Monotonic Stack Data Structure Explained

Summary

This 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

00:00

🧩 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.

05:01

🔍 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.

Keywords

💡monotonic stack

A monotonic stack is a data structure that maintains a specific order of elements, either monotonically increasing or decreasing. In the context of the video, it is used to solve the problem of finding the next greater element in an array. The stack ensures that elements are processed in a way that maintains their order relative to each other, which is crucial for the efficiency of the algorithm discussed. The video explains how a monotonic stack can be used to achieve a time complexity of O(n), as opposed to the brute force method which would result in O(n^2).

💡greedy logic

Greedy logic refers to an algorithmic strategy where at each step, the locally optimal choice is made with the hope that these local optimizations will lead to a global optimum. In the video, the monotonic stack uses greedy logic to maintain its order, allowing for efficient problem-solving in coding interviews. The script illustrates this by explaining how the stack is used to keep elements in an orderly fashion, which is key to the algorithm's success.

💡monotonic increasing

Monotonically increasing is a term used to describe a sequence where each element is either greater than or equal to the previous one. The video uses this concept to explain how the stack can be ordered such that elements either increase or stay the same as you go from left to right. This is important for the algorithm because it allows for the quick identification of the next greater element.

💡monotonic decreasing

Monotonically decreasing is the opposite of monotonically increasing, where each element is either less than or equal to the previous one. The video script uses this concept to explain the organization of the stack from right to left, ensuring that the elements are in decreasing order. This is crucial for the algorithm to efficiently find the next smaller element relative to the current element.

💡Brute Force solution

A brute force solution is a problem-solving approach that checks all possible options until the correct one is found. It is often less efficient than more sophisticated methods. In the video, the script contrasts the brute force method, which would involve iterating through the entire rest of the array for each element, with the more efficient monotonic stack approach. The brute force method is mentioned to highlight the inefficiency of checking every possibility, which the monotonic stack avoids.

💡time complexity

Time complexity is a measure of the amount of time an algorithm takes to run as a function of the length of the input. The video emphasizes the importance of time complexity in algorithm design, particularly in the context of coding interviews. The script explains how using a monotonic stack reduces the time complexity from O(n^2) to O(n), which is a significant improvement in efficiency.

💡next greater element

The 'next greater element' problem is a common interview question where the goal is to find the next element in an array that is greater than the current element. The video script provides a detailed example of how to solve this problem using a monotonic stack. The concept is central to the video's theme, as it demonstrates the practical application of the stack in solving a real-world coding problem.

💡invariant

An invariant is a condition that remains unchanged throughout the execution of an algorithm. In the context of the video, maintaining the invariant of a monotonic stack is crucial for the algorithm's correctness. The script explains how the stack is manipulated to ensure that it remains monotonically decreasing, which is the invariant needed for the algorithm to function properly.

💡popping

Popping refers to the removal of an item from the top of a stack. In the video, popping is used to manage the stack's contents, ensuring that the stack remains monotonic. The script describes how elements are popped from the stack when they are no longer needed or when they violate the monotonicity condition, which is a key operation in the algorithm.

💡pushing

Pushing is the action of adding an item to the top of a stack. The video script explains how elements are pushed onto the stack to maintain the monotonicity condition. Pushing is essential for building the stack and ensuring that it contains the necessary elements to solve the 'next greater element' problem efficiently.

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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