Asteroid Collision

Identifying Problem Isomorphism

“Asteroid Collision” requires you to simulate collisions between asteroids moving in different directions, specifically using a stack to keep track of asteroids that have not yet collided.

A simpler problem that shares the same core concept of dealing with collisions in an array is “Trapping Rain Water”. The task in this problem is to calculate how much rain water can be trapped between heights represented in an array. It shares a similar use of stacks to keep track of ‘obstacles’ (in this case, the heights), though there are no collisions involved.

An approximately isomorphic problem is “Daily Temperatures”. The problem is to find out how many days you would have to wait until a warmer temperature, which conceptually can be thought of as collisions between different ’temperatures’. It also uses the concept of a stack in a similar way.

A more complex problem is “Maximal Rectangle”. It adds extra complexity by extending the stack-based approach to a two-dimensional grid instead of a one-dimensional array.

So, the list of problems from simplest to most complex are:

1. “Trapping Rain Water” - Calculate how much rain water can be trapped between heights.
2. “Asteroid Collision” - Simulate collisions between asteroids moving in different directions.
3. “Daily Temperatures” - Find out how many days you would have to wait until a warmer temperature.
4. “Maximal Rectangle” - Find the maximal rectangle in a grid.

10 Prerequisite LeetCode Problems

Asteroid Collision involves array manipulation and use of stacks to manage collisions. Here are some simpler problems to prepare for this one:

1. 20. Valid Parentheses: The problem involves recognizing open and close pair, a simpler problem that requires using a stack.

2. 155. Min Stack: This problem helps you understand the concept of using a stack to keep track of an extra piece of information (min value).

3. 739. Daily Temperatures: This problem involves using a stack to find the next greater number.

4. 844. Backspace String Compare: A problem about manipulating two strings, and a practice in understanding when and how to effectively use a stack.

5. 456. 132 Pattern: A problem where you maintain a stack while iterating through the array to find a pattern.

6. 503. Next Greater Element II: This problem requires finding the next greater number in a circular array, where a stack can be used to maintain possible candidates.

7. 901. Online Stock Span: This problem is about maintaining a decreasing stack while iterating through the array.

8. 71. Simplify Path: This problem involves using a stack to manage the directory path.

9. 496. Next Greater Element I: This problem requires using a stack to find the next greater number.

10. 402. Remove K Digits: It uses a stack to remove digits in order to create the smallest possible number.

Clarification Questions

When dealing with problem statements, it’s important to ask clarification questions to ensure you fully understand the problem and its constraints. Here are some questions that might be relevant for this problem:

1. What should be the output if all asteroids in the array explode? Should the function return an empty array?

2. Are all elements in the array unique in absolute value, or can there be asteroids with the same size (regardless of their direction)?

3. Can the array contain zero as an element, or are they strictly positive and negative integers?

4. In the case of an asteroid of size x moving to the right and another asteroid of size x moving to the left, we assume they meet and explode. Is this always true regardless of their positions in the array?

5. How should we handle edge cases, such as when the array contains only one asteroid or when all asteroids are moving in the same direction?

6. Are we assuming the speed of all asteroids to be equal? And is the speed constant throughout the scenario?

Remember that some of these questions might already be answered in the original problem statement, but are listed here for completeness. Always make sure to read and understand the problem statement thoroughly before starting to ask clarifying questions.

Identifying Problem Isomorphism

Identifying the isomorphism involves mapping the components of the problem to a more familiar or simpler problem, or a common data structure or algorithm. Here’s the isomorphism for this problem:

1. Asteroids: These are elements in the provided input array. Each asteroid is represented as an integer, where its absolute value represents its size, and its sign (positive/negative) represents its direction of movement.

2. Collisions: This is the interaction between two asteroids moving towards each other. In terms of data structure operations, this can be seen as comparing two elements.

3. Exploding of asteroids: If two asteroids collide, the smaller one explodes. If they’re of the same size, both explode. This can be seen as a removal operation from the array.

4. Same direction movement: Two asteroids moving in the same direction will never meet. This is similar to not having an interaction or comparison between two elements in the array.

5. Asteroid field: The state of asteroids after all collisions is analogous to the final state of the array after all interactions (collisions) are processed.

6. Stack data structure: To track the interactions between asteroids, we use a stack. Each asteroid is compared with the top of the stack, and based on this comparison (the collision), the state of the stack (asteroid field) changes.

This problem is a variation of balancing symbols/parentheses problem, where we use a stack to check the balance of symbols. Here instead of checking for balance, we are checking for collisions and accordingly modifying our stack (the asteroid field).

The Asteroid Collision problem maps to the classical problem of balancing symbols, particularly parentheses, which is quite similar to the “Valid Parentheses”.

In the “Valid Parentheses” problem, we have to check whether the given string containing characters ‘(’, ‘)’, ‘{’, ‘}’, ‘[’ and ‘]’ is valid. A string is considered valid if:

1. Open brackets must be closed by the same type of brackets.
2. Open brackets must be closed in the correct order.

We solve this problem by using a stack to keep track of the opening brackets and whenever we encounter a closing bracket, we check the top of the stack to see if it matches with the closing bracket. If it does, we pop the top of the stack; if it doesn’t, or if the stack is empty, we return false. In the end, if the stack is empty, we return true (indicating a balanced string); if it’s not, we return false.

Similarly, in the Asteroid Collision problem, we use a stack to keep track of the asteroids. When a new asteroid is encountered, it is compared with the top of the stack. Depending on the result of the comparison (which mimics the asteroid collision), we manipulate the stack accordingly. The final state of the stack is our solution. The main difference is that while the “Valid Parentheses” problem focuses on balancing, the Asteroid Collision problem is more about processing collisions and determining the final state after all collisions.

The Asteroid Collision problem can be considered a variation of the classic Balancing Symbols problem, although it might not be immediately apparent.

The Balancing Symbols problem is a classic problem where you have a sequence of symbols, such as parentheses, and you want to determine if they are “balanced”. That is, for every opening symbol, there should be a corresponding closing symbol. One common way to solve this problem is by using a stack data structure.

In the case of the Asteroid Collision problem, the “opening” and “closing” symbols are replaced by positive and negative integers respectively, with positive integers moving right and negative integers moving left. We can imagine the problem as trying to determine if the movements of the asteroids are “balanced”, similar to the parentheses in the Balancing Symbols problem. That is, for every right-moving asteroid, is there a corresponding left-moving asteroid that would collide with it, and vice versa.

Therefore, the stack can be used in a similar way: When we encounter a right-moving asteroid (positive integer), we push it into the stack. When we encounter a left-moving asteroid (negative integer), we compare it with the top of the stack. If the left-moving asteroid is larger, we pop the right-moving asteroid from the stack (analogous to matching and removing a pair of opening and closing symbols), and continue the process until the stack is empty or the top of the stack is a left-moving asteroid.

Thus, while the problems are different in their specific details, they share a common structure that allows us to apply similar problem-solving strategies. It’s this underlying structure that allows us to identify the Asteroid Collision problem as a variation of the Balancing Symbols problem.

Problem Analysis and Key Insights

Analyzing the problem statement reveals several key insights:

1. Movement and Collisions: The movement of asteroids and the potential for collisions is dictated by their sign (direction) and magnitude (size).

2. Significance of Order: The order of the asteroids in the array is significant. It represents the relative positioning of the asteroids, which directly impacts which asteroids will collide.

3. Exploding of Asteroids: The rules of the collisions (the smaller one explodes, or both explode if they’re of the same size) indicate that after a collision, either none, one, or both of the colliding asteroids might survive.

4. Non-interacting Asteroids: Asteroids moving in the same direction will not interact with each other. This means that for an asteroid moving to the right, it only needs to worry about the asteroids to its right that are moving to the left, and vice versa.

5. Problem Classification: Based on the rules for asteroid interaction and movement, this problem naturally maps to a stack-based solution. Each asteroid can be pushed onto the stack, and if a collision happens (an asteroid moving to the right meets an asteroid moving to the left on top of the stack), it will interact with the asteroid on top of the stack according to the collision rules.

6. Constraint Consideration: The size of the array (up to 10^4) implies that an efficient solution is needed. A brute force approach would likely exceed time limits, reinforcing the idea that a stack-based solution (which can operate in linear time) is a good approach.

These insights help us understand how the problem operates and provide guidance towards formulating a solution, especially with the realization that we can use a stack to track and manage collisions.

Problem Boundary

The scope of the Asteroid Collision problem involves:

1. Input: The input is an array of integers, where each integer represents an asteroid. The absolute value of the integer represents the size of the asteroid, and the sign of the integer (positive/negative) represents the direction of the asteroid. The constraints on the input indicate that the length of the asteroids array will be between 2 and 10^4, and each asteroid’s size will be between -1000 and 1000, excluding 0.

2. Processing: The problem involves simulating the movement and potential collision of the asteroids. If two asteroids collide, the one with the smaller size explodes and is removed from the array. If both asteroids are of the same size, both explode and are removed. Asteroids moving in the same direction won’t collide.

3. Output: The output of the problem is the final state of the asteroid field, represented as an array of integers. This array represents the asteroids that survived all the possible collisions.

4. Constraints: The problem should be solved under the constraints that are specified, namely the sizes of the array and the asteroids. Given the size of the array can be up to 10^4, the solution needs to be reasonably efficient and likely better than a quadratic time complexity solution.

The problem does not cover persisting data, user interfaces, network operations, or other aspects that might be included in a full software application. The main focus is on the algorithmic and data structure aspects of the problem.

The boundaries of a problem are determined by the inputs, outputs, and the specific rules and constraints defined in the problem statement. For the Asteroid Collision problem, the boundaries can be established as follows:

1. Inputs: The input is an array of integers, where each integer represents an asteroid. The sign of the integer indicates the direction of the asteroid (positive for right, negative for left), and the absolute value indicates the size of the asteroid. The size of the array is limited to 2 <= asteroids.length <= 10^4, and each asteroid’s size is between -1000 and 1000, excluding 0.

2. Outputs: The output is also an array of integers, representing the final state of the asteroid field after all collisions have occurred. The output array’s size could range from 0 (if all asteroids collide and explode) to the size of the input array (if no asteroids collide).

3. Rules/Constraints: The problem’s boundaries are further defined by the rules of the asteroid collisions. If two asteroids collide, the smaller one will explode. If they are the same size, both will explode. Two asteroids moving in the same direction will never meet. These rules limit the types of operations and manipulations that can be performed on the input array.

4. Performance Constraints: Given the maximum possible size of the input array (up to 10^4), the solution needs to be efficient. A solution with a time complexity of O(n^2) or worse might not run within acceptable time limits for the maximum size inputs.

By clearly defining these boundaries, you can focus on developing a solution that operates within them. This helps to prevent overengineering the solution or considering scenarios that are beyond the scope of the problem.

Problem Classification

The problem at hand is a classical data structure problem in the Computer Science domain. It involves array manipulation and simulates a real-world situation where events (asteroid collisions) alter the state of the array.

Let’s list the ‘What’ components:

1. The problem provides an input array named asteroids, where each integer represents an asteroid.
2. The absolute value of each integer represents the asteroid’s size, and the sign represents its direction (positive means right, negative means left).
3. Asteroids move at the same speed and can collide. The result of a collision is determined by comparing the sizes of the asteroids: the smaller one explodes, and if both are of the same size, both explode.
4. The task is to return the state of the asteroids array after all possible collisions have taken place.

Based on these components, the problem can be further classified as a Stack-based problem. The rationale behind this is the nature of the interactions: each asteroid can only interact with the next one in its path, which suggests using a stack. When a new asteroid is considered, it’s compared (collides) with the topmost asteroid in the stack. The outcomes of the collision then determine whether the new asteroid gets pushed to the stack, the topmost asteroid in the stack gets removed, or both. This process continues until all asteroids have been processed and there are no more possible collisions.

This is, therefore, a problem of array manipulation using Stack data structure, involving conditions and comparisons, in the broader Computer Science and Programming domain.

Distilling the Problem to Its Core Elements

Fundamental Concept: The fundamental concept this problem is based upon is the use of a stack data structure to simulate and manage interactions in a sequence of elements. It also involves principles of array manipulation and comparison-based operations.

