01 Matrix
Identifying Problem Isomorphism
“01 Matrix” involves using a breadth-first search (BFS) to find the shortest path to a specific type of node (in this case, a 0) from every cell in a given matrix.
An approximately isomorphic problem to this is “Rotting Oranges”. In “Rotting Oranges”, you are given a grid representing a box of oranges, where each cell may contain an orange that is either fresh or rotten. The task is to find the minimum time to rot all the oranges. A rotten orange at index [i,j] can rot oranges at [i+1,j], [i-1,j], [i,j+1], [i,j-1] in one minute. If it is impossible to rot every orange, return -1.
The similarity is that both problems involve searching a grid and finding the shortest path to a specific condition. In “01 Matrix”, the condition is a cell containing 0, while in “Rotting Oranges”, the condition is to reach and rot a fresh orange. Both problems use a BFS algorithm to find these paths. However, they are not exactly the same, hence they can be seen as approximately isomorphic.
“01 Matrix” can be approximately mapped to “Shortest Bridge”.
Reasoning:
Both problems require the application of a breadth-first search (BFS) or depth-first search (DFS) algorithm to find the shortest path in a grid.
“01 Matrix” requires modifying a 2D grid, so each cell contains the shortest distance to a cell with a 0 value.
“Shortest Bridge” involves finding the shortest bridge (i.e., minimum number of adjacent steps) from one island to another in a 2D grid. It can be considered as finding the shortest distance from cells of one value (land cells of one island) to any cells of another value (land cells of the other island).
These are not exactly isomorphic because they have different settings and goals. But they are approximately isomorphic because they share the same basic underlying structure of solving a shortest path problem in a grid.
Both problems have a similar level as they both require breadth-first search traversal on the grid.
10 Prerequisite LeetCode Problems
This requires understanding of breadth-first search (BFS) and the use of a queue to traverse a matrix, while also updating distance values. Here are some simpler problems to build up to this:
733. Flood Fill: This problem involves changing the color of a 2D image, which can be solved by BFS or DFS.
200. Number of Islands: A problem that counts the number of connected components in a grid.
994. Rotting Oranges: You are asked to find the minimum time to rot all oranges by BFS traversal.
127. Word Ladder: This problem involves finding the shortest transformation sequence from one word to another word, which also uses BFS.
279. Perfect Squares: It’s a problem that uses BFS to find the least number of perfect square numbers which sum to a given number.
286. Walls and Gates: You are to fill each empty room with the distance to its nearest gate with BFS.
130. Surrounded Regions: You can use either BFS or DFS to solve this problem which requires finding regions that are not surrounded by ‘X’.
785. Is Graph Bipartite?: This problem involves checking whether a graph is bipartite or not using BFS.
207. Course Schedule: This problem involves finding a valid order to take all courses (topological sort), you can use BFS to solve it.
417. Pacific Atlantic Water Flow: It involves BFS or DFS traversal of a matrix from the borders inward.
Clarification Questions
Here are a few clarification questions we could ask about this problem:
Can we assume the matrix is always non-empty and contains at least one zero?
What should we return if the matrix is a 1x1 matrix containing a single 0? Should we return a 1x1 matrix with 0?
Are we allowed to modify the input matrix, or do we need to return a new matrix?
In terms of distance, are we only considering the four directions: up, down, left, and right? Or are diagonal movements also allowed?
For cells that contain a zero, should we return 0 as the distance to the nearest 0 (itself), or should we find the distance to the nearest other 0?
Is there a preference for a specific time or space complexity for the solution?
Problem Analysis and Key Insights
From analyzing the problem statement, we gain several key insights:
Binary Matrix: The problem involves a binary matrix, meaning the cells can only contain the values 0 or 1.
Finding Distances: The primary task is to calculate the distance from each cell to the nearest cell containing a 0. The distance between adjacent cells is 1.
Presence of Zero: The problem statement guarantees that there will be at least one cell containing a 0 in the matrix.
Adjacent Cells: The problem implies that a cell is adjacent to up to four other cells (top, bottom, left, and right).
Output Matrix: The output should be a matrix of the same size as the input, where each cell contains the shortest distance to a cell containing 0.
Size Constraints: The size of the matrix is between 1x1 and 100x100, which implies that the solution must be efficient enough to handle matrices of up to 10,000 cells.
These insights are crucial as they outline the boundaries of the problem, and understanding them is the first step towards formulating a viable solution.
Problem Boundary
The scope of this problem is defined by the following aspects:
Data Structure Scope: The problem revolves around the manipulation and analysis of a 2-dimensional matrix, which is a common data structure. The cells in this matrix can contain only binary values, 0 or 1.
Algorithmic Scope: The problem essentially requires finding the shortest path in a grid from each cell to the nearest cell containing a 0. This falls within the scope of graph theory and shortest path algorithms.
Computational Scope: The size of the input matrix can range from 1x1 to 100x100. This means that the problem’s scope includes designing an algorithm that can efficiently handle up to 10,000 cells.
Output Scope: The expected output is another 2D matrix of the same size as the input, where each cell contains the shortest distance to a 0 cell.
Constraint Scope: The problem guarantees that there will be at least one 0 in the matrix. This constraint simplifies the problem scope since we don’t need to handle cases where there are no zeros.
By defining the scope, we can better understand the problem’s boundaries, which in turn helps us to formulate a more targeted solution approach.
Establishing the boundary of a problem involves understanding the problem’s constraints and requirements. For this problem, the boundaries are defined by the following aspects:
Input Constraints: The input is a binary matrix
matof sizemxn, wheremandnare between 1 and 100. This means the total number of cells is between 1 and 10,000. The matrix is guaranteed to contain at least one 0.Output Requirements: The output is a matrix of the same size as the input. Each cell in the output matrix represents the shortest distance from the corresponding cell in the input matrix to the nearest cell containing a 0.
