Why is stack control critical in modern operating systems and
software applications?
Stack as a control mechanism rather than just for data
storage
The fundamental purpose of a stack is to provide an order of
execution and control over the sequence of instructions executed by a computer.
Stacks provide a level of predictability and safety in many real-life
circumstances because certain procedures must be executed using a last-in,
first-out (LIFO) approach. LIFO means that the most recently added stack entry
must be the first to be implemented.
The following are a few ways that modern software takes
advantage of stacks:
1.
Functional
calls and recursive functions, with the last functional call always returning
before any of the previous calls.
2.
The
use of undo and redo functionalities, available in many software applications
such as text editors and design applications.
3.
Expression
evaluation is performed by the compiler and interpreter.
4.
Validation
tasks such as matching brackets, nesting, and so on.
Stacks enable systems to stop, start and roll back the
operation of the stack and thereby provide a mechanism to perform the task with
the least amount of resources being used. Because stacks are so prevalent in
compilers, operating systems, web browsers, and many services on the backend,
we can now think of many of the challenges that we deal with daily as stemming
from the need to control executions rather than only the storage of data.
🟢 Easy Stack Problems (Foundation Builders)
1.
Valid Parentheses
🧠 Explanation
An error commonly made by developers
when validating brackets with regular expressions is only counting the number
of opening and closing brackets or checking for matching brackets without
considering the order. Brackets can be nested; therefore, it is important to
know which one was opened last to determine if the brackets are balanced
correctly. The string [)] has both a matched set of opening and closing
brackets; however, due to incorrect ordering, the parentheses within this
string are not considered to be properly nested/ordered.
⚙️Logic in Steps
1.
A
reference structure determines which closing bracket corresponds with each
opening bracket, enabling fast and accurate evaluations.
2.
An
internal stack keeps track of any opening brackets that still need to be
matched to closing brackets as the string is processed.
3.
Each
character in the string is checked individually to ensure that opening and
closing brackets are in the correct order per their defined rules.
4.
Once
an opening bracket has been added to the stack, it cannot be removed without
first locating its matching closing bracket.
5.
Whenever
a closing bracket is encountered, the last opening bracket added to the stack
will be checked against that closing bracket for validity.
6.
If
a closing bracket appears without a prior opening bracket being found, a safety
check will be made to identify this error.
7.
As
long as no mismatches are located along the way, at the end of the string, the
open brackets remaining in the stack must all have been matched by closing
brackets.
Time and Space Complexity
1. Time Complexity: O(n)
Each character in the string is processed exactly once.
2. Space Complexity: O(n)
In the worst case, all characters are opening brackets and are stored in the
stack.
📝 Key Takeaways from This
Problem and Solution
- Parentheses
validation requires maintaining the order of operations, not just balance.
- Stack
is the most efficient structure for handling nested and dependent
elements.
- Immediate
failure on mismatch improves efficiency and avoids unnecessary processing.
- Safe
handling of empty stack conditions prevents runtime errors.
- A
final validation step ensures complete correctness
2. Baseball Game
🧠 Explanation
One mistake
that's easy to make when calculating your "score" in real time is
that people often try to "just" make calculations that depend on the
history of all your prior scores. Often it doesn't work correctly because of
certain operations that depend on how your total score directly influences the
calculations of “C” (cancel), “D” (double), and “+” (adding the last two scores
together). These operations cannot be effectively completed without maintaining
a continuous score history.
⚙️Logic in
Steps
1.
The Stack class keeps track of valid scores in
the order they happen.
2.
Every operation will be executed in the order
it was called; therefore, this is a sequential process.
3.
When you enter a score, it will be added as a
number directly onto the stack.
4.
When you enter a double operation, it will
take the last element off the top of the stack (i.e., score) and put a double
(i.e., 2 times the score) on the stack.
5.
The cancel operation cancels the last score
entered by popping the last element off the stack.
6.
When you perform a sum operation, it will take
the last two numbers off the top of the stack and add them together, then push
that new score on to the top of the stack.
7.
A helper method will allow you to access the
top of the stack safely, preventing runtime exceptions when the stack is empty.
8.
Once all operations are completed, you can get
the total score by adding all scores (i.e., elements of the stack) together.
This method works well if you want to use a stack method to keep track
of history for undo/rollback features.
Time and Space Complexity
1.
Time Complexity: O(n)
Each operation is processed exactly once.
- Space Complexity: O(n)
The stack stores all valid scores generated during execution
📝Key Takeaways from This Problem and Solution
1.