Simple Description: If I were to describe this problem to someone unfamiliar with the subject, I’d say, “Imagine a row of asteroids moving in space, some towards the right, others towards the left. If two asteroids moving in opposite directions collide, the smaller one explodes, or if they’re the same size, they both explode. If they move in the same direction, they just pass by each other. Your task is to figure out which asteroids survive after all possible collisions.”

Core Problem: The core problem we are trying to solve is determining the final state of the asteroid field after all possible collisions have taken place, given the initial state and the rules of collision.

Key Components: The problem can be broken down into these key components:

1. Understanding the state of the asteroids (direction and size) from the input array.
2. Simulating the collisions based on the rules provided.
3. Keeping track of the surviving asteroids.

Minimal Set of Operations: The minimal set of operations needed to solve this problem would involve:

1. Iterating through the array of asteroids.
2. For each asteroid, checking whether a collision will occur (i.e., a positive (right-moving) asteroid on the stack and the current asteroid is negative (left-moving)).
3. If a collision happens, comparing the sizes and deciding which asteroid(s) explode.
4. Updating the stack according to the result of the collision.
5. At the end of the iteration, the remaining asteroids in the stack represent the final state after all collisions.

By following these operations, we simulate the collisions and effectively solve the problem.

Visual Model of the Problem

To visualize the problem, you can consider the asteroids as objects in a linear array. The positive integers are asteroids moving right, while negative integers are asteroids moving left. Here’s an example:

For the given array: [5, 10, -5]

You can visualize it as:

Right >     5     10    -5   < Left

The direction of the arrow indicates the direction in which the asteroids are moving. The -5 asteroid moving leftward will collide with the 10 asteroid moving rightward, because they’re moving in opposite directions. According to the collision rule, the smaller asteroid will explode. So, -5 will explode and be removed:

Right >     5     10   < Left

The remaining asteroids, 5 and 10, are moving in the same direction, so they will never collide, and thus, they both survive.

Using such visualization helps to understand the movement and possible collision of the asteroids. It also makes the processing of the array and the updating of the state after each collision easier to comprehend.

When thinking about the data structure, you can visualize a stack that keeps track of the surviving asteroids. Initially, the stack is empty. As you iterate through the array, asteroids are pushed into the stack. When a collision happens, asteroids are removed from the stack according to the collision rules. The final state of the stack represents the surviving asteroids after all collisions.

Problem Restatement

We’re given a sequence of asteroids, each represented by an integer in an array. Each asteroid’s size is the absolute value of the integer, and its direction is the sign (positive means it’s moving to the right, negative means it’s moving to the left). All asteroids move at the same speed.

The task is to figure out what asteroids would remain after all possible collisions have occurred. When two asteroids collide (which happens when a right-moving asteroid meets a left-moving asteroid), the smaller one will explode and disappear. If they’re the same size, both will explode and disappear. Asteroids moving in the same direction never collide.

The constraints on the problem are that the array’s length is between 2 and 10,000, each asteroid’s size (absolute value of the integer) is between 1 and 1000, and an asteroid can’t have a size of 0.

This means we need to devise a way to process the array following these rules and constraints, and produce the final state of the asteroid field as an array of integers.

Abstract Representation of the Problem

We’re given a sequence of objects represented by integers, where the sign represents their direction and the absolute value represents their size. These objects are moving towards each other at the same speed.

The task is to find the final sequence of objects after all possible interactions have occurred. When a right-moving object encounters a left-moving object, the smaller one is removed from the sequence. If they’re equal in size, both are removed. Objects moving in the same direction do not interact.

The constraints are that the sequence’s length is between 2 and 10,000, and the size of each object (absolute value of the integer) is between 1 and 1000, with no object having a size of 0.

The goal is to create a method to process the sequence following these rules and constraints, and return the final sequence of objects.

This abstract representation removes the specific context of asteroids and space, but keeps the essential structure and key elements of the problem. It emphasizes the interactions between objects in a sequence and the rules governing those interactions, and could potentially apply to many different situations or domains.

Terminology

There are a few specialized terms and technical concepts that are crucial for understanding this problem:

1. Array: An array is a data structure that contains a group of elements. In this problem, the array is used to represent the row of asteroids, with each element representing an asteroid’s size and direction.

2. Stack: A stack is a linear data structure which follows a particular order in which operations are performed. The order is typically LIFO (Last In First Out), where the last element added is the first one to be removed. In this problem, a stack can be used to keep track of the asteroids and handle the collisions in a systematic and efficient way.

3. Collision: In the context of this problem, a collision happens when a right-moving asteroid meets a left-moving asteroid. Depending on the sizes of the two asteroids, either one or both of them will be removed from the array.

4. Exploding: In this problem, an “exploding” asteroid is simply one that is removed from the array due to a collision.

5. Absolute Value: The absolute value of a number is its numerical value without regard to its sign. In this problem, the absolute value of an integer represents the size of an asteroid.

Understanding these concepts is crucial for understanding the problem statement and implementing a solution. The interaction and movement of the asteroids, as well as the handling of collisions, is governed by these concepts.

Problem Simplification and Explanation

Imagine you’re watching a one-way road from a bridge. Cars are entering the road from both ends and moving towards each other at the same speed. Each car has a unique number painted on it, where the number’s absolute value indicates the car’s size, and the sign of the number indicates the direction the car is moving (positive for right, negative for left).

Now, the rules of this unusual road are as follows:

1. If two cars are moving in the same direction, they’ll just pass by each other, regardless of their sizes.
2. If two cars are moving in opposite directions and they meet, the smaller car vanishes instantly. If they’re the same size, both cars vanish.

As an observer, your task is to note down the cars that will remain on the road after all possible meetings have taken place.

In simpler terms:

• You have a list (array) of cars (asteroids).
• Each car has a size and a direction (represented by an integer’s absolute value and sign, respectively).
• Cars going in the same direction will never meet (asteroids moving in the same direction never collide).
• When a car going right meets a car going left (collision), the smaller one vanishes (explodes), or both vanish if they’re the same size.

The key concepts involved are:

1. Array: The list of cars.
2. Direction: The way each car is moving (right or left).
3. Size: The size of each car, which determines what happens when two cars meet.
4. Meeting (Collision): The event where a right-moving car meets a left-moving car.
5. Vanishing (Exploding): When a car (or both cars) disappears as a result of a meeting.

These concepts interact to form the rules of the problem and to determine the final state of the road (array).

Constraints

There are several characteristics and conditions in the problem statement and constraints that can be exploited to find an efficient solution:

1. Direction of Movement: The fact that asteroids moving in the same direction don’t collide can be used to our advantage. This means we don’t need to check for collisions between asteroids moving in the same direction. We only need to handle collisions between a right-moving asteroid (positive number) and a left-moving asteroid (negative number).

2. Order of Processing: Because the asteroids are processed in a sequence (from left to right in the array), this gives a natural way to use a stack data structure, where the top of the stack always holds the last asteroid that might still collide with a future asteroid.

3. Size Comparison in Collisions: When a collision happens, the smaller asteroid always explodes, or both explode if they’re the same size. This is a simple comparison operation that can be efficiently performed.

4. Specific Numerical Range: The size of asteroids is within the range of 1 to 1000 (since the asteroid size can’t be 0). This fixed range may not directly help in manipulating the data but it prevents overflow or other related issues, making the implementation easier and more straightforward.

5. Constraints on Array Length: The array length is constrained between 2 and 10,000. This informs us that our solution should be at least O(n) or ideally better in order to be efficient for larger inputs.

Looking for these kinds of patterns and characteristics can help us in devising a more efficient solution by allowing us to make assumptions or simplify certain parts of the problem.

Analyzing the constraints gives us several key insights:

1. Problem Size: The size of the asteroids array is between 2 and 10,000. This means that our solution needs to be efficient for up to 10,000 elements. A solution with time complexity of O(n) or O(n log n) should be feasible, but anything with a higher time complexity might not perform well.

2. Range of Values: Each asteroid’s size is represented by an integer between -1000 and 1000, excluding 0. This means we’re dealing with a fixed and manageable range of values, which should simplify comparisons. The non-zero constraint means that we don’t have to worry about edge cases where the size of an asteroid might be 0.

3. Significance of Sign: The positive and negative signs of the integers don’t just represent direction; they also help in managing collisions. A positive (right-moving) asteroid on the stack can only collide with a negative (left-moving) asteroid, and the stack will only have positive numbers at any point. This helps simplify the logic for collision handling.

4. No Simultaneous Collisions: There’s an implied constraint that only two asteroids can collide at a time, as asteroids move at the same speed. This simplifies the problem, as we don’t have to consider complex scenarios where multiple collisions could occur simultaneously.

Understanding these constraints allows us to make some assumptions, simplifies the logic we have to implement, and helps guide us towards a more efficient solution.

Case Analysis

Let’s look at some different categories of test cases:

1. Category: Single Collision

Input: [8, -5]

Output: [8]

Explanation: There is only one collision, the 8 and -5 collide and the smaller asteroid, -5, explodes. Therefore, only 8 is left.

2. Category: Multiple Collisions, Same Size Asteroids

Input: [1, -1, 2, -2, 3, -3]

Output: []

Explanation: In each pair of asteroids, the two asteroids are of the same size but different directions. So they will all collide and explode, leaving no asteroids.

3. Category: Multiple Collisions, Different Size Asteroids

Input: [3, 2, -1, -2, -3]

Output: [3]

Explanation: The 2 and -1 collide resulting in 2. The 2 and -2 collide resulting in nothing. Finally, the 3 and -3 collide resulting in 3. So, 3 is the only asteroid left.

4. Category: No Collisions

Input: [4, 5, 6, 7]

Output: [4, 5, 6, 7]

Explanation: All asteroids are moving in the same direction, so there are no collisions and all asteroids survive.

5. Category: Edge Case, Minimum Input Size

Input: [1, -1]

Output: []

Explanation: The input size is the minimum possible (2 asteroids). The two asteroids are of the same size but different directions, so they collide and both explode, leaving no asteroids.

6. Category: Edge Case, All Same Asteroids

Input: [5, 5, 5, -5, -5, -5]

Output: []

Explanation: All the asteroids are of the same size. Regardless of the direction, all asteroids will collide with each other and explode, leaving no asteroids.

By examining a variety of categories and edge cases, we can ensure our solution handles all possible scenarios and nuances of the problem.

Visualizing the different scenarios can be very helpful to understand how asteroids are colliding and how the array evolves over time. Here are the visualizations for the test cases we discussed:

1. Category: Single Collision

[8, -5]

Initial state:

8 -> <- -5

After collision:

8 ->

2. Category: Multiple Collisions, Same Size Asteroids

[1, -1, 2, -2, 3, -3]

Initial state:

1 -> <- -1, 2 -> <- -2, 3 -> <- -3

After all collisions, all pairs explode and nothing is left.

3. Category: Multiple Collisions, Different Size Asteroids

[3, 2, -1, -2, -3]

Initial state:

3 ->, 2 ->, <- -1, <- -2, <- -3

After first collision:

3 ->, 2 ->, <- -2, <- -3

After second collision:

3 ->, <- -2, <- -3

After third collision:

3 ->, <- -3

After final collision:

3 ->

4. Category: No Collisions

[4, 5, 6, 7]

Initial state:

4 ->, 5 ->, 6 ->, 7 ->

There are no collisions and all asteroids survive.

5. Category: Edge Case, Minimum Input Size

[1, -1]

Initial state:

1 -> <- -1

The two asteroids collide and both explode, leaving no asteroids.

6. Category: Edge Case, All Same Asteroids

[5, 5, 5, -5, -5, -5]

Initial state:

5 ->, 5 ->, 5 ->, <- -5, <- -5, <- -5

All the asteroids collide with each other and explode, leaving no asteroids.

The arrows show the direction of each asteroid, and each step represents a unit of time in which the asteroids move and collisions happen.

Analyzing different cases provides us with several important insights:

1. Directional Movement: Asteroids moving in the same direction will never collide, regardless of their size. This can significantly reduce the number of potential collisions we need to consider.

2. Size Matters: In collisions, the smaller asteroid (in terms of absolute value) will always explode. If they are the same size, both will explode. This insight is key in determining the state of the asteroids after all collisions.

3. Sequential Processing: The asteroids are processed in a sequence (from left to right in the array). This points towards the potential use of a stack-based solution, as the problem naturally follows a Last-In-First-Out (LIFO) pattern.

