Leetcode #2732: Find a Good Subset of the Matrix
In this guide, we solve Leetcode #2732 Find a Good Subset of the Matrix 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 0-indexed m x n binary matrix grid. Let us call a non-empty subset of rows good if the sum of each column of the subset is at most half of the length of the subset.
Quick Facts
- Difficulty: Hard
- Premium: No
- Tags: Bit Manipulation, Array, Hash Table, Matrix
Intuition
Fast membership checks and value lookups are the heart of this problem, which makes a hash map the natural choice.
By storing what we have already seen (or counts/indexes), we can answer the question in one pass without backtracking.
Approach
Scan the input once, using the map to detect when the condition is satisfied and to update state as you go.
This keeps the solution linear while remaining easy to explain in an interview setting.
Steps:
- Initialize a hash map for seen items or counts.
- Iterate through the input, querying/updating the map.
- Return the first valid result or the final computed value.
Example
Input: grid = [[0,1,1,0],[0,0,0,1],[1,1,1,1]]
Output: [0,1]
Explanation: We can choose the 0th and 1st rows to create a good subset of rows.
The length of the chosen subset is 2.
- The sum of the 0th column is 0 + 0 = 0, which is at most half of the length of the subset.
- The sum of the 1st column is 1 + 0 = 1, which is at most half of the length of the subset.
- The sum of the 2nd column is 1 + 0 = 1, which is at most half of the length of the subset.
- The sum of the 3rd column is 0 + 1 = 1, which is at most half of the length of the subset.
Python Solution
class Solution:
def goodSubsetofBinaryMatrix(self, grid: List[List[int]]) -> List[int]:
g = {}
for i, row in enumerate(grid):
mask = 0
for j, x in enumerate(row):
mask |= x << j
if mask == 0:
return [i]
g[mask] = i
for a, i in g.items():
for b, j in g.items():
if (a & b) == 0:
return sorted([i, j])
return []
Complexity
The time complexity is , and the space complexity is . The space complexity is .
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.