#2732

Hard

auto_awesomeArray scanning plus hash lookup

LeetCode Problem Workspace

Find a Good Subset of the Matrix

Find a Good Subset of the Matrix is a challenging problem involving array scanning, hash lookup, and bit manipulation to find valid subsets of rows in a binary matrix.

array hash table bit manipulation matrix

Problem Statement

You are given a binary matrix grid of size m x n. A non-empty subset of rows is considered 'good' if, for each column in that subset, the sum of the column's values is at most half the number of rows in the subset (rounded down). More formally, for a subset of size k, the sum of each column should be at most floor(k / 2).

Your task is to find the smallest possible subset of rows that satisfies this condition. If no such subset exists, return an empty array. The constraints on grid ensure the number of rows is at most 10^4, and the number of columns is at most 5.

Examples

Example 1

Input: grid = [[0,1,1,0],[0,0,0,1],[1,1,1,1]]

Output: [0,1]

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.

Example 2

Input: grid = [[0]]

Output: [0]

We can choose the 0th row to create a good subset of rows. The length of the chosen subset is 1.

The sum of the 0th column is 0, which is at most half of the length of the subset.

Example 3

Input: grid = [[1,1,1],[1,1,1]]

Output: []

It is impossible to choose any subset of rows to create a good subset.

Constraints

m == grid.length
n == grid[i].length
1 <= m <= 104
1 <= n <= 5
grid[i][j] is either 0 or 1.

Solution Approach

Array Scanning

A direct approach involves scanning through all possible subsets of rows in the matrix. You can use a sliding window approach to limit the range of subsets being considered and evaluate their validity. For each subset, sum the values in each column and check if it satisfies the condition for being 'good'.

Hash Lookup Optimization

To speed up checking the column sums, hash tables can be used to track column sums dynamically as rows are added or removed from the current subset. This reduces redundant calculations and allows for efficient subset evaluation.

Bit Manipulation

Given the small number of columns (maximum of 5), bit manipulation can be employed to represent and manipulate subsets of rows more efficiently. This can simplify the problem by reducing the need for explicit row-by-row comparisons and speeding up subset generation.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity of this problem depends heavily on the method used to generate and evaluate subsets of rows. A brute force approach could take O(2^m * n), where m is the number of rows and n is the number of columns. Using hash tables and bit manipulation, the time complexity can be reduced, but the problem remains inherently hard due to the large number of subsets and evaluations needed.

What Interviewers Usually Probe

Candidates should demonstrate knowledge of subset generation techniques, such as sliding windows or bit manipulation.
Look for candidates who propose optimizing the checking of column sums using hash tables or other data structures.
A good candidate will be able to reason about the trade-offs between time complexity and the constraints of the problem.

Common Pitfalls or Variants

Common pitfalls

Overlooking the need to optimize subset generation and column sum calculations can result in inefficient solutions.
Failing to handle edge cases, such as matrices with very small sizes or matrices where no valid subset exists.
Not leveraging bit manipulation for small column sizes may lead to unnecessary complexity in generating subsets.

Follow-up variants

Consider a variant where the matrix is not binary, but contains other values that must be summed and checked.
A variant where the maximum size of the subset is restricted to a particular value could be used to challenge optimization skills.
A dynamic variant where rows are added or removed from the matrix, requiring the solution to be updated incrementally.

FAQ

What is the main approach for solving the 'Find a Good Subset of the Matrix' problem?

The main approach combines array scanning with hash lookup to evaluate valid subsets of rows in the matrix, ensuring each column sum is at most half the subset's size.

How do bit manipulation and hash tables help in solving this problem?

Bit manipulation allows efficient representation of subsets, while hash tables optimize the checking of column sums, reducing redundant calculations and speeding up the process.

What are common mistakes when solving the 'Find a Good Subset of the Matrix' problem?

Common mistakes include not optimizing the subset generation process, failing to handle edge cases, and missing the benefits of bit manipulation and hash lookups.

What should I focus on to solve this problem efficiently?

Focus on reducing the complexity of subset generation and sum checking, using techniques like bit manipulation and hash tables to speed up evaluations.

How does GhostInterview help with this problem?

GhostInterview provides hints, structured solutions, and targeted guidance to help optimize subset generation and effectively use bit manipulation and hash lookup.

terminal

Solution

Solution 1: Case Analysis

We can consider the number of rows $k$ chosen for the answer from smallest to largest.

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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 []

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