4. Multiple Collisions: An asteroid can be involved in multiple collisions. For example, in the case of [3, 2, -1, -2, -3], the 2 and -1 first collide, after which the resulting 2 and -2 collide. Understanding this can help in implementing the correct looping and comparison logic in our solution.

5. Edge Cases: Cases where there are no asteroids (after all have exploded), or where no asteroids collide at all (all moving in the same direction) need to be handled correctly in the solution.

Understanding these insights can guide us to an efficient and correct solution to the problem.

Identification of Applicable Theoretical Concepts

This problem draws heavily from a fundamental data structure concept in computer science: the stack. Stacks are Last-In-First-Out (LIFO) structures, meaning the most recent item added (pushed) is the first one out (popped).

Let’s see how this can be applied to simplify the problem:

1. Stack for Directional Movement: The problem is based on the movement of asteroids in a particular direction. In our scenario, the stack can be used to keep track of all the asteroids moving in the right direction (positive numbers). When we encounter an asteroid moving in the left direction (negative number), we can check the asteroid on the top of the stack for a potential collision.

2. Stack for Size Comparison: The size comparison can be handled efficiently by a stack. While dealing with a left-moving asteroid, we can keep popping the stack until we find an asteroid larger than the current one, or the stack is empty (which would mean the current asteroid survives), or the top asteroid and the current asteroid are the same size (in which case both explode).

3. Loop Invariant Concept: This concept from algorithm theory can be used to ensure the correctness of the solution. The loop invariant would be that at any point, the stack contains only the asteroids moving in the right direction and any upcoming asteroid moving in the left direction can only collide with the asteroid at the top of the stack.

4. Efficient Array Traversal: From an algorithmic perspective, the efficient way to traverse the array of asteroids is to do it only once, which points towards a linear time complexity solution (O(n)). As we process each asteroid, we use the stack to keep track of potential collisions.

The combination of these computer science and mathematical concepts can help to derive a clean and efficient solution to this problem.

Simple Explanation

Let’s imagine you’re playing a game.

In this game, you have some toy cars on a table. Some cars are moving to the right and some are moving to the left. All the cars are moving at the same speed.

Now, the fun part begins when two cars are moving towards each other and they crash! The rule of this game is that if a bigger car crashes into a smaller car, the smaller car will break and be removed from the game. If two cars of the same size crash into each other, both cars break and are removed from the game.

Also, if the cars are moving in the same direction, they will never crash into each other because they all are moving at the same speed. So, a small car moving ahead of a big car is safe.

The question now is, after all the crashes are over, which cars are left on the table?

This game represents our problem. The toy cars are like our asteroids. The direction the cars are moving is like the sign of the numbers (positive for right, negative for left), and the size of the cars is like the absolute value of the numbers.

So, the problem is to figure out which cars (or asteroids) would remain after all possible crashes (or collisions) have happened.

Problem Breakdown and Solution Methodology

I’ll use the metaphor of cars moving on a table from my previous explanation to illustrate how to approach this problem.

1. Step 1: Begin with the First Car: We’ll start from the left end of the table. We move from car to car, from left to right.

2. Step 2: Add Right-Moving Cars to the Stack: Whenever we see a car moving to the right, we add it to our stack (like creating a line of cars waiting at a traffic signal).

3. Step 3: Process Left-Moving Cars: When we see a car moving to the left, that’s where things get interesting, as this car might collide with the cars moving to the right (the ones in our stack).

   • If the left-moving car is smaller than the car on top of the stack, it will break and we discard it (as per our game rules).
   • If the left-moving car is the same size as the car on top of the stack, both cars break and we discard both of them. We then remove the car from the top of the stack.
   • If the left-moving car is bigger than the car on top of the stack, the car on the top of the stack will break and we remove it from the stack. But the left-moving car is still in the game and can collide with the next car in the stack. So, we repeat the collision process for this left-moving car and the next car in the stack.

   We keep repeating this process until either our left-moving car breaks or there are no more cars in the stack for it to collide with.

4. Step 4: Repeat until all Cars are Processed: We continue this process for all cars on the table.

By the end of this process, our stack contains all the surviving cars (asteroids in our original problem) after all possible collisions. The order of the cars in the stack represents their final positions from left to right on the table.

Let’s illustrate this approach with an example: asteroids = [5, 10, -5]

1. We start with the first asteroid (5), which is moving to the right. We add it to our stack.
2. We then encounter the next asteroid (10), which is also moving to the right. We add it to our stack. The stack now is [5, 10].
3. Next, we see an asteroid (-5) moving to the left. It collides with the top of the stack (10). Since -5 is smaller than 10, it breaks and we discard -5. The stack remains [5, 10].
4. As there are no more asteroids, we finish. The remaining asteroids are those in the stack: [5, 10].

The order of the remaining asteroids is the same as the original order, because they never collide due to moving in the same direction.

The overall solution complexity of this approach is linear, i.e., O(n), where n is the number of asteroids. This is because we process each asteroid once, and each operation (pushing or popping from the stack) is a constant time operation.

Inference of Problem-Solving Approach from the Problem Statement

There are a few key terms and concepts that are central to this problem:

1. Asteroids: Each asteroid is represented by an integer in the array. The absolute value of the integer indicates the size of the asteroid and the sign indicates its direction (positive for right, negative for left). When it comes to diagrams, you can represent asteroids as arrows (showing direction) with numbers (showing size).

2. Direction of Movement: Asteroids move either to the left or right. Asteroids moving in the same direction won’t collide. This is important as it indicates that we need to keep track of asteroids moving in one direction (right, in our case) until we encounter an asteroid moving in the opposite direction.

3. Collisions: Two asteroids collide if they move towards each other. In a collision, the smaller asteroid (in absolute value) explodes, or both explode if they are the same size. This means we need to handle cases where we need to compare the sizes of two asteroids.

4. Stack: This is a Last-In-First-Out (LIFO) data structure that is perfect for this problem. As we encounter asteroids, we can push the right-moving asteroids into the stack and pop the stack when we encounter a left-moving asteroid. This also informs our strategy to approach the problem in a sequential manner.

Visualizing this problem with diagrams would involve arrows (for asteroids) moving in two directions. When two arrows point towards each other, that’s a potential collision. A stack can be visualized as a vertical box where you add elements from the top and also remove them from the top.

To illustrate this, consider the example: [5,10,-5]

Here’s a possible visualization:

Right-moving asteroids (stack): |10|
                                 |5 |
                                 ---
                                  
Left-moving asteroid: -5 (going towards the stack to collide)

This kind of visualization can help you keep track of the asteroids and understand the process of pushing to and popping from the stack. It also makes it easier to see which asteroids collide and which ones survive.

The stack data structure is a Last-In-First-Out (LIFO) structure, which means the last element that’s added (pushed) to the stack is the first one to be removed (popped) from it.

The reason a stack seemed applicable to this problem is due to the behavior of the asteroids described in the problem statement:

1. Directional Movement: Asteroids are moving in two opposing directions, and only those moving in opposite directions can collide. This suggested the need for a structure that could hold on to asteroids moving in one direction (say right), until an asteroid moving in the opposite direction (left) is encountered.

2. Sequential Collisions: The way the asteroids collide is sequential. An asteroid moving in one direction can only collide with the last asteroid moving in the opposite direction that it encounters. This is exactly how a stack operates, dealing with the most recently added element first.

3. Surviving Asteroids: We’re asked to return the state of the asteroids after all collisions. This means we need to keep track of the asteroids that haven’t collided, which a stack can easily do. At the end of the process, the asteroids remaining in the stack are exactly those that survive.

Given these characteristics, a stack seems a natural fit for managing the asteroids and collisions. The use of a stack in this context is not explicitly mentioned in the problem statement, but is inferred based on the nature of the problem and the properties of a stack data structure.

Simple Explanation of the Proof

Let’s go through the working of the algorithm using simple concepts:

In this problem, we have a row of asteroids, each moving either to the right or left. If two asteroids are moving in the same direction, they will never meet and thus will never collide. However, when an asteroid moving to the right meets an asteroid moving to the left, they will collide.

Our goal is to determine the final state of all the asteroids after all possible collisions. The important insight is that the sequence of these collisions does not change the final outcome. Whether we resolve collisions at the start of the row first or the end of the row, the final result remains the same.

This is where the concept of a stack helps. We can start from the beginning of the array and use the stack to keep track of the asteroids moving to the right. Whenever we meet an asteroid moving to the left, we check the top of the stack (which represents the most recent right-moving asteroid).

If the top of the stack (right-moving asteroid) is smaller than the absolute value of the current left-moving asteroid, it means the right-moving asteroid will be destroyed in the collision, and we remove it from the stack. We then compare the left-moving asteroid with the next asteroid in the stack, repeating the process until the left-moving asteroid is destroyed or there are no more right-moving asteroids in the stack.

If the top of the stack is equal to the absolute value of the left-moving asteroid, it means both asteroids destroy each other, and we remove the top of the stack and discard the left-moving asteroid.

If the top of the stack is larger than the absolute value of the left-moving asteroid, it means the left-moving asteroid is destroyed, and we simply discard it.

In the end, the stack will contain the asteroids that survive after all collisions.

Let’s take a quick example:

Consider the array of asteroids [5, 10, -5].

We start with the first asteroid 5 and add it to the stack as it’s moving to the right.

Next, we add the asteroid 10 to the stack as it’s also moving to the right.

When we reach the asteroid -5 (moving to the left), we see that it collides with the asteroid on top of the stack, which is 10. Since 10 is larger, the -5 asteroid gets destroyed and we discard it.

So, our stack, in the end, is [5,10] which are the asteroids that survive, and is the output of our algorithm.

This is the working of our algorithm and its correctness relies on the idea that the sequence of collisions does not change the final outcome. Therefore, we can resolve them in the order we encounter them, which makes our stack-based approach valid and correct.

Stepwise Refinement

1. Stepwise Refinement:

Let’s break down the problem into more granular steps:

• Start by creating an empty stack which will hold the asteroids moving in the right direction.
• Iterate over the given array of asteroids.
  • If the current asteroid is moving to the right, push it into the stack.
  • If the current asteroid is moving to the left:
    • Continue to do the following while the stack is not empty and the top asteroid in the stack is smaller than the absolute value of the current left-moving asteroid: remove (pop) the top asteroid from the stack.
    • If the stack is now empty or the top asteroid in the stack is larger than the absolute value of the current asteroid, move on to the next asteroid.
    • If the top asteroid in the stack is equal to the absolute value of the current asteroid, remove the top asteroid from the stack and move on to the next asteroid.
• Once we’ve processed all the asteroids, the remaining asteroids in the stack are the ones that survived.

2. Granular, Actionable Steps:

We can further refine our steps into more concrete actions:

• Initialize an empty stack s.
• Loop i from 0 to the length of the asteroids array:
  • If asteroid[i] > 0, push it into s.
  • Otherwise (asteroid[i] < 0):
    • While s is not empty and the top element of s is less than the absolute value of asteroid[i], pop from s.
    • If s is empty after the above step or the top element of s is greater than the absolute value of asteroid[i], break the loop and continue to the next asteroid.
    • If the top element of s is equal to the absolute value of asteroid[i], pop from s and continue to the next asteroid.
• Return the asteroids left in the stack s.

3. Independent Parts:

The problem can be split into the following independent parts:

• Determining the direction of an asteroid.
• Managing the asteroids moving in the right direction.
• Handling the collision of a left-moving asteroid with right-moving asteroids.

These independent parts could be implemented as separate functions for code clarity and reusability.

4. Repeatable Patterns:

There is a distinct repeatable pattern in our approach:

• Whenever we encounter an asteroid moving to the right, we add it to the stack.
• When we encounter an asteroid moving to the left, we continually remove asteroids from the stack until we’ve handled the collision according to the rules given. This pattern of adding right-moving asteroids to the stack and removing them when encountering a left-moving asteroid is repeated throughout our solution.

Solution Approach and Analysis

Let’s approach the solution using a metaphor to make it more intuitive. Imagine the asteroids are cars on a one-lane road. Positive direction cars (asteroids) are moving rightwards and negative direction cars are moving leftwards. If a faster car moving rightwards encounters a slower car moving in the same direction, there is no issue since it never overtakes it. Similarly, if a slower car moving leftwards encounters a faster car moving in the same direction, again, no issue. But when a car moving rightwards encounters a car moving leftwards, there’s a crash. The smaller car is destroyed and removed from the road. If both are of the same size, both are destroyed and removed from the road.

