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So 533 bananas can be delivered to market 1000km away

Now we do have 200 + 333.3 KM covered

Remaining with 1000 bananas
And 466.6KM

So we can make this in one trip

So 1banana/KM

1000-466.6(1) = 533 bananas

Now from 200km to h km for drop Station 2

We have 2000 bananas remaining

2 Loads

2 straight trips and 1 round trip will be 3banana/km

So remaining bananas at DROP2 are 2000-3y

and we will consider it to be equal to 1000

1000+2000 = 3y ..
Y= 333.3 KM
1000 BANANAS REMAINING

So for 3000 bananas camel needs to consume 5 bananas per KM when we have 3 loads
So at STATION 1 .... WHICH IS X KM
so remaining bananas at station 1 are
3000- x*5
And when bananas left are less than 2000 we will only need 2 loads
So
2000 = 3000 - 5X
X = 200 Km

So first 200KM with 3 loads

We try to maintain total no. Of bananas to be multiple of 1000 at each point
Let's start
So initially from start point to point X
Camel move forward with bananas and come empty and repeat it 3 times since 3000> 1000
so we do have to take 5 trips

And since it's 5 trips it will be 5 🍌 per km

Let's see what we have

3000 units of bananas

Destination of 1000 KM

MAX limit of 1000 banana in one go

Camel which eats one Banana for every km it travels

So here we use intermediate DROP STATIONS --- WHERE CAMEL CAN MAKE SEEVERAL SHORT TRIPS IN BETWEEN

To deal with this we will use incremental addition approach and at last when we will weigh it
The variation from ideal will tell us which one does have contaminated pills

They sense the whole match on the fly till 2nd quarter and design their strategy and execute in last two...no wonder there is need for management and great coach

all are being literal bricks

i realized all i did was crazy brute force till 789 kata ...

Bhai apne Wale bhi knha kum h

scrimba.com/learn-react-...

fullstackopen.com/en/

This tooo

Glad my mentor survived

That explains

reference men in black ... try again with hint