Stacks are ideally suited for solving undo,
rollback, or dependency problems.
2.
Stacks are ideal because they maintain
history, so previous results can be preserved.
3.
With a stack, it is easy to access recently
added items in O(1) or constant time.
4.
Stack-based defensive checks improve the fault
tolerance of the system by preventing runtime errors and failures.
5.
Good separation of logic improves
maintainability and readability of code.
3.
Backspace String Compare
🧠 Explanation
When you process backspace inline by also comparing the strings
directly, this is often an error because backspace only deals with the last
character added. Without the previous state of the string, you will not be able
to accurately know which characters were affected by this change. This can
cause error in the actual comparisons of multiple backspaces, especially if
done in a row.
⚙️Logic in
Steps
1.
When dealing with each input string, backspace
effects are simulated independently from one another and each backspace affects
only the string from which it was applied to create a new string. The end
result is a string where, after processing all backspaces for that string, we
can see what characters are present as a result of our use of backspaces.
2.
The stack structure is what allows us to keep
track of what valid characters we ended up with, after applying backspace
operations.
3.
To add a regular character, you simply push
the character onto the stack.
4.
To remove a regular character when a backspace
character is seen, you will remove the last character added to the stack,
assuming there was one.
5.
If there was an additional backspace operation
applied to an empty stack at the very beginning, it has no effect.
6.
After processing both strings through the
stack structure, you will end up with two processed strings that can be
compared for equality.
Using the
stack structure, we can accurately simulate how backspace operations work in
modern text editors.
Time and
Space Complexity
1.
Time Complexity: O(n + m)
Each character in both strings is processed exactly once.
2.
Space Complexity: O(n + m)
Auxiliary stacks are used to store the processed characters for both strings.
📝Key Takeaways from This Problem and Solution
1.
Due to their LIFO nature, stacks implement
backspace operations very naturally.
2.
The necessity of maintaining history stems
from the possibility for recent actions to undo prior actions.
3.
By separating the processing logic from the
other components of the application, we increase clarity and support reuse.
4.
We use defensive checks to prevent invalid
removal requests and runtime errors from occurring during execution.
5.
The stack simulation will always provide valid
results for complex input patterns.
4. Implement Stack Using Queues
🧠 Explanation
A common mistake made by some developers is to use the queue's basic
operations, i.e. enqueue and dequeue, but not to alter their order at all.
Because of this, the queue behaves like a First In First Out (FIFO) queue,
whereas stacks follow a Last In First Out (LIFO) rule. Therefore, removing
elements from the front of the queue will always give you the oldest processed
item rather than the most recently added item. This leads to Stack
implementation errors.
⚙️Logic in
Steps
1.
A single queue is used to simulate stack
behavior while respecting queue constraints.
2.
Each new element is first added to the queue
using a normal enqueue operation.
3.
After insertion, the existing elements in the
queue are rotated to the back.
4.
This rotation ensures that the newly added
element moves to the front of the queue.
5.
By maintaining this order, the front of the
queue always represents the top of the stack.
6.
Removing an element from the queue directly
corresponds to popping from the stack.
7.
Accessing the front element allows retrieval
of the stack’s top without removal.
8.
An empty check is performed by verifying
whether the queue contains any elements.
This approach follows a push-heavy stack
simulation technique, where order correction is handled during insertion.
Time and Space Complexity
1.
Time Complexity: O(n)
a. Push operation: O(n) due to rotation
b. Pop operation: O(1)
c. Top operation: O(1)
d. Empty operation: O(1)
2.
Space Complexity: O(n)
The queue stores all stack elements.
📝Key Takeaways from This Problem and Solution
1.
Queues can behave like stacks by managing the
order in which elements are inserted.
2.
To have LIFO behaviour, the order of the
elements needs to be modified during the insertion process.
3.
The complexity of the "push"
operation can be shifted to allow the "pop" and "top"
operations to remain simple.
4.
An understanding of how to perform pop and top
operations is more important than how data structures are defined through their
storage formats.
5.
This concept illustrates that how data
structures function determines their classification, rather than their storage
method.
🟡 Medium Stack Problems (Pattern
Expanders)
5. Next Greater Element I
🧠 Explanation
The brute-force approach is an
example of a common mistake. In this approach, you search the entire set for
every element to find the closest bigger number. Because of this, the time
complexity will take O(n²) and won't work very well with large amounts of data
or at all! The second common issue with this method is that people do not take
into consideration how the elements are going to be compared with each other as
a result of the circular aspect of the array which leads to partial comparisons
and incorrect comparisons.