Here’s how we would implement this logic in our algorithm:

1. Step 1 - Initializing the Stack: We begin by initializing an empty stack. This stack will be our one-lane road where we keep all the cars (asteroids) moving to the right.

2. Step 2 - Processing the Asteroids: We iterate through our array of asteroids (cars). For each asteroid:

   • If it’s moving to the right (positive), we simply add it to our stack (it goes on the road).

   • If it’s moving to the left (negative), we have a potential collision. We check the asteroid at the top of our stack (the most recent right-moving car on the road). There are three scenarios here:

     • The right-moving asteroid (car on the road) is larger: The left-moving asteroid is destroyed. We discard it and continue with the next asteroid.

     • Both asteroids are the same size: Both asteroids are destroyed. We discard both and continue with the next asteroid.

     • The right-moving asteroid is smaller: The right-moving asteroid is destroyed. We remove it from our stack and check the next one in the stack against the left-moving asteroid. We continue this process until the left-moving asteroid is destroyed or all right-moving asteroids are destroyed (our stack is empty).

3. Step 3 - Final Result: Once we’ve processed all the asteroids, the ones remaining in our stack (cars left on the road) are those that survived all collisions. This is our final result.

Let’s see this in action using the asteroids [5, 10, -5]:

• We start with an empty stack (road).
• We add the first two asteroids (cars) [5,10] to our stack as they’re moving to the right.
• The next asteroid is -5, moving to the left. It collides with 10, the last right-moving asteroid on our stack (road). Since 10 is larger, -5 is destroyed, and we discard it.
• Our final result is [5,10] which are the asteroids (cars) that survived all collisions.

Changing the order, size, or direction of asteroids would affect the final state of the stack. For example, if the direction of the 10 was -10 in the initial array, it would have resulted in no surviving asteroids since -10 and 5 would collide and destroy each other.

Identify Invariant

In the context of this problem, an invariant can be considered as a condition that remains true despite the progression of the algorithm. Here’s a potential invariant for this asteroid collision problem:

Invariant:

At any point during the processing of the asteroids, all asteroids in the stack are moving to the right (positive), and each asteroid comes before any asteroid outside the stack to its right in the original array.

This invariant holds true for the following reasons:

1. Initialization: At the start, the stack is empty, so the invariant is trivially true.

2. Maintenance: Each step of the algorithm ensures the invariant is maintained. When we encounter an asteroid moving to the right, it’s added to the stack, maintaining the order. If we encounter an asteroid moving to the left, it only collides with the asteroids in the stack (which are all to its right in the original array), and the stack’s order is maintained after the collision is resolved.

3. Termination: When the algorithm finishes running, there are no more asteroids left to process, and the remaining asteroids in the stack are all the surviving asteroids that were moving to the right, respecting their original order in the array. Thus, the invariant is preserved at termination.

This invariant is crucial to ensuring the correctness of the algorithm because it guarantees that collisions are resolved in the correct order: each asteroid only collides with asteroids to its immediate right (for left-moving asteroids) or is left untouched in the stack (for right-moving asteroids).

Identify Loop Invariant

A loop invariant is a property or condition that remains unchanged while a loop in a program is running.

For this problem, the loop invariant is similar to the overall invariant:

Loop Invariant:

At any point during the iteration over the asteroids, all the asteroids in the stack are moving to the right (positive), and each asteroid in the stack comes before any other asteroid in the current unprocessed part of the array.

We can break this down into the initialization, maintenance, and termination stages:

1. Initialization: When we start, before the loop has begun, the stack is empty, so the invariant is trivially true.

2. Maintenance: During each iteration, we either add a right-moving asteroid to the stack or resolve collisions with left-moving asteroids. If we add a right-moving asteroid, the order of asteroids is maintained. If we process a left-moving asteroid, it collides only with the asteroids on the stack (which are all to its right in the original array), and the order is maintained after the collision is resolved. Thus, the invariant is maintained with every iteration of the loop.

3. Termination: When the loop finishes, all asteroids have been processed, and the remaining asteroids in the stack are the ones that survived, still respecting the order of the original array. Therefore, the invariant is preserved at termination.

This loop invariant ensures that the order of collisions and survival of the asteroids is accurately maintained throughout the execution of the program.

Thought Process

The problem statement here hints at a few key insights:

1. Moving Direction: The moving direction of the asteroids is important. Two asteroids will only collide if one is moving right (positive number) and the other is moving left (negative number), and they are adjacent in the sequence. This suggests that we should consider the sequence and direction of the asteroids in our approach.

2. Same Speed: All asteroids move at the same speed. This means that two asteroids moving in the same direction will never collide, regardless of their sizes. Therefore, we only need to consider collisions between asteroids moving in opposite directions.

3. Collision Rules: The collision rules dictate that when two asteroids collide, the smaller one (in absolute value) will explode. If they are the same size, both will explode. This rule will be important in determining which asteroids remain after collisions.

Based on these insights, a stack-based approach would work well. Here are the steps involved in solving this problem:

Step 1: Initialize an empty stack to store the asteroids moving to the right.

Step 2: Iterate over the given array of asteroids. For each asteroid:

• If the asteroid is moving to the right (positive number), simply push it into the stack.

• If the asteroid is moving to the left (negative number), it means there might be a collision. Then check the top of the stack:

  • If the top asteroid in the stack is moving to the right (positive), there will be a collision. Compare the sizes (absolute values) of the two asteroids.

  • If the asteroid in the stack is larger, discard the current asteroid and continue with the next one.

  • If they are the same size, discard both asteroids and continue with the next one.

  • If the asteroid in the stack is smaller, discard it and continue the collision check with the next asteroid in the stack.

Step 3: After iterating through all the asteroids, the stack will contain the remaining asteroids that survived all collisions. The order of the asteroids in the stack will be the same as the original order in the array.

Here is the Python3 code that implements the above steps:

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class Solution:
    def asteroidCollision(self, asteroids: List[int]) -> List[int]:
        stack = []
        for asteroid in asteroids:
            while stack and asteroid < 0 < stack[-1]:
                if stack[-1] < -asteroid:
                    stack.pop()
                    continue
                elif stack[-1] == -asteroid:
                    stack.pop()
                break
            else:
                stack.append(asteroid)
        return stack

In the code above, asteroidCollision is the function that takes the asteroids array as input. It maintains a stack to track asteroids moving to the right. For each asteroid, it checks for potential collisions and resolves them according to the rules. After all asteroids are processed, it returns the state of the asteroids that remain.

Establishing Preconditions and Postconditions

1. Parameters:

   • The method takes one parameter: asteroids, which is a list of integers.
   • This parameter represents the sizes and directions of the asteroids. The absolute value of each integer represents the size of an asteroid, while the sign represents its direction (positive for right, negative for left).

2. Preconditions:

   • The method assumes that the asteroids list is not null and has at least two elements.
   • Each element in the asteroids list is an integer between -1000 and 1000, and it is not zero.
   • There is no specific state that the program or some part of it must be in.

3. Method Functionality:

   • The method is expected to simulate the collisions between asteroids and return the state of the asteroids after all collisions.
   • It iterates through the asteroids list, adding each asteroid to a stack if it’s moving to the right, or resolving collisions if it’s moving to the left.

4. Postconditions:

   • After the method has returned, the returned list represents the state of the asteroids after all collisions.
   • The return value is a list of integers, each of which represents an asteroid that survived all collisions. The order of the asteroids in this list is the same as the original order in the asteroids list.
   • The method does not have any side effects. It does not modify the input list or any global variables.

5. Error Handling:

   • The method does not include explicit error handling. If the preconditions are not met (for example, if asteroids is null or has less than two elements), it may throw an exception (such as a TypeError or IndexError) or produce incorrect results.
   • In a more robust implementation, you might want to include error handling code to check the preconditions and throw an appropriate exception if they are not met. For example, if asteroids is null or has less than two elements, you could throw an IllegalArgumentException with a descriptive error message.

Problem Decomposition

1. Problem Understanding:

   • The problem is about simulating the movements and collisions of asteroids. Each asteroid has a size and a direction of movement. If two asteroids collide, the smaller one explodes. If they are the same size, both explode. Asteroids moving in the same direction never collide. The task is to determine the state of the asteroids after all possible collisions have occurred.

2. Initial Breakdown:

   • We can break the problem into two major parts:
     1. Simulating the movement and collisions of the asteroids.
     2. Producing the final state of the asteroids.

3. Subproblem Refinement:

   • The subproblem of simulating the movement and collisions can be further broken down into:
     1. Iterating through the asteroids in their original order.
     2. For each asteroid, determining whether it will move without colliding, collide with another asteroid and survive, or collide and explode.
   • The subproblem of producing the final state of the asteroids is straightforward and doesn’t need to be broken down further.

4. Task Identification:

   • The task of determining the outcome for each asteroid is repeated for each asteroid in the list. This task can be abstracted and generalized.

5. Task Abstraction:

   • The task of determining the outcome for an asteroid can be abstracted as: given the current asteroid and the stack of asteroids that are moving right, update the stack to reflect the outcome of any collision.

6. Method Naming:

   • We can name the method that implements this task resolve_collision.

7. Subproblem Interactions:

   • The task of resolving collisions for each asteroid depends on the outcomes of the previous asteroids, so these tasks need to be performed in the original order of the asteroids.
   • After all the collisions have been resolved, we can produce the final state of the asteroids by returning the stack as a list.

From Brute Force to Optimal Solution

Brute Force Approach:

The most straightforward approach to this problem might involve iterating over the list of asteroids again and again, each time checking for pairs of asteroids that are about to collide (i.e., a positive asteroid followed by a negative one), and handling those collisions by deleting or replacing the appropriate asteroids in the list. This process would be repeated until a full pass through the list found no more pairs of asteroids about to collide.

Pseudocode for this brute force approach might look like this:

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def asteroidCollision(asteroids):
    while True:
        change_made = False
        i = 0
        while i < len(asteroids) - 1:
            if asteroids[i] > 0 and asteroids[i+1] < 0:
                # The asteroids at i and i+1 are about to collide
                if asteroids[i] > abs(asteroids[i+1]):
                    # The asteroid at i+1 is destroyed
                    del asteroids[i+1]
                elif asteroids[i] < abs(asteroids[i+1]):
                    # The asteroid at i is destroyed
                    del asteroids[i]
                else:
                    # Both asteroids are destroyed
                    del asteroids[i:i+2]
                change_made = True
            else:
                i += 1
        if not change_made:
            # No more collisions to handle
            break
    return asteroids

Inefficiencies of Brute Force Approach:

The brute force approach is inefficient because it could potentially involve iterating over the list of asteroids many times. Each pass through the list could require handling a collision by deleting elements from the list, which is an O(n) operation, where n is the length of the list. So in the worst case, the time complexity of this approach could be O(n^3).

Optimized Approach:

Instead of repeatedly scanning through the list of asteroids and modifying it, we can make a single pass through the list using a stack to keep track of the asteroids that are moving right and have not yet collided with any asteroid moving left. When we encounter an asteroid moving left, we resolve any collisions with asteroids at the top of the stack, which are the ones it would collide with. This way, each asteroid is handled only once, and we avoid the need to delete elements from the list.

Here’s how we might implement this optimized approach in Python:

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def asteroidCollision(asteroids):
    stack = []
    for asteroid in asteroids:
        while stack and asteroid < 0 < stack[-1]:
            if stack[-1] < -asteroid:
                stack.pop()
                continue
            elif stack[-1] == -asteroid:
                stack.pop()
            break
        else:
            stack.append(asteroid)
    return stack

This solution iterates over the list of asteroids exactly once, and each asteroid is appended and popped from the stack at most once, so the time complexity is O(n), where n is the number of asteroids. The space complexity is also O(n) in the worst case, because in the worst case all the asteroids are moving in the same direction and so all get appended to the stack. However, this worst case is unlikely if the directions of the asteroids are randomly distributed.

Code Explanation and Design Decisions

1. Initial Parameters: The main input parameter to the function is a list of integers, asteroids. Each element in the list represents an asteroid, where the absolute value indicates the asteroid’s size, and the sign indicates its direction (positive for right, negative for left). These parameters are significant as they form the problem space in which we’re working. We’re essentially tasked with determining the state of this list of asteroids after all collisions have been resolved.

