To solve Maximal Square, use dynamic programming to determine the largest square of 1's. By iterating through the matrix and updating a state matrix, the problem reduces to checking for square size at each point. The final result will be the area of the largest square found.
Problem Statement
Given a binary matrix of size m x n filled with 0's and 1's, find the largest square that can be formed by 1's. Return the area of this square.
For example, consider the matrix [["1","0","1","0","0"], ["1","0","1","1","1"], ["1","1","1","1","1"], ["1","0","0","1","0"]]. The largest square of 1's has an area of 4.
Examples
Example 1
Input: matrix = [["1","0","1","0","0"],["1","0","1","1","1"],["1","1","1","1","1"],["1","0","0","1","0"]]
Output: 4
Example details omitted.
Example 2
Input: matrix = [["0","1"],["1","0"]]
Output: 1
Example details omitted.
Example 3
Input: matrix = [["0"]]
Output: 0
Example details omitted.
Constraints
- m == matrix.length
- n == matrix[i].length
- 1 <= m, n <= 300
- matrix[i][j] is '0' or '1'.
Solution Approach
State Transition Dynamic Programming
The problem is solved by maintaining a dynamic programming (DP) matrix where each cell stores the size of the largest square that can end at that position. Transition between states involves checking the left, top, and diagonal cells and taking the minimum of the three plus one, which ensures the square grows optimally.
Iterating through the Matrix
Start iterating through each cell in the matrix. If the cell contains a 1, update the DP matrix by using the values from the previous row, column, and diagonal. If the cell contains a 0, continue with the next iteration, as no square can end at that position.
Final Result Computation
The maximum value in the DP matrix represents the side length of the largest square. Squaring this value gives the area, which is the final answer.
Complexity Analysis
| Metric | Value |
|---|---|
| Time | Depends on the final approach |
| Space | Depends on the final approach |
Time complexity is O(m * n) where m and n are the dimensions of the matrix, as each cell is processed once. Space complexity can be optimized to O(n) by keeping only the current and previous rows of the DP matrix, reducing the overall space usage from O(m * n).
What Interviewers Usually Probe
- Assess the candidate's ability to optimize space in dynamic programming solutions.
- Evaluate how well the candidate explains and uses state transitions in dynamic programming.
- Check if the candidate can identify the trade-offs between time and space complexities.
Common Pitfalls or Variants
Common pitfalls
- Forgetting to initialize the first row and first column of the DP matrix correctly.
- Using a brute force approach to check every potential square without leveraging dynamic programming.
- Not optimizing space by reducing the DP matrix to only necessary rows.
Follow-up variants
- Modify the problem to find the largest square of 1's within a submatrix of the matrix.
- Consider a version where the matrix size can vary during runtime or input.
- Challenge the candidate to solve the problem without using dynamic programming for a brute force solution.
How GhostInterview Helps
- GhostInterview's dynamic programming suggestions ensure the candidate can approach similar matrix problems with an optimized solution.
- The tool helps in checking edge cases, like small matrices or matrices full of 0's, to make sure the candidate handles all scenarios.
- GhostInterview offers guidance on explaining time and space trade-offs in real-time, which can be valuable for interview scenarios.
Topic Pages
Related GhostInterview Pages
- LeetCode Interview Copilot - Use GhostInterview as a live solver when you want direct help with LeetCode-style coding questions.
- Coding Interview Assistant - See how GhostInterview supports array, string, linked list, graph, and tree interview workflows.
- How GhostInterview Works - Review the screenshot, reasoning, and answer flow before using the solver in a live interview.
FAQ
What is the core pattern used to solve the Maximal Square problem?
The problem is primarily solved using state transition dynamic programming, where each matrix cell tracks the largest square that can end at that position.
Can the Maximal Square problem be solved without dynamic programming?
While it's possible to solve it with brute force, dynamic programming offers an efficient solution, reducing time complexity from O(m * n^2) to O(m * n).
How do you optimize space in Maximal Square's solution?
Space can be optimized by storing only the current and previous rows of the DP matrix, reducing space complexity from O(m * n) to O(n).
What happens if the matrix contains all 0's in the Maximal Square problem?
If the matrix consists solely of 0's, the largest square will have an area of 0, as no 1's are available to form a square.
What is the expected output for a 1x1 matrix in the Maximal Square problem?
For a 1x1 matrix, the output will either be 0 or 1, depending on whether the single cell contains a 0 or 1.
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