Distance Calculation: The distance between two adjacent cells is 1. Here, a cell is considered adjacent if it’s directly to the left, right, above, or below another cell. This implies a 4-connected grid, excluding diagonal connections.
Computational Boundaries: The solution needs to be computationally efficient to handle matrices with up to 10,000 cells. This means the algorithm’s time complexity should ideally be linear or close to linear with respect to the number of cells.
Manipulation Constraints: The problem statement doesn’t specify whether we are allowed to modify the input matrix. If this is not allowed, our solution will need to generate a new matrix for the output without altering the input matrix.
By understanding these boundaries, we can ensure that our solution meets the problem’s requirements and operates within the specified constraints.
Problem Classification
The problem falls into the category of Graph Theory and Dynamic Programming.
The ‘What’ components of the problem are:
Input: A binary matrix
matof sizemxn, where each cell is either 0 or 1.Output: A matrix of the same size, where each cell contains the distance to the nearest 0 in the original matrix.
Constraints: The size of the matrix is bounded by 1 and 10,000. The matrix will have at least one cell with the value 0.
Distance Measurement: The distance between two adjacent cells is 1. This implies a cell can be adjacent to up to 4 other cells (top, bottom, left, and right).
The problem can be classified as a Shortest Path problem in Graph Theory. Because we can view the matrix as a grid-like graph, where each cell is a node connected to its adjacent cells, and the goal is to find the shortest path from each cell to the nearest 0. It can also be seen as a Dynamic Programming problem, as the optimal solution for each cell depends on the solutions for its neighboring cells.
Distilling the Problem to Its Core Elements
Fundamental Concept: This problem is fundamentally based on the concept of shortest path finding in a graph. The binary matrix can be viewed as a grid, where each cell is a node, and it’s connected to its adjacent cells. The task is to find the shortest path (minimum distance) from each cell to the nearest cell containing a 0.
Simplified Description: Suppose you are in a room filled with numbered boxes, where each box is either labeled ‘0’ or ‘1’. You can move from one box to an adjacent box (up, down, left, or right) with a step count of 1. If you’re at a ‘1’ box, you want to know the fewest number of steps you need to take to reach a ‘0’ box. This problem is about calculating these step counts for all ‘1’ boxes given the room layout.
Core Problem: The core problem we’re trying to solve is finding the minimum distance from each cell to the nearest 0 in a binary matrix.
Key Components: The key components of the problem are:
- The binary matrix input.
- The concept of ‘distance’ between cells.
- The condition that the distance between adjacent cells is 1.
- The task of calculating the minimum distance from each cell to the nearest 0.
Minimal Set of Operations: The minimal set of operations to solve this problem involves:
- Iterating over each cell in the matrix.
- For each cell, if it’s a 0, no further action is needed.
- If it’s a 1, we need to calculate the shortest distance to a 0. This involves exploring the neighboring cells in an orderly manner, keeping track of the distance traversed until we reach a 0. This can be effectively achieved using a breadth-first search (BFS) algorithm.
Visual Model of the Problem
Visualizing this problem involves thinking about the binary matrix as a grid or a map. Each cell in the grid represents a location that contains either a 0 or a 1. The task is to find the shortest path from each cell to the nearest cell containing a 0.
Let’s take the example from the problem statement:
Input: mat = [[0,0,0],[0,1,0],[1,1,1]]
Output: [[0,0,0],[0,1,0],[1,2,1]]
You can visualize the input matrix as a 3x3 grid:
0 0 0
0 1 0
1 1 1
Here, cells with 0 are like “safe zones”. Cells with 1 are like “start points” from where you want to find the shortest path to the nearest safe zone. You can only move up, down, left, or right, and each move counts as a distance of 1.
The output matrix then represents the minimum distance from each cell to the nearest 0:
0 0 0
0 1 0
1 2 1
For cells that already contain a 0, the distance to the nearest 0 is 0 (as it’s already a safe zone). For the cell at position (2,1), the nearest 0 is one step away. For the cell at position (3,2), the nearest 0 is two steps away, and so on.
Visualizing the problem in this way can help you better understand the problem and guide your approach to solving it.
Problem Restatement
You’re given a rectangular grid, where each cell in the grid contains either a 0 or a 1. The task is to calculate, for each cell, the shortest distance to the nearest cell that contains a 0. The distance is calculated in terms of the number of cells you’d need to traverse to reach a 0 cell, moving only up, down, left, or right (not diagonally). If a cell already contains a 0, the distance to the nearest 0 is simply 0.
The grid can be of any size, but it will always contain at least one cell and at most 10,000 cells. The problem guarantees that there will be at least one cell in the grid that contains a 0.
Your goal is to return a new grid of the same size, where each cell contains the shortest distance to a 0 from the corresponding cell in the input grid.
Abstract Representation of the Problem
An abstract representation would focus on the core elements of the problem, stripping away any unnecessary details. Here’s an abstract representation of the problem:
You are given a two-dimensional array (M x N) filled with binary values (0 or 1). Your task is to transform this array into a new array of the same dimensions. In the transformed array, each element represents the minimum number of steps required to reach a cell containing the value 0 from the corresponding cell in the original array. You can move to an adjacent cell in the original array in one step, considering only vertical or horizontal movements, not diagonal.
This abstract representation emphasizes the structure of the problem (a two-dimensional array with binary values) and the task at hand (transforming this array based on the distance to the nearest 0). It refrains from specifying any real-world context, focusing solely on the problem’s mathematical and algorithmic nature.