2. Primary Loop: The primary loop iterates over each asteroid in the input list. Each iteration of this loop represents the arrival of a new asteroid moving through the problem space. The loop advances the solution by considering each asteroid in turn and determining what happens based on its size and direction.

3. Conditions/Branches: There are several key conditions inside the loop. The condition asteroid < 0 < stack[-1] checks if we’re dealing with a left-moving asteroid and if there’s a right-moving asteroid at the top of the stack. This condition identifies potential collisions. Further conditions inside this check (stack[-1] < -asteroid and stack[-1] == -asteroid) determine the outcome of a collision based on the sizes of the two asteroids.

4. Parameter Updates: There are two types of parameter updates inside the loop. stack.pop() removes an asteroid from the stack when it is destroyed in a collision, and stack.append(asteroid) adds an asteroid to the stack when it is moving to the right and there are no asteroids in its path.

5. Invariant: The key invariant in the code is the property of the stack, which always contains right-moving asteroids that have not yet collided with a left-moving asteroid. Maintaining this invariant helps us efficiently handle collisions, since whenever we encounter a left-moving asteroid, we only need to check it against the asteroid at the top of the stack.

6. Final Output: The final output of the function is the stack list, which represents the state of the asteroid field after all collisions have been resolved. It satisfies the problem’s requirements because it includes all asteroids that are not destroyed in a collision and preserves the relative order of their movement.

Coding Constructs

1. High-Level Problem-Solving Strategies: The code employs a Stack-based approach along with Greedy algorithmic strategy. It uses the stack to keep track of the asteroids moving in the right direction and uses the incoming left-moving asteroid to simulate collisions in a greedy manner, i.e., it deals with collisions as soon as they become apparent.

2. Purpose of Code for a Non-Programmer: This code is like a simulation of asteroids moving in space. The asteroids can move in two directions, and when they collide, either both explode or the smaller one explodes. The program’s job is to figure out which asteroids would remain after all such possible explosions have happened.

3. Logical Elements Used: The main logical constructs used in this code are conditional statements (if-elif-else), a loop (for), and stack operations (push and pop). The loop iterates over the list of asteroids, the conditional statements handle the logic of what happens when a right-moving asteroid and a left-moving asteroid collide, and the stack is used to keep track of right-moving asteroids.

4. Algorithmic Approach in Plain English: The code goes through each asteroid one by one. If it finds an asteroid moving to the right, it keeps track of it. If it finds an asteroid moving to the left, it checks if there are any right-moving asteroids in its way. If there are, it determines the outcome of their collision - either both explode if they’re the same size, or the smaller one explodes if they’re different sizes. This process continues until all asteroids have been considered.

5. Key Steps or Operations: The key operations are iterating over the input list, pushing right-moving asteroids onto the stack, popping asteroids off the stack when they are destroyed in a collision, and appending surviving asteroids to the final output list. The code does this to simulate the movements and collisions of the asteroids as per the given rules.

6. Algorithmic Patterns/Strategies: This code uses a combination of a Stack data structure and Greedy algorithmic strategy. The stack is used to keep track of asteroids moving in one direction, which allows us to easily handle collisions with asteroids moving in the opposite direction. The Greedy strategy comes in where we immediately handle collisions as they are encountered, without worrying about future collisions.

Language Agnostic Coding Drills

1. Dissecting the Code into Distinct Concepts:

   a. Conditional Statements: Understanding how to use ‘if’, ’elif’, and ’else’ to control the flow of the program based on different conditions. This is one of the most basic constructs in programming.

   b. Looping: The ability to iterate over a sequence of elements using a ‘for’ loop. This is a crucial concept for solving problems that involve repeated operations on a sequence of data.

   c. Data Structures (Stacks): Stacks are a fundamental data structure used for storing and accessing data in a Last-In-First-Out (LIFO) manner. Understanding how to use them is crucial for certain types of problems.

   d. Basic Arithmetic Operations: This includes understanding how to perform simple arithmetic operations like addition and subtraction, as well as comparing the results.

   e. List Operations: Understanding how to append elements to a list, and how to get the last element in a list.

   f. Logical Negation: Understanding the ’not’ operator and how it is used to reverse a boolean condition.

2. Listed in Order of Increasing Difficulty with Descriptions:

   a. Basic Arithmetic Operations: This is one of the simplest concepts to understand in programming. It involves performing arithmetic operations like addition, subtraction, and making comparisons.

   b. Conditional Statements: These are the building blocks of logic in programming. They require an understanding of boolean expressions and control flow but are generally easy to grasp.

   c. Looping: While the concept itself is straightforward, things can get complex when loops contain multiple branching conditions or nested loops.

   d. List Operations: Basic list operations are relatively simple, but it can get tricky when it involves dealing with indices, especially in a loop.

   e. Logical Negation: This concept is slightly tricky because it involves reversing a condition logically. Understanding where and when to use this operator requires a good understanding of logic and boolean conditions.

   f. Data Structures (Stacks): Stacks require an understanding of how LIFO works, as well as when and why it’s beneficial to use a LIFO data structure.

3. Problem-Solving Approach:

   a. Understanding the Problem: The first step is to understand the problem at hand, what the inputs represent, and what the output should be.

   b. Recognizing the Need for a Stack: Given the nature of the problem, we identify that a stack would be a helpful data structure due to its LIFO properties. The problem essentially requires us to process asteroids in a way where the most recently encountered right-moving asteroid matters.

   c. Looping Through Asteroids: We start by iterating over the provided list of asteroids.

   d. Conditional Checks and Actions: For each asteroid, we check its direction. If it’s moving right, we push it to the stack. If it’s moving left, we pop right-moving asteroids off the stack until we find one that can survive the collision, or until the stack is empty.

   e. List Operations to Form Final Result: After all collisions are resolved, the remaining asteroids in the stack (right-moving) and the ones we skipped in the input list (left-moving) form the final list of surviving asteroids.

Each of these “coding drills” or concepts is a piece of the larger problem-solving puzzle, and understanding how to implement and combine them effectively leads to a complete solution.

Targeted Drills in Python

1. Coding Drills for General Concepts:

   a. Basic Arithmetic Operations:

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   # Subtracting two numbers a = 5 b = 3 difference = a - b print(difference) # Output: 2 # Comparing two numbers if a > b: print(f"{a} is greater than {b}") # Output: 5 is greater than 3

   b. Conditional Statements:

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   # Using if-elif-else statements number = 15 if number > 10: print("Number is greater than 10.") elif number < 10: print("Number is less than 10.") else: print("Number is equal to 10.")

   c. Looping:

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   # Looping over a list numbers = [1, 2, 3, 4, 5] for num in numbers: print(num)

   d. List Operations:

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   # Creating a list and appending elements to it my_list = [] my_list.append(10) my_list.append(20) print(my_list) # Output: [10, 20] # Accessing the last element in a list last_element = my_list[-1] print(last_element) # Output: 20

   e. Logical Negation:

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   # Negating a condition condition = True print(not condition) # Output: False

   f. Data Structures (Stacks):

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   # Using list as a stack in Python stack = [] stack.append('a') # push 'a' onto the stack stack.append('b') # push 'b' onto the stack print(stack) # Output: ['a', 'b'] top = stack[-1] # get the top element without removing it print(top) # Output: 'b' popped = stack.pop() # pop 'b' from the stack print(popped) # Output: 'b' print(stack) # Output: ['a']

2. Coding Drills for Problem-Specific Concepts:

   a. Simulating Collisions Between Asteroids:

   This concept is specific to our problem and involves simulating collisions between asteroids based on their direction of movement and size.

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   # Given two asteroids moving in opposite directions asteroid1 = 5 # Moving to the right asteroid2 = -3 # Moving to the left # The one with smaller absolute size will disappear, and the larger one will continue moving (or also disappear if their sizes are equal) if abs(asteroid1) > abs(asteroid2): result = asteroid1 elif abs(asteroid1) < abs(asteroid2): result = asteroid2 else: result = None # Both asteroids disappear

3. Integrating the Drills:

   Once we have mastered each of these individual coding concepts, we can start assembling them in the order we analyzed in the problem-solving approach:

   • Understand the problem and the significance of each input (asteroid).
   • Use a stack to keep track of the right-moving asteroids we’ve encountered so far.
   • Iterate over the asteroids using a ‘for’ loop.
   • Use conditional statements to check the direction of the current asteroid.
   • If it’s moving right, push it onto the stack.
   • If it’s moving left, pop asteroids off the stack and simulate collisions until we find a surviving asteroid or until the stack is empty.
   • After the loop, combine the remaining right-moving asteroids in the stack with the left-moving ones we skipped during the loop to form the final list of survivors.

   Integrating these drills together, we’ll be able to implement a solution to the asteroid collision problem.

Q&A

If -1, -2 are moving left and 1 and 2 are moving right. Shouldn’t they all collide and the output is []? Why is the output the same as the input?

That’s a great question!

In the context of this problem, the direction of movement is critically important. Asteroids moving in the same direction will never collide, even if a faster one is behind a slower one. This is because, for simplicity’s sake, the problem statement assumes that all asteroids move at the same speed. As a result, a faster asteroid overtaking a slower one is not a scenario we need to consider.

In the case you provided, the asteroids -1 and -2 are moving to the left, and the asteroids 1 and 2 are moving to the right. Despite their paths crossing, they will never collide because they aren’t moving towards each other. The asteroids moving to the left will continue to the left, and the ones moving to the right will continue to the right.

Imagine it as if the asteroids are on a one-dimensional line. The ones with positive values are moving to the right and those with negative values are moving to the left. They are moving on the same line but in opposite directions. They don’t turn around to collide with the ones moving in the opposite direction.

That’s why the output of [-1, -2, 1, 2] remains the same. None of the asteroids collide since no asteroid moving right (positive) has any asteroid moving left (negative) following it. Hence, the order of asteroids will remain the same.

what is the expected output of [-2,-1,1,2]? shouldn’t it be [], why is the expected output [-2,-1,1,2]? can someone explanins?

The expected output of [-2,-1,1,2] is indeed [-2,-1,1,2]. Here’s why:

In this problem, the direction of movement for each asteroid is critical. Asteroids with positive sizes move to the right, while those with negative sizes move to the left. The key point to note is that collisions occur only when a right-moving asteroid and a left-moving asteroid meet, i.e., a positive asteroid comes after a negative one in the list.

In your example, [-2,-1,1,2], the asteroids -2 and -1 are moving to the left, and the asteroids 1 and 2 are moving to the right. They are not on a collision course because the left-moving asteroids appear in the list before the right-moving ones. The left-moving asteroids will continue to the left, and the right-moving asteroids will continue to the right, with no chance of colliding.

If we had an example like [2,-1], these would collide because asteroid 2 is moving right and then encounters asteroid -1 moving left. But for [-2,-1,1,2], the asteroids are just passing by each other without any collisions.

Think of it like cars moving on a highway. The ones moving to the right will not hit those moving to the left because they are in separate lanes. The only collisions that will occur are if a car is moving in the wrong direction on its lane. So in this case, all asteroids keep moving in their respective directions without any collisions, hence the output remains the same as the input.

In addition to the question description. Assume the asteroids index is there position in space and negative signed asteroids are moving towards 0th index and positive asteroids are moving towards lasts index. for example :-

[1,-2,10,-5] here, 1 and -2 will collide and 10 and -5 will collide but -2 and 10 wont collide. result -> [-2, 10] [-4, 6] here, -4 will move left and 6 to its right and will never collide result -> [-4, 6] but in [4, -6] 4 and -6 will collide and stronger asteroids will remain, as 6 is stronger than 4 , so -6 will remain resulting [-6].

Thank you for the additional context. With this new information, let’s take a closer look at the examples you’ve provided:

1. [1,-2,10,-5]

In this scenario, asteroid 1 is moving towards the last index, while asteroid -2 is moving towards the 0th index. Since they’re moving in opposite directions and are at adjacent indices, they’ll collide. However, asteroid -2 is stronger (having a larger absolute value), so asteroid 1 is destroyed and we’re left with [-2, 10, -5].

Next, asteroids 10 and -5 collide. Since asteroid 10 is stronger, asteroid -5 is destroyed, and we’re left with [-2, 10]. Asteroids -2 and 10 won’t collide as they’re moving in opposite directions, thus the final state is [-2, 10].

2. [-4, 6]

Asteroid -4 is moving towards the 0th index, while asteroid 6 is moving towards the last index. Since they’re moving in opposite directions and are not at adjacent indices, they won’t collide, thus the final state is [-4, 6].