⚙️Logic in Steps
1.
The default value for all elements of the result
array is -1, indicating that the item has not yet been solved.
2.
A stack is used to keep track of the indexes of
the elements that do not have an associated next greater value.
3.
To achieve this circular behavior, the input
array is traversed twice without any changes to its original structure.
4.
In each traversal, the current element of the
array is checked against the element in the index at the top of the stack.
5.
If the current element is greater than the
element at the index on the top of the stack, then the current element becomes
the next greater element for that index.
6.
Once an index has been found to have its next
greater value, it will be removed from the stack.
7.
This process continues until either the stack is
empty or the current element is less than or equal to the current value in the
index at the top of the stack.
8.
To prevent duplicate computations, the only time
a new index will be pushed onto the stack is during the first traversal.
9.
Any index remaining in the stack after all
traversals have concluded will remain as -1 to indicate that no elements were
found that were greater than that index.
This method guarantees optimal performance because each
element is processed with the use of a monotonically decreasing stack.
Time and Space Complexity
1. Time
Complexity: O(n)
Each element is pushed onto and popped from the stack at most once, resulting
in linear time complexity despite the nested loop structure.
2. Space
Complexity: O(n)
Additional space is used for the stack and the result array.
📝Key Takeaways from This
Problem and Solution
1.
The brute force approach is straightforward, but
it becomes inefficient as the size of the input increases.
2.
Using a monotonic stack allows us to perform the
same comparisons without worrying about duplicates.
3.
Circular arrays are easily managed through
modular indexing.
4.
By storing indices instead of values, we can
assign results directly without intermediate variables.
5.
Amortization can change how we interpret nested
loops, which may not always be O(n²).
6. Daily
Temperatures
🧠 Explanation
One error many people make is
implementing a brute force solution where you search future days for a warmer
day for each day, leading to an O(n²) time complexity which does not scale well
when the input size is large. Another error people commonly make is making
their solution overly complex by using auxiliary data structures that are not
needed instead of utilizing the properties of the stack data structure.
⚙️Logic in Steps
1.
Initialize an answer array to hold the number of
days you’ll have to wait for a warmer temperature.
2.
To track days where you cannot find a warmer day
yet, use a stack to maintain the index of those days.
3.
The stack is maintained in a monotonic
decreasing temperature order.
4.
While processing through the array (left to
right), assume that each day can potentially be a warmer day for all previous
days than its index.
5.
If the temperature of today is greater than
yesterday (at the top of the stack), then we have found out how many days we
were waiting for a warmer day for the previous day.
6.
The difference in index values is the number of
days waited for the previous day to have a warmer temperature.
7.
After you figure out how many days to wait for
previous days, you remove the indexes from the stack.
8.
After today has been processed, you add the
index value of today back onto the stack.
9.
Any remaining days in the stack at this point
should all have a value of zero because a warmer day does not exist.
10. By
using the monotonic stack pattern, you can efficiently find the answer to many
waiting days at once.
Time and Space Complexity
1. Time
Complexity: O(n)
Each index is pushed to and popped from the stack at most once, ensuring linear
time complexity.
2. Space
Complexity: O(n)
Additional space is used for the stack and the result array.
📝Key Takeaways from This
Problem and Solution
1.
Brute-force approaches are obvious solutions but
are very low in efficiency.
2.
Monotonic Stack removes the need to compare
elements repeatedly.
3.
Storing the results using an index allows for
simplified result algorithms.
4.
This method will be frequently employed to
resolve range and comparison issues.
5.
Recognizing how the stack operates can help
improve the efficiency of array-related issues.
7. Evaluate Reverse Polish Notation
🧠 Explanation
Evaluating the expressions based
on infix rules or operator precedence is a common mistake. Instead of a simpler
evaluation process, this creates complexity in the logic and may produce
incorrect results. A common mistake is to use floor division to divide.
Dividing floor-style will give incorrect answers when negative numbers are
used. Therefore, Reverse Polish Notation must always be evaluated using a
stack, and does not use operator precedence.
⚙️Logic in Steps
1.
A stack is used for storing operands while
performing an operation with an expression. When evaluating an expression, each
token will be processed left to right.
2.
NUMBERS are treated as OPERANDS (pushed onto the
stack), and when we encounter an OPERATOR (e.g., +, -, *, / etc.), we pop the
TWO operands off the STACK and then perform the OPERATION (i.e. left to right).
3.
After performing the OPERATION, we push the
RESULT back onto the STACK.
4.
