Design a stack that supports getMin() in O(1) time and O(1) extra space

An approach that uses O(1) time and O(n) extra space is discussed here.

In this article, a new approach is discussed that supports minimum with O(1) extra space. We define a variableminEle that stores the current minimum element in the stack. Now the interesting part is, how to handle the case when minimum element is removed. To handle this, we push “2x – minEle” into the stack instead of x so that previous minimum element can be retrieved using current minEle and its value stored in stack. Below are detailed steps and explanation of working.

Push(x) : Inserts x at the top of stack.

  • If stack is empty, insert x into the stack and make minEle equal to x.
  • If stack is not empty, compare x with minEle. Two cases arise:
    • If x is greater than or equal to minEle, simply insert x.
    • If x is less than minEle, insert (2*x – minEle) into the stack and make minEle equal to x. For example, let previous minEle was 3. Now we want to insert 2. We update minEle as 2 and insert 2*2 – 3 = 1 into the stack.

Pop() : Removes an element from top of stack.

  • Remove element from top. Let the removed element be y. Two cases arise:
    • If y is greater than or equal to minEle, the minimum element in the stack is still minEle.
    • If y is less than minEle, the minimum element now becomes (2*minEle – y), so update (minEle = 2*minEle – y). This is where we retrieve previous minimum from current minimum and its value in stack. For example, let the element to be removed be 1 and minEle be 2. We remove 1 and update minEle as 2*2 – 1 = 3.

Important Points:

  • Stack doesn’t hold actual value of an element if it is minimum so far.
  • Actual minimum element is always stored in minEle

Illustration

Push(x)
stack_insert

  • Number to be Inserted: 3, Stack is empty, so insert 3 into stack and minEle = 3.
  • Number to be Inserted: 5, Stack is not empty, 5> minEle, insert 5 into stack and minEle = 3.
  • Number to be Inserted: 2, Stack is not empty, 2< minEle, insert (2*2-3 = 1) into stack and minEle = 2.
  • Number to be Inserted: 1, Stack is not empty, 1< minEle, insert (2*1-2 = 0) into stack and minEle = 1.
  • Number to be Inserted: 1, Stack is not empty, 1 = minEle, insert 1 into stack and minEle = 1.
  • Number to be Inserted: -1, Stack is not empty, -1 < minEle, insert (2*-1 – 1 = -3) into stack and minEle = -1.

Pop()
stack_removal

  • Initially the minimum element minEle in the stack is -1.
  • Number removed: -3, Since -3 is less than the minimum element the original number being removed is minEle which is -1, and the new minEle = 2*-1 – (-3) = 1
  • Number removed: 1, 1 == minEle, so number removed is 1 and minEle is still equal to 1.
  • Number removed: 0, 0< minEle, original number is minEle which is 1 and new minEle = 2*1 – 0 = 2.
  • Number removed: 1, 1< minEle, original number is minEle which is 2 and new minEle = 2*2 – 1 = 3.
  • Number removed: 5, 5> minEle, original number is 5 and minEle is still 3

Implementation:

 

Result:

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