Question

NO NEED TO WRITE CODE,EXPLAIN IN C++ PLEASE.

Suppose we have a Stack that can grow indefinitely (for example, the push method has been fixed to double the size of the array when at capacity instead of throwing a StackFullException). We want to create a second Stack data structure, which I promise will always contain only comparable items (e.g., integers, strings). We also want to add a function findMin to the Stack interface that will return the smallest element currently in the stack. We could implement this by searching the array, but that takes time linear in the number of elements in the Stack. Explain how you would change the Stack data structure to allow for this function to run in O(1) time. If you are storing additional private member data, state what else you are storing. If you are changing existing functions push, pop, or top (or the constructor/size
functions), explain briefly how you are changing them. Their running times must still be O(1); for example, you cannot search the full stack for the newest min at every push and pop.

You may assume that there will never be a duplicate item pushed to the Stack.

Answer 1

For finding the minimum element in stack in O(1) time and O(1) space, we need to declare a variable which will always store the minimum element value. Let the variable be minimumElement.

Let us first see for push(x) operation:

• If the stack is empty push x and set minimumElement = x
• If the stack is not empty and x is greater than the minimumElement then insert x directly into the stack as at this time we don't need to update the minimumElement
• If the stack is not empty and x is lesser than the minimumElement then insert (2*x - minimumElement) and set minimumElement = x. Example if the previous minimumElement was 5 and now we need to insert 1 then insert (2*1-5) ,i.e., insert -3 in the stack and set minimumElement = 1.

For pop(x) operation:

Remove one element from the top of the stack and let us call it as z

• If the removed element is greater than or equal to the minimumElement then no need to change anything as the minimum element is in the stack only.
• If the removed element is smaller than the minimumElement then we set minimumElement = (2*minimumElement-z). This is the place where the previous minimum element was found using the current minimum element. Example if z=2 and minimumElement = 4 then we update minimumElement = (2*4-2) = 6. So new minimumElement = 6

Example:

Push(x) Operation

Element Inserted	Elements in Stack	minimumElement
4	4	4
7	4, 7	4
1	4, 7, -2	1

• First stack is empty and we inserted 4 so minimumElement = 4. Minimum element in stack is 4.
• Then we insert 7 and as 7>minimumElement then it will be inserted directly without changing anything. Minimum element in stack is still 4
• Then we want to insert 1 and as 1<minimumElement then we will insert 2*1-minimumElement ,.i.e, -2 and minimumElement=1. Now minimum element in stack is 1.

Pop() Operation

Number popped	Original Number	Elements in stack	minimumElement
		4, 7, -2	1
-2	1	4, 7	4
7	7	4	4

• Currently minimumElement = 1 which is the minimum element.
• When we pop -2 then we find that -2<1 so the element to be popped is the minimum element in the stack ,i.e., minimumElement which is 1. So 1 which was the minimum element will be removed and new minimumElement=(2*1-(-2)) = 4.
• So after removing 1 we can see that stack has two elements 4 and 7 and 4 is the minimum.
• Now when 7 will be popped we will see 7>minimumElement so no need to change anything.

Key Points:

• The minimum Element is always stored in minimumElement and not in stack.
• The value stored in stack are used to find the element which was previously minimum.

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