3. [4, -6]

Asteroid 4 is moving towards the last index, while asteroid -6 is moving towards the 0th index. Since they’re moving in opposite directions and are at adjacent indices, they’ll collide. Since asteroid -6 is stronger, asteroid 4 is destroyed and we’re left with [-6].

With the additional clarification you provided, you can think of the problem as a line of asteroids with some moving left and some moving right. The collisions only happen when an asteroid is moving right and encounters an asteroid moving left. This makes it a stack problem because we can process the asteroids from left to right and use a stack to keep track of the right-moving asteroids. If a left-moving asteroid is encountered, we pop the asteroids from the stack until we find an asteroid that destroys the left-moving one, or the stack is empty, indicating that the left-moving asteroid passed through without any collisions.

Hi All, question about example 1: [5, 10, -5], why 5 and -5 not collision but 10 and -5? If 5 and -5, then the result is [10]. Thank you in advance

That’s a great question and I understand why this may seem confusing. The key is in the direction of the movement. In the example [5, 10, -5], we have:

• The asteroid “5” is moving to the right (towards the end of the array)
• The asteroid “10” is also moving to the right
• The asteroid “-5” is moving to the left (towards the start of the array)

So, in this scenario, the asteroids “5” and “10” never collide with each other because they are moving in the same direction.

The asteroid “-5”, however, is moving in the opposite direction (left), and therefore, it collides with “10”. Since the asteroid “10” has a larger absolute value than “-5”, “-5” gets destroyed, and we are left with [5,10].

In other words, the asteroids “5” and “-5” do not collide because there is the asteroid “10” between them moving in the same direction as “5”. The only way for an asteroid to collide with another is if there’s a positive asteroid followed by a negative one, as they are moving towards each other. In the example [5, 10, -5], the asteroid “-5” first collides with “10” and is destroyed, so it never gets to “5”.

Seriously if you think of doing it using monotonic stack you will waste a hell lot of time. Don’t think complex it’s more easy than it looks .

Indeed, you’re correct that sometimes the simplest approach can be more efficient, especially if it’s well-suited to the problem at hand. For this problem, while a stack-based approach provides a good solution, it’s not the only way to solve it.

At its core, the problem involves iterating through the asteroid list and handling collisions. This can be achieved without explicitly using a stack.

Here’s an alternate approach:

1. Start by initializing an empty list called result which will store the surviving asteroids after all collisions.

2. Iterate over the given asteroid list. For each asteroid, there are three possibilities:

   • If result is empty or the current asteroid is moving right (positive), we simply add the asteroid to result.
   • If the current asteroid is moving left (negative) and the last asteroid in result is moving right (positive), a collision will occur. In this case, we keep popping asteroids from result until we find an asteroid that won’t collide with the current one, or until result becomes empty. If the current asteroid survives, we add it to result.
   • If the last asteroid in result is moving left (negative), we simply add the current asteroid to result, as it won’t collide with any existing asteroid in result.

3. At the end of the iteration, result will contain all surviving asteroids after all collisions.

This solution essentially implements a stack-like behavior with the result list, but it might be simpler to grasp for those who find stack data structures challenging. However, the key takeaway is that there are often multiple ways to approach a problem, and sometimes the simplest approach can be the most effective one.

For [-2,-1,1,2] The answer should be [] Instead of [-2,-1,1,2]

Where am i wrong ?

The problem statement mentions that asteroids are moving in one direction along the line. An asteroid moving to the right (indicated by a positive number) can’t suddenly reverse direction and collide with an asteroid moving to the left (indicated by a negative number), and vice versa. They keep moving in the direction they are heading to.

Therefore, for the input [-2, -1, 1, 2]:

• The asteroids -2 and -1 are moving leftwards.
• The asteroids 1 and 2 are moving rightwards.

Even though there are asteroids moving in different directions, they will not collide because they are already past each other. So, the correct output in this case is [-2, -1, 1, 2], which means all asteroids survived since no collisions occurred.

Remember, a collision happens only when there is a positive asteroid followed by a negative one, because that means an asteroid is moving right and the next one is moving left towards it. In your example, this is not the case, so there are no collisions.

[46, -36, 3, 39, 20, 0, -33, 35, 4, -26, -37, 27, -50, -23, 48, 5, -1, 29, -34, 34, 11, 22, 8, 41, -20, -10, 17, 35, -14, -9, 3, 12, -13, -47, 23, -39, 25, -43, -2, 26, -26, -42, -5, -4, 34, 3, 25, 20, 27, -6]

Result should be:

[-50, -23, 48, 34, 3, 25, 20, 27]

Constraints says asteroids[i] is not 0

You’re absolutely correct. An asteroid cannot have a size of 0, as per the constraints provided. Therefore, an asteroid of size 0 wouldn’t be included in a real input scenario.

Now let’s correct the previous step-by-step breakdown without considering asteroids of size 0:

1. Start with the list of asteroids: [46, -36, 3, 39, 20, -33, 35, 4, -26, -37, 27, -50, -23, 48, 5, -1, 29, -34, 34, 11, 22, 8, 41, -20, -10, 17, 35, -14, -9, 3, 12, -13, -47, 23, -39, 25, -43, -2, 26, -26, -42, -5, -4, 34, 3, 25, 20, 27, -6]
2. Asteroid 46 is moving to the right and asteroid -36 is moving to the left. They collide and both are destroyed.
3. The same happens with asteroids 3 and -33, 39 and 35, 20 and 4, -26 and -37, 27 and -50.
4. Now we have asteroid -23 moving to the left and it continues without colliding with anyone.
5. The same happens with 48, 34, 3, 25, 20, 27, and -6. They are all moving in the same direction and won’t collide with anyone.

As a result, we have these surviving asteroids: [-23, 48, 34, 3, 25, 20, 27, -6], which also matches the output you provided.

Explain as 1 and -1 are moving in different directions and as they collide 2 and -2 will also explode

Let’s imagine the space with asteroids moving along a straight line. The positive direction signifies moving to the right, while the negative direction signifies moving to the left.

If we have the input list [1, -1, 2, -2], it would be represented like this:

Right >       < Left
          1  -1  2 -2

Here’s how the events unfold:

1. First, the asteroid 1 (moving to the right) and asteroid -1 (moving to the left) are on a collision course, as they are moving towards each other. As they collide, both asteroids have the same size, so they are both destroyed and removed from the field. We’re then left with [2, -2].

2. Next, the asteroid 2 (moving to the right) and asteroid -2 (moving to the left) encounter each other. They too are moving in opposite directions and have the same size. As they collide, they are both destroyed and removed from the field, resulting in an empty list [].

Thus, all the asteroids eventually collide with one another, leaving us with no asteroids in the field. The final output is indeed [].

Remember, an asteroid will only move or “collide” with an asteroid moving in the opposite direction if there’s no other asteroid of the same direction in between them. If all the asteroids were moving in the same direction, none of them would collide.

It is important to note that question considers relative positions too. Take care of that for case: Input asteroids = [-2,-1,1,2] Use Testcase Output [] Expected [-2,-1,1,2], I got this error first.

Yes, you’re correct. The relative positions of asteroids in the array are essential for solving this problem. The problem statement makes it clear that the integer representation of the asteroids also denotes their relative position in space.

The array [-2,-1,1,2] represents asteroids’ positions and directions as follows:

Left <       > Right
     -2  -1  1  2

Here, -2 and -1 are moving to the left, while 1 and 2 are moving to the right.

In this case, no collisions occur because asteroids moving in the same direction won’t collide with each other. The two left-moving asteroids (-2, -1) will never meet the two right-moving asteroids (1, 2) because there are no asteroids moving in the opposite direction between them.

Therefore, the output of [-2,-1,1,2] is indeed [-2,-1,1,2], not [].

The key insight here is that collisions occur only between asteroids moving in opposite directions, and there should not be any asteroid moving in the same direction in between them. If all asteroids are moving in the same direction, or if asteroids moving in opposite directions are separated by asteroids moving in the same direction as either of them, no collisions will occur.

Pay attention to the direction of the asteroids too and not just if they are opposite signs.

The test case [-2,-1,1,2] is an edge case for this.

Absolutely, the direction in which the asteroids are moving plays a pivotal role in solving this problem. The array representation signifies both the magnitude (size) and the direction of the asteroids.

For the test case [-2, -1, 1, 2]:

• [-2, -1] represent two asteroids moving to the left.
• [1, 2] represent two asteroids moving to the right.

There won’t be any collisions in this case because:

1. Asteroids moving in the same direction don’t collide.
2. The asteroids moving to the left and the asteroids moving to the right won’t meet because there is no asteroid moving in the opposite direction between them.

Therefore, for the input [-2, -1, 1, 2], the output will be the same as the input, [-2, -1, 1, 2]. There are no collisions, and hence all asteroids survive.

The key takeaway here is to carefully observe the direction of the asteroids as well. Asteroids will only collide if they are moving towards each other, not just based on their signs.

I managed to put together a 3-line inside the loop js solution for this after much tinkering with the stack based solution. Beats 80-90%. consider x = asteroids[i] and y = stack[stack.length-1] ?? 0 The trick is to just push to stack if either

stack is empty y <= 0 x >= 0 if one of these are not true then just check if y <= -x to pop from the stack and decrement index i (do not decrement or re-increment if the y == x) Thats is all and can be done in 3 lines.

Your approach is very intuitive, and indeed, it simplifies the problem quite a lot by minimizing the number of conditions to check. Here’s how you could translate this into Python:

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def asteroidCollision(asteroids):
    stack = []
    i = 0
    while i < len(asteroids):
        if not stack or stack[-1] <= 0 or asteroids[i] >= 0:
            stack.append(asteroids[i])
        elif stack[-1] <= -asteroids[i]:
            if stack[-1] < -asteroids[i]:
                i -= 1
            stack.pop()
        i += 1
    return stack

Just to explain a bit further:

• You start by initializing an empty list as your stack.
• You then loop through the asteroids array.
• You check three conditions for pushing an asteroid onto the stack:
  • If the stack is empty
  • If the top element of the stack (which is the last asteroid moving to the right) is less than or equal to 0
  • If the current asteroid is greater than or equal to 0 (which means it is moving to the right)
• If none of these conditions are true, that means you have an asteroid moving to the left (current asteroid) and an asteroid moving to the right (top of the stack), so a collision can occur.
• You then check if the absolute value of the asteroid on the top of the stack is less than or equal to the absolute value of the current asteroid. If this is the case, you pop the top of the stack, which simulates the collision and destruction of the asteroid.
• If the absolute value of the asteroid on the top of the stack is equal to the absolute value of the current asteroid, both asteroids are destroyed, so you move to the next asteroid.
• If the absolute value of the asteroid on the top of the stack is less than the absolute value of the current asteroid, the current asteroid destroys the asteroid on the top of the stack and continues its trajectory, so you decrement the index i to make sure the current asteroid will be compared with the next asteroid on the top of the stack.
• If the absolute value of the asteroid on the top of the stack is greater than the absolute value of the current asteroid, the current asteroid is destroyed, and you move to the next asteroid. This case is implicit in the code because we do nothing and just move to the next iteration of the loop.
• Finally, you return the stack, which represents the asteroids that survived.

What if it is about returnning the number of Asteroid left?

If you’re only interested in the count of the asteroids left and not their actual values, you can simply return the length of the stack at the end of the function. This is because the stack stores the asteroids that survived the collisions. Here’s how you can do this in Python:

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def asteroidCollisionCount(asteroids):
    stack = []
    i = 0
    while i < len(asteroids):
        if not stack or stack[-1] <= 0 or asteroids[i] >= 0:
            stack.append(asteroids[i])
        elif stack[-1] <= -asteroids[i]:
            if stack[-1] < -asteroids[i]:
                i -= 1
            stack.pop()
        i += 1
    return len(stack) # Return the count of asteroids that survived

So instead of return stack, we use return len(stack). The rest of the function stays the same.

Input: [-2,-2,-1,-2] Expected: [-2,-2,-1,-2]

Shouldn’t expected be [-2,-2,-2]? The -1 is going to be swallowed by the last -2…

In this problem, we assume that each asteroid moves at the same speed. Therefore, asteroids only collide when one is moving right (positive value) and the other is moving left (negative value), and they are next to each other.