Terminology
There are a few technical concepts that are crucial to understanding this problem:
Binary Matrix: A binary matrix is a two-dimensional array where each element is either 0 or 1. In this problem, each cell in the given matrix is either a ‘0’ cell or a ‘1’ cell.
Graph: A graph in computer science is a data structure that consists of a set of nodes (or vertices) and a set of edges that connect these nodes. In this problem, the binary matrix can be thought of as a graph where each cell is a node and is connected to its adjacent cells.
Shortest Path: The shortest path between two nodes in a graph is the path with the minimum total distance (or weight). Here, the distance is the number of steps between cells. In this problem, we’re finding the shortest path from each ‘1’ cell to the nearest ‘0’ cell.
Breadth-First Search (BFS): BFS is a graph traversal algorithm that explores all the vertices of a graph in breadth-first order, i.e., it explores all the vertices at the present “depth” before going to the next level of depth. BFS is typically used to find the shortest path in a graph. In this problem, BFS can be used to find the shortest path from each ‘1’ cell to the nearest ‘0’ cell.
Understanding these concepts is crucial to comprehending the problem statement and devising an efficient solution.
Problem Simplification and Explanation
Let’s simplify this problem:
Think of the given binary matrix as a large field divided into small square blocks. Each block is either marked as ‘safe’ (represented by 0) or ‘start’ (represented by 1). Now, you’re standing on a ‘start’ block, and your task is to reach the nearest ‘safe’ block by moving either up, down, left, or right. Each move from one block to an adjacent block is considered a single step.
The problem wants us to calculate, for each ‘start’ block, the fewest number of steps required to reach a ‘safe’ block. If a block is already ‘safe’, the number of steps is 0.
Here’s an analogy to illustrate the problem:
Imagine you’re playing a video game where you’re in a maze filled with empty spaces (0s) and traps (1s). You can move up, down, left, or right, and each move counts as a step. When you’re on a trap, your task is to find the quickest route to an empty space. If you’re already on an empty space, you don’t need to move.
Key concepts involved are:
Binary Matrix: This is our game field or maze, where each cell is a block or space in the game.
Distance Measurement: We’re measuring the distance in terms of the number of steps between blocks in the game.
Shortest Path: We need to find the quickest route from each ‘start’ block (trap) to a ‘safe’ block (empty space), which is essentially the shortest path in the maze.
These concepts interact together to define the rules of the game, and the problem is asking for the result of the game for each block.
Constraints
There are a few specific characteristics in the problem statement that we can exploit to find an efficient solution:
Binary Values: The matrix contains only binary values (0s and 1s). This allows us to apply different strategies for 0s and 1s, simplifying the process of finding the shortest path to 0.
Existence of At Least One Zero: The problem statement guarantees that there is at least one 0 in the matrix. This ensures that there’s always a destination for the shortest path calculation and we don’t need to handle the edge case where there are no zeros in the matrix.
Adjacent Cell Movement: We can only move to an adjacent cell (up, down, left, right). This restricts the possible movements and simplifies the path exploration.
Uniform Distance: The distance between two adjacent cells is always 1. This uniformity simplifies the distance calculation as we don’t need to consider varying distances.
Matrix Size Constraints: The maximum size of the matrix is 100x100, or 10,000 elements. This constraint gives us an upper limit on the problem size, allowing us to consider algorithms with a time complexity that scales linearly or quadratically with the number of elements.
Looking for these patterns or characteristics helps us understand the problem better and guides our approach towards finding an efficient solution.
The constraints provide several important insights that help guide the approach to solving the problem:
Presence of Zero: The guarantee that there is at least one 0 in the matrix is a significant insight. It ensures that a valid solution always exists for each cell containing a 1. This allows us to safely assume that we will always find a 0 while calculating the shortest distance.
Binary Matrix: The matrix only contains 0s and 1s. This simplifies the problem as we only have to deal with two types of cells, which makes the decision process more straightforward.
Size of the Matrix: The size of the matrix is up to 100x100, or 10,000 cells. This constraint provides an upper limit on the problem size, helping us understand the expected computational complexity of our solution. We need an algorithm that can handle this amount of data efficiently.
Uniform Distance: The fact that the distance between two adjacent cells is always 1 greatly simplifies the calculation of distances. This uniformity means that we only need to count the number of steps, without worrying about different distances between cells.
Adjacency Definition: The problem defines adjacent cells as those directly above, below, left, or right of a cell, excluding diagonals. This constraint narrows down the possible movements from each cell, reducing the complexity of the path exploration.
These insights help narrow down the possible solution space and lead us towards an efficient solution. They provide valuable information about what we can expect from the input data and how we should handle it.
Case Analysis
Let’s consider a few additional examples or test cases that cover various scenarios:
Case 1: Single Cell Matrix (Edge Case)
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This is the smallest possible input, a 1x1 matrix. Since there’s only one cell and it’s a 0, the output is also a 1x1 matrix with 0.
Case 2: All Zero Matrix (Boundary Case)
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In this case, all cells are 0s. So, the shortest distance from each cell to a 0 is 0 itself.
Case 3: All One Matrix (Edge Case)
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Here, all cells are 1s except one. The cells adjacent to the 0 have a shortest distance of 1, and the cell at the top left has a shortest distance of 2.
Case 4: Zig-Zag Pattern (Normal Case)
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Here, 0s and 1s are arranged in a zig-zag pattern. Each 1 cell is surrounded by 0 cells, so the shortest distance is always 1.
