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B reads i (stored value 0)
A reads i (0)
A stores i + 1 (1)
A reads i (1)
A stores i + 1 (2)
B stores i + 1 (1)
A reads i (1)
B completes (5)
A stores i + 1 (2).

So, min = 2, max = 8.

max should be 8 reason all steps are followed sequentially
considering no synchronization issue

min would be 3

Well... I guess this is the Caller dependency...

For Normal Cases we can Say if Thread A only executes then the min value will be there.
and For both thread executes the max value will get generated, But Still there are more complexity involved if we keep discussing.

But i n Laymen Term.

Min=3 ,
Max =8

Min 3
Max 8

Min: 0 (neither thread are schedule when you output them).
Max; 8.

Well its straight forward , whether its thread A (min .i.e 3) or both Thread A and Thread B which is 8(Max). Keeping in mind that thread doesnt garuantee order, A and B executes randomly.

min 3
Max 8

min = 5,max=8

i=0;
threadA
{ i++;
i++;
i++;
}
threadB
{ i++;
i++;
i++;
i++;
i++;
}