Leetcode #85: Maximal Rectangle
In this guide, we solve Leetcode #85 Maximal Rectangle in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
Given a rows x cols binary matrix filled with 0's and 1's, find the largest rectangle containing only 1's and return its area. 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: 6 Explanation: The maximal rectangle is shown in the above picture.
Quick Facts
- Difficulty: Hard
- Premium: No
- Tags: Stack, Array, Dynamic Programming, Matrix, Monotonic Stack
Intuition
The problem breaks into overlapping subproblems, so caching results prevents exponential repetition.
A carefully chosen DP state captures exactly what we need to build the final answer.
Approach
Define the DP state and recurrence, then compute states in the correct order.
Optionally compress space once the recurrence is clear.
Steps:
- Choose a DP state definition.
- Write the recurrence and base cases.
- Compute states in the correct order.
Example
Input: matrix = [["1","0","1","0","0"],["1","0","1","1","1"],["1","1","1","1","1"],["1","0","0","1","0"]]
Output: 6
Explanation: The maximal rectangle is shown in the above picture.
Python Solution
class Solution:
def maximalRectangle(self, matrix: List[List[str]]) -> int:
heights = [0] * len(matrix[0])
ans = 0
for row in matrix:
for j, v in enumerate(row):
if v == "1":
heights[j] += 1
else:
heights[j] = 0
ans = max(ans, self.largestRectangleArea(heights))
return ans
def largestRectangleArea(self, heights: List[int]) -> int:
n = len(heights)
stk = []
left = [-1] * n
right = [n] * n
for i, h in enumerate(heights):
while stk and heights[stk[-1]] >= h:
stk.pop()
if stk:
left[i] = stk[-1]
stk.append(i)
stk = []
for i in range(n - 1, -1, -1):
h = heights[i]
while stk and heights[stk[-1]] >= h:
stk.pop()
if stk:
right[i] = stk[-1]
stk.append(i)
return max(h * (right[i] - left[i] - 1) for i, h in enumerate(heights))
Complexity
The time complexity is , where is the number of rows in and is the number of columns in . The space complexity is O(n·m) or optimized.
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.