Case 5: One Zero at the Center (Normal Case)
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In this case, there’s only one 0 at the center of the matrix. The cells adjacent to the 0 have a shortest distance of 1, and the corner cells have a shortest distance of 2.
These cases cover a variety of scenarios, including edge and boundary conditions, that help to ensure our solution handles all possible scenarios. They highlight the importance of correctly calculating the shortest distance and considering all possible movement directions.
Visualizing these cases can help better understand the problem and the expected results. Let’s visualize each case with the given binary matrix and the output matrix:
Case 1: Single Cell Matrix (Edge Case)
Given Matrix:
0
Output Matrix:
0
This is the simplest case, with only one cell. Since it’s a 0, the distance to the nearest 0 is also 0.
Case 2: All Zero Matrix (Boundary Case)
Given Matrix:
0 0
0 0
Output Matrix:
0 0
0 0
Every cell in this matrix is a 0. Therefore, the distance from each cell to the nearest 0 is 0.
Case 3: All One Matrix (Edge Case)
Given Matrix:
1 1
1 0
Output Matrix:
1 1
1 0
In this case, all cells but one are 1s. The cells adjacent to the 0 have a shortest distance of 1, and the top left cell has a shortest distance of 2.
Case 4: Zig-Zag Pattern (Normal Case)
Given Matrix:
0 1 0
1 0 1
0 1 0
Output Matrix:
0 1 0
1 0 1
0 1 0
In this zig-zag pattern, each 1 cell is surrounded by 0 cells, so the shortest distance is always 1.
Case 5: One Zero at the Center (Normal Case)
Given Matrix:
1 1 1
1 0 1
1 1 1
Output Matrix:
2 1 2
1 0 1
2 1 2
In this case, there’s only one 0 at the center of the matrix. The cells adjacent to the 0 have a shortest distance of 1, and the corner cells have a shortest distance of 2.
Visualizing these cases provides a more intuitive understanding of the problem and the expected results.
Analyzing the different cases provides several key insights:
Handling Zeros: For any cell that already contains a 0, the distance to the nearest 0 is always 0. This is clear from all cases where the input contains 0s.
Single Cell Matrix: In the case of a single-cell matrix, the cell is itself the nearest 0 or 1. This edge case shows that the solution needs to handle matrices of all valid sizes, including the smallest possible size.
All Zero or All One Matrix: In the case of all zeros, the output is the same as the input. If the matrix contains all ones except for one zero, the cells adjacent to the 0 have a shortest distance of 1, while others have larger distances.
Zig-Zag Pattern: In a zig-zag pattern where each 1 cell is surrounded by 0 cells, the shortest distance from each 1 to a 0 is always 1, highlighting that the solution needs to correctly calculate distances in all directions.
One Zero at the Center: When there’s only one 0 at the center of the matrix, the cells adjacent to the 0 have a shortest distance of 1, and the corner cells have a shortest distance of 2. This case demonstrates that the solution needs to handle the scenario where 0s and 1s are not evenly distributed.
These insights are crucial for understanding the problem’s requirements and constraints, which in turn helps in developing a robust and efficient solution.
Identification of Applicable Theoretical Concepts
There are several mathematical and algorithmic concepts that can be applied to simplify this problem:
Breadth-First Search (BFS): BFS is a graph traversal algorithm that can be used to find the shortest path between two nodes in an unweighted graph, which is our case here. In the context of this problem, BFS can be used to find the shortest path from each cell to the nearest 0 cell. BFS is more suitable than Depth-First Search (DFS) for this problem because BFS guarantees finding the shortest path in an unweighted graph, while DFS does not.
Distance Metric: The problem uses Manhattan Distance (also known as Taxicab geometry or L1 norm), which is the distance between two points in a grid based on a strictly horizontal and/or vertical path (as opposed to the diagonal or “as the crow flies” distance of the Euclidean distance).
Graph Theory: The problem can be viewed as a graph problem where each cell in the matrix is a node in the graph and there is an edge between every pair of adjacent cells. The weight of each edge is 1, representing one step of movement.
Dynamic Programming (DP): Though not as straightforward as BFS for this particular problem, DP can also be used to solve this problem. The idea is to perform two passes of the matrix: one top-left to bottom-right and one bottom-right to top-left. During each pass, we update each cell to be either its current value or the minimum value of its neighbors plus one (representing one step of movement).
Applying these concepts makes the problem more manageable by providing proven methods to calculate the shortest path in a graph (BFS and DP) and by defining a distance metric (Manhattan Distance) that suits the movement rules in the problem.
Simple Explanation
Imagine you’re in a large garden divided into many small squares. Some squares have a beautiful flower (we’ll call these ‘0’ squares), and some squares are empty (we’ll call these ‘1’ squares). Now, you’re standing on an empty square, and you want to walk to the nearest square with a flower.
You can only walk to a square that’s directly next to the one you’re standing on - you can go to the left, right, or move forward or backward, but you can’t move diagonally. Every step you take from one square to another counts as one step.
Now, your task is to figure out, for each empty square, the least number of steps you need to take to reach a square with a flower. For squares that already have a flower, you don’t need to move at all, so the number of steps is zero.
The problem we’re solving is basically this - but instead of a garden with flowers, we have a grid of numbers. Our job is to replace each ‘1’ with the number of steps to the nearest ‘0’.
Problem Breakdown and Solution Methodology
Let’s break down the problem-solving process into smaller steps:
Step 1: Identify the problem components
We’re given a grid with some cells marked as ‘1’ (starting points) and some as ‘0’ (destinations). The task is to find the shortest path from each ‘1’ to the nearest ‘0’, moving up, down, left, or right.
