http://tech-queries.blogspot.com/
On Fri, Mar 4, 2011 at 2:16 PM, Akash Agrawal
akash.agrawa...@gmail.comwrote:
First try urself with thinking longest common subsequence if u r
still
not able to see below algo. I will write a blog post soon for the
same.
Funda is to see what is the max time I can spend on TV if I watch
program
X.
1. Merge all the programs of diff channels in sorted order of
their
end-time in prog[] array. (need
O(nlogk))
2. n ← length(prog) - 1
3. for i ← 1 to n
a. do prog[i].cnt ← 0
b. max ← 0
c. j ← 0
d. while prog[j].end prog[i].start
i. if max
prog[j].cnt
1. do max ← proj[j].cnt
ii. j ← j + 1
e. do prog[i].cnt ← max + prog[i].end – prog[i].start
4. do res ← 0
5. for i ← 1 to n
a. if res prog[i].cnt
i. do res
←prog[i].cnt
6. return res
PS: Hope it saved indentation.
Regards,
Akash Agrawal
http://tech-queries.blogspot.com/
On Tue, Feb 1, 2011 at 2:46 PM, Vikas Kumar dev.vika...@gmail.com
wrote:
@Snehal
just a hint: there is no need of that channel 1 channel 2.
just treat each program as independent program.
On Mon, Jan 31, 2011 at 10:44 PM, snehal jain
learner@gmail.comwrote:
or provide some link
On Mon, Jan 31, 2011 at 10:44 PM, snehal jain
learner@gmail.comwrote:
@ juver++
can you please share
