Monotonic Stack Data Structure Explained
Summary
TLDRThis video introduces the monotonic stack, a powerful tool for problem-solving in coding interviews. It's a stack that maintains order, either increasing or decreasing, and is used to find the next largest height in a list. The script explains how a brute force approach would be inefficient, and instead, a monotonically decreasing stack is used to solve the problem in O(n) time. The video walks through an example with five people's heights, demonstrating how the stack works and why it's more efficient than checking every element. It concludes by explaining the code implementation and the stack's ability to store both position and magnitude of values, which is key to its efficiency.
Takeaways
- 📚 The monotonic stack is a data structure used for problem-solving in coding interviews, particularly useful for maintaining order in a stack with greedy logic.
- 🔄 It can be either monotonically increasing, where elements stay the same or increase from left to right, or monotonically decreasing, where elements stay the same or decrease.
- 👥 An example problem discussed is finding the height value of the next tallest person given an array of heights, with a brute force solution having a time complexity of O(n^2).
- 🔄 The monotonic stack approach iterates from right to left, maintaining a monotonically decreasing order to efficiently find the next greater element.
- 🔍 If the stack is empty when a new element is encountered, it means there is no greater element on the right, so negative one is assigned as the next greater height.
- 🧩 The stack invariant is maintained by popping elements that are less than or equal to the current element being considered, ensuring the stack remains monotonically decreasing.
- 💡 The top element of the stack represents the next greatest height greater than the current element, providing the solution for the current position.
- 💻 The code implementation involves looping from right to left, using a stack to keep track of the next greater elements, and pushing the current element onto the stack after processing.
- 🚀 The monotonic stack achieves a time complexity of O(n) by storing both the position and magnitude of values, which is a significant improvement over the brute force method.
- 🎯 The final implementation results in an efficient algorithm for the next greater element problem, demonstrating the power of the monotonic stack in solving such interview questions.
Q & A
What is a monotonic stack?
-A monotonic stack is a data structure that maintains a specific order of elements, either monotonically increasing or decreasing, and uses greedy logic to keep elements orderly as new elements are added.
How does the monotonic stack differ from a regular stack?
-A monotonic stack differs from a regular stack in that it enforces a specific order (increasing or decreasing) among its elements, whereas a regular stack follows a last-in, first-out (LIFO) principle without any order requirement.
What is the purpose of using a monotonic stack in coding interviews?
-In coding interviews, a monotonic stack is used to solve problems efficiently by leveraging its ordered structure to quickly find next greater elements, which can lead to solutions with better time complexity compared to brute force methods.
What is the time complexity of the brute force solution for finding the next greater element in an array?
-The time complexity of the brute force solution for finding the next greater element in an array is O(n^2), where n is the number of elements in the array.
How does the monotonic stack help in reducing the time complexity for finding the next greater element?
-The monotonic stack helps in reducing the time complexity by maintaining a decreasing order of elements, allowing for the next greater element to be found in constant time, thus achieving a linear time complexity of O(n).
What is the strategy for maintaining a monotonically decreasing stack in the given example?
-In the given example, the strategy for maintaining a monotonically decreasing stack is to pop elements from the stack if they are less than or equal to the current element being considered, ensuring that the stack always contains elements greater than the current element.
How does the monotonic stack handle the case when there is no greater element to the right?
-When there is no greater element to the right, the stack is empty, and the value assigned is negative one, indicating that there is no greater height on the right side of the current element.
What is the significance of the position and magnitude in the context of the monotonic stack?
-In the context of the monotonic stack, the position is significant as it determines the order of elements, and the magnitude is significant as it is used to determine the next greater element by popping elements from the stack.
Can you provide a step-by-step explanation of how the monotonic stack processes the example array of heights?
-The monotonic stack processes the example array of heights by iterating from right to left, maintaining a decreasing order. For each height, it pops elements from the stack that are less than or equal to the current height. If the stack is empty, it assigns negative one. Otherwise, the top element of the stack is the next greater height. Each height is then pushed onto the stack to maintain the decreasing order.
What is the final output of the monotonic stack after processing the example array of heights?
-The final output of the monotonic stack after processing the example array of heights is an array where each element represents the next greater height to the right or negative one if there is no greater height.
Outlines
🧩 Monotonic Stack for Problem Solving
This paragraph introduces the concept of a monotonic stack, a data structure used for problem-solving in coding interviews. It explains that a monotonic stack maintains a specific order of elements, either monotonically increasing or decreasing. The paragraph sets up an example where the goal is to find the next largest height among five people, given their heights. It contrasts a brute force solution with a more efficient approach using a monotonic stack. The stack is used to keep track of the heights in a decreasing order from right to left, allowing for quick identification of the next greater height without iterating through the entire array. The process involves pushing elements onto the stack while ensuring it remains monotonically decreasing and popping elements when they are less than or equal to the current height being considered. The example demonstrates how the stack operates with specific numbers to solve the problem efficiently.
🔍 Implementing the Monotonic Stack Algorithm
The second paragraph delves into the implementation of the monotonic stack algorithm for finding the next greater element in an array. It describes the process of iterating from right to left, using a stack to maintain the heights in a monotonically decreasing order. The algorithm involves popping elements from the stack if they are less than or equal to the current height being processed, ensuring that the stack always contains the next greater height for each element. If the stack is empty for a particular element, it means there is no greater height to the right, and the answer is recorded as negative one. The paragraph concludes by explaining the time complexity advantage of using a monotonic stack, which is O(n) compared to the O(n^2) of the brute force method. The explanation emphasizes how the monotonic stack efficiently stores both the position and magnitude of values, allowing for quick retrieval of the next greater element.