Step 2: Choose an appropriate method to solve the problem
Given the problem’s nature, Breadth-First Search (BFS) is a good fit. BFS is a graph traversal algorithm typically used to find the shortest path in an unweighted graph, which is what we have here if we consider the grid as a graph with each cell as a node.
Step 3: Implement BFS
We initialize a queue with all ‘0’ cells since they are our starting points for BFS. We also initialize a directions list representing the four possible movements (up, down, left, right). Then, we start the BFS process:
- While there are still nodes in our queue, we take the first node.
- For each direction, we calculate the new coordinates.
- If the new coordinates are valid (inside the grid) and it’s a ‘1’ cell, we update it with the current distance (which is the distance to the original node plus 1) and add it to the queue.
Step 4: Return the updated grid
Once we’ve traversed all the nodes, we return the updated grid where each ‘1’ cell has been replaced with the shortest distance to the nearest ‘0’.
How changes in the problem’s parameters affect the solution:
- If the grid gets larger, the BFS will take longer because it has to traverse more nodes.
- If there are more ‘0’ cells, the BFS might finish faster because there are more starting points.
Example Case:
Let’s demonstrate this approach using the following grid:
Input: mat = [[1, 1, 1], [1, 0, 1], [1, 1, 1]]
First, we add the ‘0’ cell to our queue and set its distance to 0. Then, we start BFS. We visit all the adjacent ‘1’ cells and update their distance to 1, then add them to the queue. We continue this process until all the ‘1’ cells have been visited and their shortest distance to a ‘0’ has been calculated.
The resulting grid would be:
Output: [[2, 1, 2], [1, 0, 1], [2, 1, 2]]
This approach ensures that each ‘1’ cell is updated with the shortest distance to the nearest ‘0’, thus solving the problem.
Inference of Problem-Solving Approach from the Problem Statement
Here are the key terms in this problem and how they inform the approach to solving it:
Binary Matrix: The problem presents us with a binary matrix, where cells are either ‘0’ or ‘1’. This allows us to treat ‘0’ cells as starting points and ‘1’ cells as cells that need to find the shortest path to a ‘0’.
Distance Between Cells: The problem asks for the shortest distance from ‘1’ cells to the nearest ‘0’. This concept guides us towards using shortest path algorithms such as Breadth-First Search (BFS).
Adjacent Cells: We’re allowed to move to adjacent cells in the matrix (up, down, left, or right). This informs us that we need to consider all four directions when exploring the matrix.
Shortest Path: The goal is to find the shortest path from ‘1’ cells to the nearest ‘0’. This directs us to use BFS, a common algorithm for finding shortest paths in unweighted graphs.
Visualizing these concepts can be done by drawing the given binary matrix as a grid and marking ‘0’ cells as starting points and ‘1’ cells as endpoints. Then, you can draw arrows between cells to represent the allowed movements. You can also use a number matrix to represent the shortest distance from each cell to the nearest ‘0’. This can help you understand the problem better and can serve as a visual check for your solution.
The problem can be interpreted as finding the shortest path from ‘1’ cells to the nearest ‘0’ cell in a grid, which is a classic scenario where Breadth-First Search (BFS) can be applied. Here are the key elements that guided this inference:
Shortest Path: The problem asks for the shortest path from each ‘1’ to the nearest ‘0’. BFS is a common algorithm for finding the shortest path in an unweighted graph, where the distance between all pairs of neighboring vertices is the same, as in this case.
Grid Navigation: We are navigating through a grid (or a 2D array), which can be seen as a graph where each cell is a node and edges connect to neighboring cells. BFS is often used in grid navigation problems, especially when the goal is to find the shortest path from one point to another.
Equal Weights: We can move to any neighboring cell with an equal “cost” of 1 step. This means our graph is unweighted (or equivalently, all edges have the same weight). BFS works particularly well and guarantees the shortest path in unweighted graphs.
All Paths: We need to find paths from ‘1’ cells to any ‘0’ cell. BFS, starting from all ‘0’ cells, naturally explores all possible paths to ‘1’ cells, ensuring we find the nearest ‘0’ for each ‘1’.
These elements of the problem align well with the characteristics and strengths of BFS, leading to the inference that BFS is a suitable method to solve this problem.
Simple Explanation of the Proof
The algorithm is called Breadth-First Search (BFS). It’s a commonly used algorithm for traversing or searching through a graph (or in our case, a grid), and it’s particularly good for finding the shortest path between two nodes in an unweighted graph. The reason it works well for this is because it explores all the neighbors of a node before moving on to their neighbors, ensuring that it finds the shortest path.
Let’s break down the proof of why BFS works for our problem:
Correctness: BFS starts from each ‘0’ cell and explores its neighbors, marking the distance for each ‘1’ cell it encounters as the current distance plus one. Because BFS explores all neighbors at the current distance before moving on to cells at the next distance, it guarantees that the first time a ‘1’ cell is visited, the marked distance is the shortest.
Completeness: BFS continues until all reachable cells have been visited, ensuring that it finds the shortest distance for all ‘1’ cells in the grid. The algorithm starts from each ‘0’ cell and ends when there are no more ‘1’ cells to visit, ensuring that every ‘1’ cell’s distance to the nearest ‘0’ is calculated.
Optimality: The algorithm doesn’t stop until all ‘1’ cells have been visited, so it explores all possible paths to each ‘1’ cell. Since it always explores shorter paths before longer ones, the first path it finds to a ‘1’ cell is guaranteed to be the shortest. Therefore, the algorithm finds the optimal (shortest) path to each ‘1’ cell.
