Question

How would you maintain a stack so that whenever you pop from the stack, you know the minimum element in the stack? The algorithm should have a constant complexity

Thanks in advance

Answer 1

Ok, here’s what you need to do..

You need to maintain some data structure containing information of each element you insert regarding, what was the minimum and second minimum value at the time when you inserted that element..

So, you want to have information like this: –

For each Element pushed ->

• minimum value after insertion
• minimum value before insertion

This information will be needed when you pop that element from stack.. So, you would know that, whether you are popping minimum value or not.. If yes, then you can replace the current minimum value with the minimum value before pushing this element.. If not, then there will be no change in minimum value at that time..

For E.g: –

Suppose currently you have following element in stack: –

[3, 5, 7, 9, 2, 4]

• And you push a value 8.. Then you have two values to maintain..
  (minimum value before 8 in pushed, i.e. 2, and miminum value after 8
  is inserted, i.e. 2): –

  min_value_before_push = min(stack)
  push 8
  min_value_after_push = min(stack)
  

• And if you push a value 1, then minimum_before_insertion is 2,
  and minimum_after_insertion is 1: –

  min_value_before_push = 2
  min_value_after_push = 1
  

Now your stack is :-

[1, 8, 3, 5, 7, 9, 2, 4]

Now, if you pop, You will see that for value 1: –

    min_value_before_push = 2
    min_value_after_push = 1

So, popping will alter the minimum value, So, you change the current minimum value with the minimum_value_before_push of 1.. So, again your minimum is 2..

Your current stack becomes : –

[8, 3, 5, 7, 9, 2, 4]

Now, let’s check whether this algorithm works for duplicate element : –

Suppose you want to push a value 2 again..

min_value_before_push = 2
min_value_after_push = 2

Then you go on and pop, you see that for value 2, min_value_after_push is 2, so this means popping it will alter the minimum value.. So you replace this value with min_value_before_push, which is also 2.. Which is what we wanted..

NOTE: – One Benefit of this Algorithm is that, you won’t need to do much comparison.. Just a comparison with current_minimum_value while pushing.. And a comparison with current_minimum_value when you pop..

You can try to proceed to think what data structure can you have..