CareerCup

JAR A(5L)   JAR B(3L)
    5           0
    -> (Pour to B)
    2           3
                -> (Empty the B)
    2           0
    -> (Pour to B)
    0           2
->(Fill 5 L into A )
    5           2
    ->(Pour 1L to B to get 3 L in the B)
    4           3

Step 1: Fill the 5-liter jar.
Step 2: Empty 3 liters out of it into the 3-liter jar.
Step 3: Discard the contents of 3-liter jar.
Step 4: Empty the 2 liters in the 5-liter jar into the 3-liter jar.
Step 5: Fill the 5-liter jar again.
Step 6: Empty a liter from the 5-liter jar into the partially filled 3-liter jar (which has 2 liters in it).
Step 7: Discard the contents of the 3-liter jar.
Step 8: The 5 liter jar has 4 liters left out in it.

(5-3) + (5-3)

Take water in 5 lit pour them in to 3 lts and through the 3 out. now get the 2 lits from the 5 ltr can to another storage. repeat the same you will get 4. One draw back is you require another container for holding 2lts.

>>5 3
0 3
3 0
3 3
5 1
0 1
1 0
1 3
4 0
Got it!

See Die Hard starring Bruce Willis :)

Fill both containers entirely and then pour out half of each container. You will then have 2 1/2 liters in the one jar and 1 1/2 liters in the second. Combine them and you get 4 liters.

kgyouphlog

Fill 5 litres jar A and pour all in to 3 litre jar B.Now you 'l be left with 2 litres in A.
Empty B and transfer those 2 litres to B.Now Fill A with 5 litres and form A ,transfer one litre to B.
Thats all,It gives 4 Litres in Jar A.

using 5 fill 3 -> 2 in 5.
empty 3 ,
fill 3 using 5 -> 2 in 3.
fill 5.
fill 3 with 5 -> 3 in 3,4 in 5

plz tell how to solve this equation 2A3M10D2S4 using BODMAS rule in java

plz tell how to solve this equation 2A3M10D2S4 using BODMAS rule in java

Another method: fill 3 liter and pour it into 5 lit can. Again fill 3 liter and pour into 5 lit. Can.
Now you have 1 liter in 3lit can. Now empty 5 lit Can. Pour 1 lit into 5lit can, again fill 3lit completely and pour it into 5litcan.

How would one write code for the same?
I was asked the same question in an interview today
I was supposed write code to find the minimum number of steps.

one countainer holds 51/4 litre of oil

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