Leetcode #1727: Largest Submatrix With Rearrangements
In this guide, we solve Leetcode #1727 Largest Submatrix With Rearrangements 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
You are given a binary matrix matrix of size m x n, and you are allowed to rearrange the columns of the matrix in any order. Return the area of the largest submatrix within matrix where every element of the submatrix is 1 after reordering the columns optimally.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Greedy, Array, Matrix, Sorting
Intuition
A locally optimal choice leads to a globally optimal result for this structure.
That means we can commit to decisions as we scan without backtracking.
Approach
Sort or preprocess if needed, then repeatedly take the best available local choice.
Maintain the minimal state necessary to validate the greedy decision.
Steps:
- Sort or preprocess as needed.
- Iterate and pick the best local option.
- Track the current solution.
Example
Input: matrix = [[0,0,1],[1,1,1],[1,0,1]]
Output: 4
Explanation: You can rearrange the columns as shown above.
The largest submatrix of 1s, in bold, has an area of 4.
Python Solution
class Solution:
def largestSubmatrix(self, matrix: List[List[int]]) -> int:
for i in range(1, len(matrix)):
for j in range(len(matrix[0])):
if matrix[i][j]:
matrix[i][j] = matrix[i - 1][j] + 1
ans = 0
for row in matrix:
row.sort(reverse=True)
for j, v in enumerate(row, 1):
ans = max(ans, j * v)
return ans
Complexity
The time complexity is . The space complexity is O(1) to O(n).
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.