Problem with code logic

#1

So what i understud is dp[i]= min(dp[i-1],dp[i-2],dp[i-3])
Is this right? So i need to keep summing them ?
dp[i] will store total minimum sum upto that point ?

#2

Hey @pranjalarora98
Yes its almost correct

Realtion is
dp[i] = min({ dp[i-1] , dp[i-2] , dp[i-3] } ) + a[i];

#3

Just 1 issue i had in undesrtanding.
The school’s rules say that no student can go three days in a row without any SUPW duty.

This is written so it means that every continious 3 days atleast 1 student should be there? They can be 2 or 3 also so we have to find minimum whatever included is it like this?

#4

3 2 1 1 2 3 1 3 2 1

if this is the data then that means
Sorry these are minutes

Now this means that nikhil have to assign himself a duty without breaking that 3 day rule

#5

And suppose i am writing base case so for dp[0]=arr[i]
dp[1] should be arr[0] + arr[1] coz we can select 2 right?
and dp2 should be max of arr0+arr1,arr1+arr2,arr0+arr2

#6

is this fine?for base cases

#7

Wait let me reread the question :slight_smile:
I think I misunderstood it maybe

#8

Base case is
dp[1] = a[1];
dp[2] = min(a[2],a[1]);
dp[3] = min(a[2],a[1],a[3]);

Because only 1 day he have to work
if 2 days then he will not work on either first day or 2nd day
if 3 days he will on any of the first 3 days

#9

The school’s rules say that no student can go three days in a row without any SUPW duty.
So it means that atleast 1 should be there from every continious 3 right? there can be 2 or there can be 3 also?

#10

But we have to minimize so we will be seeing only 1 of the minimum ?

#11

Yes correct from what I am understanding u are saying

#12

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