In simpler terms, imagine you’re throwing stones in a pond from a point (these are our ‘0’ cells). The ripples spread out evenly in all directions, reaching nearby points first before moving on to further points (this is our BFS). The first ripple to reach each point is the shortest path to that point, just like our BFS finds the shortest path to each ‘1’ cell.
Stepwise Refinement
Let’s refine our approach and break it down into granular, actionable steps:
Step 1: Initial Setup
- Create a queue for the BFS algorithm.
- Iterate over the grid. For every cell, if it’s a ‘0’, add its coordinates to the queue. If it’s a ‘1’, change it to a large number (to differentiate from ‘0’ cells and to be updated later).
Step 2: Breadth-First Search (BFS)
- While the queue is not empty:
- Dequeue a cell from the queue.
- For each of its four neighboring cells (up, down, left, right):
- If the neighboring cell is within the grid and its value is larger than the current cell’s value + 1:
- Update the neighboring cell’s value to current cell’s value + 1.
- Add the neighboring cell’s coordinates to the queue.
- If the neighboring cell is within the grid and its value is larger than the current cell’s value + 1:
Step 3: Return the Updated Grid
- After the BFS completes, return the updated grid.
This problem can be broken down into these independent parts:
- Iterating over the grid to identify ‘0’ cells and initializing ‘1’ cells.
- Applying the BFS algorithm to update the ‘1’ cells with their shortest distance to a ‘0’.
There is a repeatable pattern in the solution:
- The BFS algorithm, which repeatedly dequeues a cell, checks its neighbors, and enqueues valid neighbors, is a loop with a repeating pattern.
- The way we explore each cell’s neighbors is also a repeating pattern.
This stepwise refinement and identification of independent parts and repeatable patterns helps to understand the problem better and simplify the solution.
Solution Approach and Analysis
Here’s a detailed breakdown of how to approach solving this problem:
Step 1: Understand the Problem
We’re given a grid with cells marked as either ‘0’ or ‘1’. Our task is to replace each ‘1’ cell with the shortest distance to the nearest ‘0’ cell. We can only move horizontally or vertically from one cell to another.
Step 2: Map the Problem to Known Structures/Algorithms
The grid can be viewed as a graph, where each cell is a node connected to its neighbors. The problem then becomes finding the shortest path in an unweighted graph from ‘1’ nodes to the nearest ‘0’ node, which can be solved using the Breadth-First Search (BFS) algorithm.
Step 3: Apply the BFS Algorithm
First, initialize a queue with all ‘0’ cells, since these are our starting points. We also initialize the distance for all ‘1’ cells as infinity (∞) since we have not yet calculated their distances.
Then, perform the BFS:
- While there are cells left in the queue, dequeue the first cell.
- For each of its neighboring cells, if the cell is a ‘1’ and its distance is greater than the current cell’s distance plus 1, update its distance and add it to the queue.
Step 4: Return the Result
Once we’ve visited all the cells, the distances for all ‘1’ cells have been updated with the shortest distance to the nearest ‘0’. We return this updated grid.
Effect of Changes in Parameters
If the size of the grid increases, the BFS would take longer as it has more cells to process. If there are more ‘0’ cells, it could potentially speed up the BFS as these serve as starting points for our search.
Example Case
Let’s take an example to illustrate this process:
Input: mat = [[1, 1, 1], [1, 0, 1], [1, 1, 1]]
We start BFS from the ‘0’ cell and explore its neighboring cells. For each neighboring ‘1’ cell, we update its distance as 1 and add it to the queue. We repeat this process until the queue is empty.
The updated grid will be:
Output: [[2, 1, 2], [1, 0, 1], [2, 1, 2]]
This output shows that each ‘1’ cell has been replaced with the shortest distance to the nearest ‘0’.
Identify Invariant
An invariant is a condition that remains true during the execution of an algorithm. In this problem, the invariant is:
“At the start of each iteration of the Breadth-First Search (BFS), the queue contains only cells that have been visited, and the distance of each cell in the queue represents the shortest path from that cell to the nearest ‘0’ cell in the grid.”
This invariant holds because:
We initially populate the queue with ‘0’ cells, representing that the shortest path from a ‘0’ cell to itself is 0.
During each iteration of the BFS, we dequeue a cell, examine its neighbors, and if the neighbor is a ‘1’ cell and its current distance is larger than the dequeued cell’s distance plus 1, we update the neighbor’s distance and enqueue it. This ensures that each ‘1’ cell’s distance is updated with the shortest path to a ‘0’ cell when it is first enqueued.
This process continues until the queue is empty, ensuring that we visit every cell and calculate the shortest path to a ‘0’ cell.
This invariant is essential for the correctness of the BFS algorithm in solving this problem. It ensures that we correctly calculate the shortest distance from each ‘1’ cell to the nearest ‘0’ cell.
Identify Loop Invariant
A loop invariant is a condition that is initially true and remains true after each iteration of a loop. In this problem, the loop invariant within the Breadth-First Search (BFS) algorithm is:
“At the start of each iteration of the BFS loop, every cell in the queue has been updated with the shortest distance to the nearest ‘0’ cell, and no cell outside the queue can reach a ‘0’ cell with a shorter path through any cell inside the queue.”
This invariant holds because:
At the start of the BFS, the queue contains only ‘0’ cells, and the shortest distance from any ‘0’ cell to itself is, indeed, 0.
During each iteration of the BFS, we dequeue a cell and check its neighbors. If a neighbor is a ‘1’ cell and its current distance is larger than the dequeued cell’s distance plus 1, we update the neighbor’s distance and enqueue it. This ensures that when a ‘1’ cell is enqueued, its distance is updated with the shortest path to a ‘0’ cell.