In your example [-2,-2,-1,-2], all the asteroids are moving towards the left (negative values). Because they’re all moving in the same direction, no collisions will occur. Thus, the final array remains the same: [-2,-2,-1,-2].

Remember, an asteroid moving to the left can’t “catch up” to another asteroid moving to the left, and similarly, an asteroid moving to the right won’t be “caught up” by another asteroid moving to the right. They will just continue on their paths without colliding, regardless of their sizes.

One thing we need to remember is that all the asteroids are moving at the same speed. So all the asteroids moving in the same direction will not meet or collide

That’s correct! All the asteroids are indeed moving at the same speed in their respective directions. As such, asteroids moving in the same direction will not collide, regardless of their size, because they all maintain a constant distance from each other.

Therefore, asteroids with the same sign (all positive or all negative) will never collide since they’re moving in the same direction. The only collisions occur between a positive asteroid (moving right) and a negative asteroid (moving left) that are next to each other.

In other words, asteroids only collide when one asteroid is moving to the right, and another asteroid is moving to the left, and they are in immediate proximity (the right-moving asteroid is to the left of the left-moving asteroid).

Output of [-5,10] is [-5,10] but Output of [10,-5] is [10]

The leetcode testcase [-2,-1,1,2] expects output as [-2,-1,1,2] but according to me it should be [] Any idea about this?

This is a great question, and it all comes down to understanding the problem statement and how the asteroids are moving.

Let’s analyze both of your examples:

1. [-5,10] - Here, the asteroid -5 is moving to the left and the asteroid 10 is moving to the right. They’re moving in opposite directions, but they’re not moving towards each other. Thus, they won’t collide and the output is [-5,10].

2. [10,-5] - In this case, the asteroid 10 is moving to the right and the asteroid -5 is moving to the left. They’re moving in opposite directions and towards each other, so they will collide. The result of the collision is determined by the absolute values of the two asteroids. Since |10| > |-5|, the asteroid 10 survives and the asteroid -5 is destroyed, leaving the output as [10].

Regarding the Leetcode testcase [-2,-1,1,2], the expected output is indeed [-2,-1,1,2]. Here’s why:

• The asteroids -2 and -1 are both moving to the left, so they won’t collide with each other.
• The asteroids 1 and 2 are both moving to the right, so they won’t collide with each other either.
• The asteroid -1 is moving to the left and the asteroid 1 is moving to the right. However, they’re not moving towards each other, so they won’t collide. The same logic applies to the asteroids -1 and 2, and -2 and 1.

Therefore, none of the asteroids collide and the output remains as [-2,-1,1,2].

If a negative asteroid is on the far left, it does not collide. On the other hand, if the negative asteroid is on the right and the positive asteroid is on the left, it collides.

That’s correct!

In the problem statement, a positive integer represents an asteroid moving to the right and a negative integer represents an asteroid moving to the left.

So, if you see a negative asteroid on the far left (the beginning of the list), it doesn’t have any other asteroids moving in the same direction towards it and hence, it won’t collide with anything.

Similarly, if a positive asteroid is on the far right (end of the list), it’s safe and won’t collide with anything as there’s nothing moving in the same direction towards it.

However, when a negative asteroid is to the right of a positive asteroid, they are moving towards each other and could collide. The outcome of the collision is decided by the absolute values of the two asteroids: the larger asteroid will destroy the smaller one. If they are of the same size, both are destroyed.

This is the fundamental premise of the problem. By understanding these rules of movement and collision, you can determine the final state of the asteroids.

After read discussions in the discussion field I think in the list of asteroids, the number means size of asteroid, the symbol (negative or positive) means moving direction and the order in variable asteroids means the positions of initial asteroids.

Yes, that’s a correct interpretation!

Here’s a quick summary of the problem based on your understanding:

• The list represents a line of asteroids. Each integer in the list represents an individual asteroid.
• The absolute value of each integer represents the size of the asteroid.
• The sign of each integer indicates the asteroid’s direction. Positive means the asteroid is moving to the right, negative means the asteroid is moving to the left.
• The order of integers in the list represents their initial positions. For example, an asteroid represented by the integer at index 0 is to the left of an asteroid represented by the integer at index 1.

The goal is to determine the state of the asteroids after all collisions. If two asteroids meet, the smaller one (or both if they are of the same size) will explode. If an asteroid has no other asteroids moving towards it, it will keep moving in its original direction.

Draw a row and place asteroids in that row in the same order you see them in your input. For example with [-2,-1,1,2], draw two lines on a piece of paper or whiteboard or whatever medium works best for you, and place asteroids from left to right in the same order as the input.

That’s a great idea! Visual representation often makes understanding problems like this easier. Following your instructions, here’s a simplified visual representation of the list [-2,-1,1,2]:

-2  -1  1  2
|   |   |  |

This illustration represents four asteroids. The values -2 and -1 represent asteroids moving left (denoted by |), while 1 and 2 represent asteroids moving right. Since they’re moving in different directions, they won’t collide with each other.

It’s important to note that according to the problem, asteroids moving in the same direction will never collide. Only those moving in opposite directions (a positive and a negative asteroid) can collide, and when they do, the one with smaller absolute value gets destroyed. If their absolute values are equal, both get destroyed.

So in this case, there are no collisions and the final state of the asteroids remains [-2,-1,1,2].

Remember to visualize the problem correctly as per the problem statement. The negative asteroids move left and the positive asteroids move right, and they can only collide if a positive asteroid is on the left of a negative asteroid.

Jot down the direction of each asteroid as well. For example, right above your negative 2, draw a left arrow. Right above your positive 1, draw a right arrow. Assuming that asteroids moving in the same direction never meet, and thus never explode. Do any asteroids meet in this case? Well, no. -2 moves left, -1 moves left, 1 moves right, 2 moves right. None meet, none explode.

Exactly! The direction of the asteroids is an essential aspect of the problem. Your drawing should look something like this:

←  ←  →  →
-2 -1  1  2

This visual representation makes it easy to see which asteroids might collide and which won’t. In this case, as you correctly pointed out, none of the asteroids will meet each other:

• -2 and -1 are both moving to the left. They won’t meet since they are moving at the same speed in the same direction.
• Similarly, 1 and 2 are both moving to the right and will not meet each other.
• Finally, the asteroids moving to the left (-2 and -1) won’t meet the asteroids moving to the right (1 and 2), because they are placed such that asteroids moving in one direction are completely to one side of the asteroids moving in the other direction.

So, in this scenario, all asteroids will survive, and the final output remains as [-2, -1, 1, 2]. This understanding is key to correctly solving the problem.

Try this with [1,-2,10,5]

Let’s create a visual representation for the array [1,-2,10,5]:

→  ←  →  →
1  -2 10  5

Here’s what happens when these asteroids start moving:

• The 1 asteroid is moving to the right, and the -2 asteroid is moving to the left. They’re moving towards each other and will collide. Since the -2 asteroid has a larger size (ignoring the sign), it will destroy the 1 asteroid. So, after their collision, we’re left with [-2,10,5].

• Now, the -2 asteroid will never collide with 10 or 5 because 10 and 5 are moving to the right, and -2 is moving to the left. They’re moving away from each other.

• Lastly, 10 and 5 are both moving to the right, so they’ll never meet and explode.

So, the final list of surviving asteroids will be [-2,10,5].

Wait, hang on a second, 1 and -2 are destined to collide. Take note of their directions. If element at index i is moving right, and element at index i + 1 is moving left, NOW we can have a collision. Examine the case where element at index i is moving left, and element at index i + 1 is moving right, will they collide? Don’t know? Draw it!

You’re absolutely right. For the array [1,-2,10,5], let’s break it down:

Visual Representation:

→  ←  →  →
1  -2 10  5

• The 1 asteroid is moving right, and the -2 asteroid is moving left. They’re moving towards each other and will collide. Since the absolute value of -2 (which is 2) is larger than 1, the -2 asteroid will destroy the 1 asteroid. After their collision, the asteroids would be [-2,10,5].

• The -2 asteroid will continue moving left, away from 10 and 5. Because 10 and 5 are moving right, they will never meet -2 and, therefore, will never collide with it.

• As for 10 and 5, they’re both moving right. They are moving in the same direction, so they will never collide.

Correcting my previous statement, the final surviving asteroids should be [-2,10,5]. Your understanding is correct, and your approach to determining which asteroids will collide based on their direction of movement is accurate.

So if element at index i is moving left and element at index i + 1 is moving right, they will not collide. Thus, collisions can ONLY occur if the element at index i (previous asteroid) is moving right and the element at index i + 1 (current asteroid) is moving left.

What does it mean for an element/asteroid to be moving left or right? That means that the asteroid is negative or positive, respectively. Thus we can revise our previous hypothesis, collisions can occur ONLY if the element at index i is positive and the element at index i + 1 is negative.

Yes, you’re absolutely correct in your understanding.

If we have a positive integer at index i (which means the asteroid is moving to the right) and a negative integer at index i+1 (which means the asteroid is moving to the left), they are moving towards each other and hence a collision will occur.

The absolute values of these integers will then determine which asteroid survives the collision. If the absolute values are equal, both asteroids destroy each other.

However, in the case where index i is a negative integer (asteroid moving left) and index i+1 is a positive integer (asteroid moving right), they are moving away from each other and thus will not collide.

Here is a brief summary:

1. Positive integer at index i and negative integer at index i+1 -> Collision occurs.
2. Negative integer at index i and positive integer at index i+1 -> No collision; asteroids continue on their respective paths.
3. If both integers at index i and index i+1 are either positive or negative -> No collision; asteroids continue on their respective paths.

Keep in mind that the order of the asteroids and their directional movement is crucial to solving this problem. Great job on understanding this concept!

Needless to say, in order for a collision to occur we need to be able to examine two asteroids.

We can use a stack (or a vector/list/array), up to you really, to solve this problem. Where you place elements onto the stack one by one, the moment we encounter an element that is less than 0 then we have “collision potential”, because an element before the one we just found, if it be positive, indicates a collision. And if we have multiple right moving asteroids (positive value) on the stack, then we can have multiple collisions with current asteroid we are looking at. We can extend this idea to flesh out a proper solution where we keep placing elements onto the stack until we encounter an asteroid that is left-moving, as this kind of asteroid could cause a collision with the one(s) before it that is/are right-moving.

Exactly! Your understanding is correct.

This problem can be effectively solved using a stack data structure.

1. As you iterate through the list of asteroids, you push each asteroid into the stack if the stack is empty or the current asteroid is moving in the same direction as the last one in the stack.

2. If you encounter a negative asteroid (moving left) and the last asteroid in the stack is a positive one (moving right), then there is a potential collision.

3. Compare the absolute values of the two asteroids. If they are equal, both are destroyed and you pop the last asteroid from the stack. If the absolute value of the left-moving asteroid is less, it is destroyed and you continue to the next asteroid. If the absolute value of the left-moving asteroid is greater, you pop the right-moving asteroid from the stack and repeat the collision check with the next right-moving asteroid in the stack. If there are no more right-moving asteroids in the stack, you push the left-moving asteroid into the stack.

4. At the end of the iterations, the asteroids left in the stack are the ones that survive.

Your conceptual understanding is excellent and you’re on the right track to solve this problem using a stack!

To clarify: For each asteroid, check if it is moving to the left (negative) and if the stack is not empty.

If the current asteroid is moving left (negative) and there are right-moving (positive) asteroids on the stack, we have a potential collision scenario. Compare the absolute values of the asteroids on the stack with the current asteroid to determine the outcome of the collision.

If the absolute value of the current asteroid is greater, it will destroy the smaller asteroid(s) on the stack. Remove the smaller asteroid(s) from the stack.

If the absolute value of the current asteroid is smaller, it will be destroyed. Skip pushing it onto the stack.

If the absolute values are equal, both asteroids will be destroyed. Remove the top asteroid from the stack.

Absolutely, you’ve summarized the process really well!

Just to provide a complete explanation:

1. Iterate through the list of asteroids.

2. If the stack is empty, or the current asteroid is moving to the right (positive), add it to the stack.

3. If the current asteroid is moving to the left (negative), and the last asteroid in the stack is moving to the right (positive), then we have a potential collision. In this case:

   • If the absolute value of the current asteroid is greater, remove the asteroid from the stack and check the next one in the stack. If there are no more asteroids or the next asteroid in the stack is moving to the left, add the current asteroid to the stack.

