Maximum Size Square Sub Matrix
Detailed guide and Python implementation for the 'Maximum Size Square Sub Matrix' problem.
1. Concept Overview
The 'Maximum Size Square Sub Matrix' problem is a key challenge in the Dynamic Programming section.
This implementation focuses on hard-level logic in Python.
We prioritize technical accuracy and code readability in our provided solutions.
2. Real-World Applications
3. Visual Intuition
Visualizing the logic flow for Maximum Size Square Sub Matrix.
4. Prerequisites
5. Step-by-Step Thinking
1. Understand the problem
Read the problem statement for Maximum Size Square Sub Matrix carefully.
2. Formulate brute force
Draft a simple iterative solution.
3. Identify inefficiency
Look for redundant calculations.
4. Optimize search path
Use hashing or sorting to speed up the process.
5. Final Implementation
Clean up the code for production standards.
Problem Statement
Write a function max_square_submatrix(matrix) that finds the side length of the maximum square sub-matrix composed entirely of 1s in a given binary matrix.
- •1 <= len(matrix), len(matrix[0]) <= 500
- •matrix[i][j] is either 0 or 1
Examples
max_square_submatrix([[0, 1, 1, 0, 1], [1, 1, 0, 1, 0], [0, 1, 1, 1, 0], [1, 1, 1, 1, 0], [1, 1, 1, 1, 1], [0, 0, 0, 0, 0]])
3
The maximum size square sub-matrix of 1s has size 3x3, located from row 2 to 4 and column 1 to 3.
max_square_submatrix([[1, 1], [1, 1]])
2
The entire matrix is a 2x2 square of 1s.
Need a Hint?
Edge Cases to Watch
- Empty input structures
- Single element inputs
- Large numerical bounds
Ready to Solve?
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Interview Insights & Variations
Complexity Analysis Breakdown
Why Time: Directly evaluates all possibilities.
Why Space: Uses standard local memory.
Why Time: Optimized paths reduce total operations.
Why Space: May trade memory for speed.
Optimized Solution Python Code
Optimized Solution Python Code
def max_square_opt(matrix):
if not matrix: return 0
ROWS, COLS = len(matrix), len(matrix[0])
dp = [[0] * (COLS + 1) for _ in range(ROWS + 1)]
max_side = 0
for r in range(1, ROWS + 1):
for c in range(1, COLS + 1):
if matrix[r-1][c-1] == 1:
dp[r][c] = 1 + min(dp[r-1][c], dp[r][c-1], dp[r-1][c-1])
max_side = max(max_side, dp[r][c])
return max_sideBrute Force Code (Spoiler Guarded)
Brute Force Code (Spoiler Guarded)
def max_square_brute(matrix):
ROWS, COLS = len(matrix), len(matrix[0])
res = 0
for r in range(ROWS):
for c in range(COLS):
for k in range(1, min(ROWS-r, COLS-c)+1):
all_ones = True
for i in range(r, r+k):
for j in range(c, c+k):
if matrix[i][j] == 0: all_ones = False; break
if not all_ones: break
if all_ones: res = max(res, k)
return resAlgorithm Pattern Checklist
When dealing with Dynamic Programming data patterns.
Core Prerequisites
Revision Key Notes
Common Mistakes & Pitfalls
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