Because BFS proceeds level by level, any ‘1’ cell outside the queue cannot reach a ‘0’ cell with a shorter path through any cell inside the queue. This is because any such path would have been already discovered by the BFS and the ‘1’ cell would have been added to the queue.
The loop invariant helps us understand and ensure the correctness of the BFS algorithm. It gives us the confidence that our algorithm correctly calculates the shortest distance from each ‘1’ cell to the nearest ‘0’ cell.
In this specific problem, the terms “invariant” and “loop invariant” are closely related and refer to the same condition that is maintained throughout the execution of the algorithm. Both refer to the property that, at the start of each iteration of the Breadth-First Search (BFS), every cell in the queue has been updated with the shortest distance to the nearest ‘0’ cell, and no cell outside the queue can reach a ‘0’ cell with a shorter path through any cell inside the queue.
The term “invariant” is a broader concept and refers to a condition that remains true during the execution of an algorithm. On the other hand, a “loop invariant” is an invariant that remains true from one iteration of a loop to the next.
In the context of this problem, since our algorithm primarily consists of a loop (the BFS), the invariant and the loop invariant refer to the same condition. However, in a more complex algorithm with multiple loops or recursive calls, there could potentially be different invariants for different parts of the algorithm.
Thought Process
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This Solution class has a updateMatrix method that takes a 2D list mat as input and returns the updated 2D list. It uses the Breadth-First Search (BFS) algorithm to find the shortest distance from each ‘1’ cell to the nearest ‘0’ cell in the grid.
Time Limit Exceeded
The “Time Limit Exceeded” message usually means that your solution is correct but it isn’t efficient enough for larger inputs or the worst-case scenario.
The provided code uses Breadth-First Search (BFS) to solve the problem, which should have a time complexity of (O(mn)), where (m) and (n) are the number of rows and columns in the matrix, respectively. This is usually efficient enough for this type of problem.
However, in our current approach, we are initializing ‘1’ cells as infinity (∞) which is not necessary and adds additional time complexity. We can optimize our code by avoiding this initialization and only visiting ‘1’ cells that have not been visited before. We can achieve this by leaving ‘1’ cells as they are initially, and then during the BFS process, we only visit a cell if its value is ‘1’, which indicates that it hasn’t been visited before. This way, we only visit each cell once, which can significantly improve the performance.
Let’s modify our solution:
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This optimized solution should help avoid the “Time Limit Exceeded” error.
Wrong Answer Runtime: 63 ms Case 1 Case 2 Input mat = [[0,0,0],[0,1,0],[1,1,1]] Output [[0,0,0],[0,3,0],[3,2,3]] Expected [[0,0,0],[0,1,0],[1,2,1]]
The “Wrong Answer” message indicates that the returned result of the algorithm does not match the expected output.
Looking at the output, it appears that the ‘1’ cells are not updated correctly. The issue comes from the fact that we are updating the ‘1’ cells with mat[i][j] + 1, where mat[i][j] is the value of the current cell. This works when the current cell is a ‘0’, but not when it’s a ‘1’ that has been updated previously.
To fix this, we need to update the ‘1’ cells with mat[i][j] + 1, where (i, j) are the coordinates of the dequeued cell. This will ensure that we are always adding 1 to the distance of the closest ‘0’ cell.
Let’s correct this in the code:
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This corrected code should return the correct output for the given test case.
Wrong Answer Runtime: 60 ms Case 1 Case 2 Input mat = [[0,0,0],[0,1,0],[1,1,1]] Output [[0,0,0],[0,3,0],[3,2,3]] Expected [[0,0,0],[0,1,0],[1,2,1]]
There is an error in the logic of the BFS algorithm. The issue arises from the part where we’re checking if a cell is a ‘1’.
The check mat[new_i][new_j] == 1 is not sufficient. This is because once a ‘1’ cell is updated, it’s no longer a ‘1’. So, in later stages of the BFS, this check will fail for cells that have already been updated, and these cells won’t be processed again if needed.
Instead, we should check if the new cell’s current distance is greater than the dequeued cell’s distance plus 1. If it is, this means we’ve found a shorter path to the new cell, so we should update its distance and add it to the queue.
Here’s the corrected code:
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This corrected code should give the correct output for the given test case.
Wrong Answer Runtime: 46 ms Case 1 Case 2 Input mat = [[0,0,0],[0,1,0],[1,1,1]] Output [[0,0,0],[0,1,0],[1,1,1]] Expected [[0,0,0],[0,1,0],[1,2,1]]
The issue arises from the order in which we’re processing the neighbors of each cell. By default, we’re processing the top, bottom, left, and right neighbors in that order. However, this can lead to incorrect results if a ‘1’ cell gets updated before all of its neighbors have been processed.
To ensure that all neighbors of a ‘0’ cell are processed before any ‘1’ cell is updated, we need to fully process each level of the BFS before moving on to the next level. We can achieve this by processing the neighbors of each ‘0’ cell in the order of their distances from the ‘0’ cell.
Here’s the corrected code:
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This corrected code should give the correct output for the given test case.
from typing import List from collections import deque
class Solution: def updateMatrix(self, mat: List[List[int]]) -> List[List[int]]: rows, cols = len(mat), len(mat[0])
# Initialize all '1' cells as infinity and '0' cells as 0
dist = [[0 if mat[i][j] == 0 else float('inf') for j in range(cols)] for i in range(rows)]
# Initialize the queue with all '0' cells
queue = deque((i, j) for i in range(rows) for j in range(cols) if mat[i][j] == 0)
# Start the BFS process
while queue:
i, j = queue.popleft()
for di, dj in [(-1, 0), (1, 0), (0, -1), (0, 1)]:
new_i, new_j = i + di, j + dj
# If the new cell is within the grid and its current distance is greater than the dequeued cell's distance + 1
if 0 <= new_i < rows and 0 <= new_j < cols and dist[new_i][new_j] > dist[i][j] + 1:
# Update the cell's distance with the dequeued cell's distance + 1
dist[new_i][new_j] = dist[i][j] + 1
# Add the cell to the queue
queue.append((new_i, new_j))
return dist
Establishing Preconditions and Postconditions
Parameters:
- What are the inputs to the method?