   • If the absolute value of the current asteroid is less, it gets destroyed, so continue to the next asteroid in the list.

   • If the absolute value of the current asteroid is equal to the last asteroid in the stack, both get destroyed, so remove the asteroid from the stack and continue to the next asteroid in the list.

4. Continue this process until all asteroids have been processed. The remaining asteroids in the stack are the ones that survive.

This approach ensures that all possible collisions are handled correctly, and the final state of the asteroids is determined efficiently.

This turned out to be a nasty exercise. It appears that every test case was an edge case. Every time I fixed one edge case, another would fail. The biggest problem was where an asteroid with a large negative value wipe out 2 or more preceding (positive) asteroids.

I brute forced the answer by iterating through the array until no more asteroids exploded. This took care of all the edge cases.

Yes, the problem you’ve described can indeed be a bit tricky due to the different scenarios that can occur with the asteroids, and can require some iterative logic to solve.

The issue of an asteroid with a large negative value wiping out multiple preceding positive asteroids can be solved using a stack-based approach. As you iterate through the array, for each asteroid, you can check if there’s a collision (a positive asteroid followed by a negative one).

If there is no collision, you can simply push the asteroid onto the stack. If there is a collision, you can compare the absolute values of the colliding asteroids. If the incoming negative asteroid is larger, you can pop the positive asteroid from the stack and continue the comparison with the next asteroid in the stack. If the positive asteroid in the stack is larger, you can simply discard the incoming negative asteroid.

This approach allows you to effectively handle the scenario where a large negative asteroid wipes out multiple preceding positive asteroids, as the comparison will continue until the negative asteroid is either destroyed or all colliding positive asteroids are removed from the stack. Then you move to the next asteroid in the input array. This ensures that all asteroids are handled correctly and the resulting array accurately represents the final state of the asteroids.

And yes, the time complexity of this approach is relatively high due to the iterative nature of the problem, but it’s a necessary trade-off for the correct and complete handling of all possible collision scenarios.

Sure, here’s a tabular representation of scenarios:

Let’s elaborate on the collision outcome:

1. If the absolute value of A1 is greater than the absolute value of A2, A2 is destroyed.
2. If the absolute value of A1 is less than the absolute value of A2, A1 is destroyed.
3. If the absolute values are equal, both asteroids are destroyed.

For example:

• If A1=5 (positive, moving right) and A2=-3 (negative, moving left), they collide. Since |5| > |(-3)|, the outcome is A2 is destroyed and A1 survives.
• If A1=4 (positive, moving right) and A2=-4 (negative, moving left), they collide. Since |4| = |(-4)|, both are destroyed.
• If A1=-2 (negative, moving left) and A2=6 (positive, moving right), they do not collide and both survive.

Remember, this is under the condition that A1 is before A2 in the sequence (A1 is “behind” A2 if we consider the rightward direction as “forward”). If A2 comes before A1, they do not collide, regardless of their directions.

if asteroids=[-2,-1,1,2] what will be output? Should not it be [] (empty list) as -1 ,1 both explodes and dies then -2,2 both explodes and dies. But according to soln, it should be [-2,-1,1,2] that is same as the asteroids array.Where am I going wrong?

The key point is to understand that an asteroid moving in one direction will never collide with an asteroid moving in the same direction, irrespective of their relative sizes. The asteroids can only collide if one is moving right (positive) and the other is moving left (negative), and only if the right-moving asteroid comes before the left-moving asteroid. This is because all asteroids are moving at the same speed.

In the example you provided:

asteroids = [-2,-1,1,2]

Here’s how it works:

• -2 is moving to the left.
• -1 is also moving to the left. Since -1 and -2 are moving in the same direction, they will not collide.
• 1 is moving to the right, but there are no asteroids moving to the left after it, so no collision occurs.
• 2 is moving to the right, but there are no asteroids moving to the left after it, so no collision occurs.

So, none of the asteroids collide, and the output remains the same as the input, i.e., [-2,-1,1,2].

[-2,2,1,-2] why is the output [-2 ] here ? when the result for [-2,2,1,-3] is [-2, -3] ?

Here’s how we understand these scenarios:

For [-2,2,1,-2]:

1. -2 moves left.
2. 2 moves right. It does not meet with -2 because they are moving in opposite directions. But it meets with 1, which is smaller and moving in the same direction. However, they don’t collide because asteroids moving in the same direction never collide.
3. -2 moves left. It meets 1 and 2 both moving right. All three asteroids collide. The asteroid -2 survives because it is larger (in absolute value) than 1 and equal to 2.

So the output is [-2].

For [-2,2,1,-3]:

1. -2 moves left.
2. 2 moves right. It doesn’t meet -2 because they are moving in opposite directions. It meets with 1, which is smaller and moving in the same direction. But they don’t collide because asteroids moving in the same direction never collide.
3. -3 moves left. It meets 1 and 2 both moving right. All three asteroids collide. The asteroid -3 survives because it is larger (in absolute value) than 1 and 2.

So the output is [-2, -3].

In the first testcase, the 2 and -2 annihilate each other leaving nothing while in the second testcase, the -3 annihilates the 2 but because it is bigger, it doesn’t get annihilated.

That’s correct. The logic here is that when an asteroid moving to the right (positive number) encounters an asteroid moving to the left (negative number), they collide. The asteroid with the smaller absolute magnitude is destroyed. If both have the same absolute magnitude, they both get destroyed.

In the first testcase [-2,2,1,-2], the right-moving 2 collides with the left-moving -2. They have the same absolute magnitude so both are destroyed, and only the first -2 is left, hence the result is [-2].

In the second testcase [-2,2,1,-3], the right-moving 2 collides with the left-moving -3. The absolute magnitude of -3 is larger, so the 2 gets destroyed. Hence, -3 survives and the output is [-2, -3], since the first -2 never collides with any other asteroid.

[-2,-1,1,2] In this case shouldnt the answer be [], but its showing expected answer to be [-2,-1,1,2] !! ….. How man ???

and same goes with test case : [-15,15] !!

can anyone pls explain !

The order of the numbers in the array and their direction of motion matter in this problem.

An asteroid is represented by an integer in the list. The absolute value represents the size of the asteroid, while the sign represents the direction (positive values are moving right, negative values are moving left).

So in the array [-2,-1,1,2], the -2 and -1 are moving left, and the 1 and 2 are moving right. There is no case where an asteroid moving right will encounter an asteroid moving left (which is the only time a collision would occur), so none of the asteroids in this array would collide.

Similarly, in [-15,15], -15 is moving left and 15 is moving right. They won’t collide either because they’re moving in opposite directions.

So in both of these cases, the asteroids would continue on their paths without colliding, which is why the expected output matches the input.

Asteroid goes either right or left. If it is positive it goes right. If it is negative it goes left. So in the condition you mentioned, asteroid will not meet each other and collide. Positive number 1,2 goes right and negative number -2,-1 goes left

Yes, you’re correct. The direction of the asteroid is important in this problem.

In the example [-2,-1,1,2], -2 and -1 are asteroids moving to the left, while 1 and 2 are asteroids moving to the right. Since no right-moving asteroid (positive) will encounter a left-moving asteroid (negative), no collisions will occur.

Therefore, all the asteroids will continue moving in their respective directions and the final state of the asteroids will be the same as the initial state, which is [-2,-1,1,2].

This understanding is crucial for correctly solving the problem.

[-2,-1,1,2] this test actual answer is EMPTY , but result show here is - [-2,-1,1,2] why is this possible ?

Actually, the expected answer for [-2,-1,1,2] is indeed [-2,-1,1,2] due to the way the asteroids are moving.

Here’s why: the sign of the asteroid’s value indicates its direction. Negative asteroids move to the left, while positive asteroids move to the right.

In the case of [-2,-1,1,2], you have two asteroids moving to the left (-2 and -1), and two moving to the right (1 and 2).

Since there are no asteroids moving towards each other, no collisions occur. Thus, the final output is the same as the original array [-2,-1,1,2].

Input: [10,2,-5] The output can also be [10,2] as 10 will collide with -5 returning 10 and then [10,2] will be left. Why is only [10] (though it is also correct) given as only one possible output?

In the scenario [10,2,-5], the collisions actually occur in the following order:

1. The asteroid 2 (moving right) will not collide with anything as there are no asteroids moving left in front of it.

2. Then we reach -5 (moving left), which does have an asteroid moving right before it (2). Here 2 and -5 will collide. The larger asteroid (-5) survives.

3. The surviving asteroid -5 continues moving left and collides with 10. Since 10 is larger, it survives and -5 is destroyed.

This results in a final array of [10].

The reason we don’t stop after the first collision and return [10,2] is because the problem statement implies that the collisions continue until there are no more possible collisions.

That’s why the expected output is [10], not [10,2].

[-4,-1,10,2,-1,8,-9,-6,5,2] How is the expected output [-4,-1,10,5,2]? shouldnt it be [-4,-1,10,-9,-6,5,2]?

The logic behind the asteroid collision rules in this problem is that they are moving at the same speed and they only collide if a right-moving asteroid encounters a left-moving one. In your example [-4,-1,10,2,-1,8,-9,-6,5,2], the asteroids move in the following manner:

• -4 and -1 move to the left, so they never collide with other asteroids.
• 10 and 2 move to the right and they do not collide with -4 and -1 as they are moving in the opposite direction.
• -1 will collide with 2 and both will explode, leaving 10 moving to the right.
• 8 moves to the right and will not collide with anything.
• -9 moves to the left and collides with 8, resulting in both exploding as they are of equal size.
• -6 moves to the left and collides with 10, resulting in both exploding as they are of equal size.
• 5 and 2 move to the right and don’t collide with anything.

So the final list of asteroids after all collisions is [-4,-1,10,5,2], not [-4,-1,10,-9,-6,5,2].

Large positive asteroids will take out all smaller negative asteroids in their path. So the 10 will take out both the -9 and -6 leaving the expected output as it is.

Yes, that’s correct. In the problem’s context, if an asteroid moves right (with a positive number), it will collide and destroy all the smaller asteroids moving left (with a negative number) in its path.

In your case, asteroid 10 is moving to the right and it will collide with -9 and -6, which are moving to the left. Because the size of asteroid 10 is larger than the absolute values of -9 and -6, it will destroy these two asteroids. As a result, these two asteroids will not appear in the final array.

So, it’s important to understand the size and direction of each asteroid when solving this problem. If the size of the right-moving asteroid is larger, it will continue its path and destroy all the smaller left-moving asteroids it encounters. If it’s smaller, it will be destroyed instead. And if the sizes are equal, both asteroids will be destroyed.

Similar Problems

Here are some problems that share similar underlying concepts (like the use of stack, collision or dealing with ordering):

1. 20. Valid Parentheses: Similar to the asteroids problem, you use a stack to keep track of open and close parentheses. This problem also deals with collision-like events (when you have a close parenthesis that corresponds to an open one).

2. 155. Min Stack: In this problem, you need to implement a stack that can quickly return the minimum value. The stack is used to keep track of elements in a specific order, just like the asteroids problem.

3. 394. Decode String: This problem also involves stack usage. It can be thought of as another type of collision problem, where the ‘collision’ happens when a closing bracket meets its corresponding opening bracket.

4. 739. Daily Temperatures: This is another problem where you use a stack to remember certain conditions (warmer temperatures) and deal with them later, like we did with the moving asteroids.

5. 503. Next Greater Element II: This is a circular array problem which uses stack to keep track of the elements. The ‘collision’ is the situation when a greater element is found.

6. 456. 132 Pattern: Here, you have to find a pattern in an array using a stack, similar to how you kept track of asteroids in the original problem.

7. 227. Basic Calculator II: This problem involves using a stack to process operations in a specific order, similar to the order in which asteroids collide.

8. 316. Remove Duplicate Letters: In this problem, you maintain a stack of letters while ensuring that the top of the stack is always the smallest possible character.

9. 321. Create Maximum Number: Here you create the maximum number of size k from two arrays. Similar to the asteroid problem, you have to maintain an order and handle the array elements properly.

10. 402. Remove K Digits: This problem also involves maintaining the relative order of digits and using a stack to keep track of the smallest possible number, similar to how you kept track of asteroids and their collisions.

All of these involve using a stack to keep track of some order or condition, which is a key concept in the asteroids problem. Also, many of them can be seen as dealing with ‘collisions’ of some sort, similar to how the asteroids collided in the original problem.