As DIVISIONS should be calculated towards zero,
thus careful consideration must be taken when handling DIVISION when using this
method.
5.
When all tokens have been processed, you will
end up with ONLY one value remaining on the Stack which will be your FINAL
Result of your Evaluation of the EXPRESSION.
6.
This method represents how POSTFIX EXPRESSION as
used by CALCULATOR and COMPILER are evaluated.
Time and Space Complexity
1. Time
Complexity: O(n)
Each token is processed exactly once.
2. Space
Complexity: O(n)
The stack stores operands during evaluation.
📝Key Takeaways from This
Problem and Solution
1.
Reverse Polish Notation is a natural fit for a
stack-based approach to the evaluation of expressions.
2.
In a postfix expression, operator precedence
does not exist.
3.
The order of operands must be maintained for the
correct operation of subtraction and division.
4.
A failure to handle division correctly will
introduce the possibility of subtle logical errors into the application.
5.
Code using the stack-based evaluation approach
is cleaner and more maintainable.
8. Mini Stack
🧠 Explanation
One of the more frequently made
errors while determining the minimum value of a Stack when using the getMin()
method is that the developer will iterate through the entire Stack each time
this method is called to find the lowest value within the Stack Collection.
This practice results in O(n) Time Complexity for getMin() and violates the
requirement of having a constant-time operation for this method. Another error
developers often make is they remove or clear the Stack to find the minimum. By
doing this, the data will no longer exist resulting in incorrect Stack
functionality.
⚙️Logic in Steps
1.
Two stacks are used to separate the different
functions of pushing and returning items from the stack.
2.
The first stack is the main stack for all values
that the user has pushed onto it.
3.
The second stack is a supporting stack that
stores the current minimum value at each point in time.
4.
When the user pushes a new item onto the primary
stack, the supporting stack will be updated only under certain conditions (if
the new value is less than or equal to the existing minimum value).
5.
If the user performs a pop operation and the
item removed equals the minimum value, then it will also be removed from the
supporting stack.
6.
The top operation will return the value of the
item that was pushed onto the main stack most recently.
7.
The minimum operation will return the value on
top of the supporting stack and will always be done in constant time.
8.
This design allows the user to have immediate
access to the minimum value without performing an entire scan of the main
stack.
Time and Space Complexity
1. Time
Complexity: O(1)
a. Push:
O(1)
b. Pop:
O(1)
c. Top:
O(1)
d. GetMin:
O(1)
2. Space
Complexity: O(n)
Additional space is used for the auxiliary minimum stack.
📝Key Takeaways from This
Problem and Solution
1.
Auxiliary data structure for constant-time
constraints.
2.
Creating separation of responsibilities improves
clarity and performance.
3.
Record keeping of historical minimums prevents
repeated calculations.
4.
Incrementation of stack problems might require
creative use of space and time in order to solve them.
5.
This pattern is common in real-life systems and
in interviews.
Conclusion
This blog discussed how
stack-based concepts grow from straightforward validation challenges into more
sophisticated optimization methods. Through eight distinct examples, we
understood how stacks can effectively record and sequence the history, arrangement,
or direct correlation of dependent objects within different contexts. Most
notably, through recognising patterns such as ‘monotonic stack’ and ‘auxiliary
stack’, we were able to derive optimal solutions to initially thought
‘brute-force’ techniques. Also, acquiring these patterns helps build a robust
platform for solving complex interview-level questions and the same level of
problem-solving ability in the workplace.
If you missed the Previous part
If you enjoyed reading this
stack-oriented tutorial, please take a moment to look at the foundation upon
which it is built on. In the last blog, I discussed array patterns and
solutions with practical experience, and real-world LeetCode examples. To review
that blog, please follow this link:
Array Problem-Solving Patterns Explained Using LeetCode Examples
Together, these two blogs form a
solid base for tackling core data structure problems with confidence.
Coming Up
In our upcoming blog we will
examine how queues can be utilized to solve problems through problem-solving
patterns based upon queues. This will include examples of the use of FIFO or,
First-In-First-Out, to schedule, buffer and manage processes or events
occurring in real-time. Additionally, we will examine a series of examples of
queue or deque based problems that arise both during technical interviews and
actual production cases. Once we've completed this discussion we will have a
solid foundation for approaching the more advanced patterns found in working
with data structures listed above.
Master Queue & Deque Problems in Python
Closing Whispers
Some problems wait, stacked quietly in time,
Answers rise when order starts to rhyme.
What we push today may return tomorrow’s key,
In layers of logic, clarity learns to be.
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