- What types are these parameters?
- What do these parameters represent in the context of the problem?
Preconditions:
- Before this method is called, what must be true about the state of the program or the values of the parameters?
- Are there any constraints on the input parameters?
- Is there a specific state that the program or some part of it must be in?
Method Functionality:
- What is this method expected to do?
- How does it interact with the inputs and the current state of the program?
Postconditions:
- After the method has been called and has returned, what is now true about the state of the program or the values of the parameters?
- What does the return value represent or indicate?
- What side effects, if any, does the method have?
Error Handling:
- How does the method respond if the preconditions are not met?
- Does it throw an exception, return a special value, or do something else?
Problem Decomposition
Problem Understanding:
- Can you explain the problem in your own words? What are the key components and requirements?
Initial Breakdown:
- Start by identifying the major parts or stages of the problem. How can you break the problem into several broad subproblems?
Subproblem Refinement:
- For each subproblem identified, ask yourself if it can be further broken down. What are the smaller tasks that need to be done to solve each subproblem?
Task Identification:
- Within these smaller tasks, are there any that are repeated or very similar? Could these be generalized into a single, reusable task?
Task Abstraction:
- For each task you’ve identified, is it abstracted enough to be clear and reusable, but still makes sense in the context of the problem?
Method Naming:
- Can you give each task a simple, descriptive name that makes its purpose clear?
Subproblem Interactions:
- How do these subproblems or tasks interact with each other? In what order do they need to be performed? Are there any dependencies?
From Brute Force to Optimal Solution
Could you please begin by illustrating a brute force solution for this problem? After detailing and discussing the inefficiencies of the brute force approach, could you then guide us through the process of optimizing this solution? Please explain each step towards optimization, discussing the reasoning behind each decision made, and how it improves upon the previous solution. Also, could you show how these optimizations impact the time and space complexity of our solution?
Code Explanation and Design Decisions
Identify the initial parameters and explain their significance in the context of the problem statement or the solution domain.
Discuss the primary loop or iteration over the input data. What does each iteration represent in terms of the problem you’re trying to solve? How does the iteration advance or contribute to the solution?
If there are conditions or branches within the loop, what do these conditions signify? Explain the logical reasoning behind the branching in the context of the problem’s constraints or requirements.
If there are updates or modifications to parameters within the loop, clarify why these changes are necessary. How do these modifications reflect changes in the state of the solution or the constraints of the problem?
Describe any invariant that’s maintained throughout the code, and explain how it helps meet the problem’s constraints or objectives.
Discuss the significance of the final output in relation to the problem statement or solution domain. What does it represent and how does it satisfy the problem’s requirements?
Remember, the focus here is not to explain what the code does on a syntactic level, but to communicate the intent and rationale behind the code in the context of the problem being solved.
Coding Constructs
Consider the following piece of complex software code.
What are the high-level problem-solving strategies or techniques being used by this code?
If you had to explain the purpose of this code to a non-programmer, what would you say?
Can you identify the logical elements or constructs used in this code, independent of any programming language?
Could you describe the algorithmic approach used by this code in plain English?
What are the key steps or operations this code is performing on the input data, and why?
Can you identify the algorithmic patterns or strategies used by this code, irrespective of the specific programming language syntax?
Language Agnostic Coding Drills
Your mission is to deconstruct this code into the smallest possible learning units, each corresponding to a separate coding concept. Consider these concepts as unique coding drills that can be individually implemented and later assembled into the final solution.
Dissect the code and identify each distinct concept it contains. Remember, this process should be language-agnostic and generally applicable to most modern programming languages.
Once you’ve identified these coding concepts or drills, list them out in order of increasing difficulty. Provide a brief description of each concept and why it is classified at its particular difficulty level.
Next, describe the problem-solving approach that would lead from the problem statement to the final solution. Think about how each of these coding drills contributes to the overall solution. Elucidate the step-by-step process involved in using these drills to solve the problem. Please refrain from writing any actual code; we’re focusing on understanding the process and strategy.
Targeted Drills in Python
Now that you’ve identified and ordered the coding concepts from a complex software code in the previous exercise, let’s focus on creating Python-based coding drills for each of those concepts.
Begin by writing a separate piece of Python code that encapsulates each identified concept. These individual drills should illustrate how to implement each concept in Python. Please ensure that these are suitable even for those with a basic understanding of Python.
In addition to the general concepts, identify and write coding drills for any problem-specific concepts that might be needed to create a solution. Describe why these drills are essential for our problem.
Once all drills have been coded, describe how these pieces can be integrated together in the right order to solve the initial problem. Each drill should contribute to building up to the final solution.
Remember, the goal is to not only to write these drills but also to ensure that they can be cohesively assembled into one comprehensive solution.
Q&A
Similar Problems
Can you suggest 10 problems from LeetCode that require similar problem-solving strategies or use similar underlying concepts as the problem we’ve just solved? These problems can be from any domain or topic, but they should involve similar steps or techniques in the solution process. Also, please briefly explain why you consider each of these problems to be related to our original problem.