# 2732. Find a Good Subset of the Matrix

## Description

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.

More formally, if the length of the chosen subset of rows is k, then the sum of each column should be at most floor(k / 2).

Return an integer array that contains row indices of a good subset sorted in ascending order.

If there are multiple good subsets, you can return any of them. If there are no good subsets, return an empty array.

A subset of rows of the matrix grid is any matrix that can be obtained by deleting some (possibly none or all) rows from grid.

Example 1:

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.


Example 2:

Input: grid = [[0]]
Output: [0]
Explanation: 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: []
Explanation: 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.

## Solutions

• class Solution {
public List<Integer> goodSubsetofBinaryMatrix(int[][] grid) {
Map<Integer, Integer> g = new HashMap<>();
for (int i = 0; i < grid.length; ++i) {
for (int j = 0; j < grid[0].length; ++j) {
}
return List.of(i);
}
}
for (var e1 : g.entrySet()) {
for (var e2 : g.entrySet()) {
if ((e1.getKey() & e2.getKey()) == 0) {
int i = e1.getValue(), j = e2.getValue();
return List.of(Math.min(i, j), Math.max(i, j));
}
}
}
return List.of();
}
}

• class Solution {
public:
vector<int> goodSubsetofBinaryMatrix(vector<vector<int>>& grid) {
unordered_map<int, int> g;
for (int i = 0; i < grid.size(); ++i) {
for (int j = 0; j < grid[0].size(); ++j) {
}
return {i};
}
}
for (auto& [a, i] : g) {
for (auto& [b, j] : g) {
if ((a & b) == 0) {
return {min(i, j), max(i, j)};
}
}
}
return {};
}
};

• class Solution:
def goodSubsetofBinaryMatrix(self, grid: List[List[int]]) -> List[int]:
g = {}
for i, row in enumerate(grid):
for j, x in enumerate(row):
return [i]
for a, i in g.items():
for b, j in g.items():
if (a & b) == 0:
return sorted([i, j])
return []


• func goodSubsetofBinaryMatrix(grid [][]int) []int {
g := map[int]int{}
for i, row := range grid {
for j, x := range row {
}
return []int{i}
}
}
for a, i := range g {
for b, j := range g {
if a&b == 0 {
return []int{min(i, j), max(i, j)}
}
}
}
return []int{}
}

• function goodSubsetofBinaryMatrix(grid: number[][]): number[] {
const g: Map<number, number> = new Map();
const m = grid.length;
const n = grid[0].length;
for (let i = 0; i < m; ++i) {
for (let j = 0; j < n; ++j) {
}
return [i];
}
}
for (const [a, i] of g.entries()) {
for (const [b, j] of g.entries()) {
if ((a & b) === 0) {
return [Math.min(i, j), Math.max(i, j)];
}
}
}
return [];
}


• use std::collections::HashMap;

impl Solution {
pub fn good_subsetof_binary_matrix(grid: Vec<Vec<i32>>) -> Vec<i32> {
let mut g: HashMap<i32, i32> = HashMap::new();
for (i, row) in grid.iter().enumerate() {
for (j, &x) in row.iter().enumerate() {
}
return vec![i as i32];
}
}

for (&a, &i) in g.iter() {
for (&b, &j) in g.iter() {
if (a & b) == 0 {
return vec![i.min(j), i.max(j)];
}
}
}

vec![]
}
}