Mindmap
Keywords
💡monotonic stack
💡greedy logic
💡monotonic increasing
💡monotonic decreasing
💡Brute Force solution
💡time complexity
💡next greater element
💡invariant
💡popping
💡pushing
Highlights
Monotonic stack is a useful concept for problem-solving in coding interviews.
It uses greedy logic to maintain order in the stack.
The stack can be monotonically increasing or decreasing.
Example problem: find the height value of the next tallest person.
Brute force solution has a time complexity of O(n^2).
Iterate from right to left using a monotone stack.
Maintain a monotonically decreasing stack for the problem.
If stack is empty, assign negative one for no greater height.
Popping elements from the stack to maintain the invariant.
The stack stores greater values than the current height.
Push elements into the stack to maintain a decreasing order.
The top element of the stack is the next greater element.
The algorithm efficiently finds the next greater element in O(n) time complexity.
The monotonic stack stores both position and magnitude of values.
Position is stored by the order in the stack.
Magnitude is stored by popping elements off the stack.
The monotonic stack avoids the O(n^2) time complexity of the brute force approach.
Transcripts
possibly one of the most confusing yet
useful ideas for problem solving in
coding interviews is the monotonic stack
in this video we're going to show you an
example and summarize it so make sure to
stick to the end the monotonic stack is
as you might have guessed a stack which
uses some greedy logic to keep the
elements in the stack orderly after each
new element is put onto the stack what
kind of order are we talking about it
could be either monotonically increasing
which means the elements either increase
or stay the same as we go left to right
or monotonically decreasing which means
that it either decreases or stays the
same as we go left to right let's look
at an example given the heights of five
people find the height value of the
person with the next largest height if
there is no larger height store negative
one let's think about the Brute Force
solution first for each height iterate
through the rest of the array to find
out the next greater height the time
complexity would be o of n square but do
we really need to Loop through the
entire rest of the array we're going to
iterate from right to left and use a
monotone stack to keep the higher people
on your right in this case we're going
to be maintaining a monotonically
decreasing stack every time we put an
element into the stack we still want to
make sure that the stack is
monotonically decreasing we have two one
two four and three so of course once
again we're going to be moving from
right to left and we're going to have
our stack which maintains the greater
values than the current height so for
example if we were to deal with this
three here and this 3 pertains to this
person in the blue on the right here we
see that the stack is empty if the stack
is empty that basically means that okay
nothing is greater than three so far so
what that means is we can just assign
this value negative one which means that
the greatest height so far or the
greatest height on the right side of
this person is nothing meaning it's
negative one
so in that case let's just put the three
into our monotonic stack and remember
this is a decreasing monotonic stack
next let's deal with this number four
and as four pertains to this red person
here and so if we were trying to put in
this 4 here we're once again we're
trying to maintain a monotonically
decreasing stack meaning the smaller
elements are at the top and the larger
elements are at the bottom if we were
trying to put this 4 into here that
would break our invariant so what we
need to do is we need to take this 3 out
and then finally we see that the stack
is empty meaning that nothing is greater
than this 4 on the right side so
put the negative 1 here meaning that
nothing to the right is greater than
this form
so let's put this 4 into our monotonic
stack
next let's deal with this 2 here which
pertains to this green person
obviously we can see that the answer for
this is going to be 4 because the person
of height 4
is greater than this percent of height 2
and so what we can do is we realize okay
4 is the top of our stack which means
that this is the next greatest element
greater than this two so we can easily
just put the 4 here which is our answer
and we can put 2 into our monotonic
stack
because then when we deal with the next
person which is a person height one
we can see okay the person at the top of
the stack is the first person that's
greater than this right that's why we're
keeping these elements in here in a
monotonically decreasing order so let's
put this element into here and we see
that the top elements is a two so that's
our answer
for two
and we maintain our invariance which is
just going to be decreasing so
we can just put the one into our
monotonic stack
and finally let's deal with this yellow
person here which is of height 2. we
want to put this into our stack but we
want to find the first person that's
strictly greater than this 2 right so
what we're going to do is we want to
maintain the invariant so we take out
the one okay but this is still not
decreasing strictly decreasing so we
need to take this 2 out and then finally
we see that we can successfully put the
4 or this 2 inside and the top element
is going to be the four meaning that the
answer in this case is going to be 4.
so we put the 2 in
and that's our final stack we're done
here and we see that we finished all the
items and this is the answer for our
specific case
so what does the code look like for the
next greater element problem what our
function does is takes in a array of
heights and what we're going to do is
create an array of integers as well as a
monotonic stack this could just be
implemented as a regular stack
we're going to Loop through the elements
from the right side all the way to the
left side
and what we're going to say is while our
stock is not empty and the element at
the top of the stack is less than or
equal to the current height that we want
to insert then pop the stack
next what we want to do is check if the
current stack is empty if it is empty
then we just say that the answer is
negative 1 for the current element
because nothing is greater than it and
otherwise the element
at the top of the stack is going to be
the answer for the ith element
then simply we want to push our element
into the stack because we know it's
going to maintain the invariant
finally return the answer
the reason that all of this works is
because the monotonic stack stores both
the position and magnitude of the values
the position is stored by the order in
the stack and the magnitude is stored by
popping stuff off of the stack and
that's exactly why we're able to achieve
this o of n time complexity otherwise we
would have an O of N squared that's it
for the monotonic stack if you like this
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