# 2123. Minimum Operations to Remove Adjacent Ones in Matrix

## Description

You are given a 0-indexed binary matrix grid. In one operation, you can flip any 1 in grid to be 0.

A binary matrix is well-isolated if there is no 1 in the matrix that is 4-directionally connected (i.e., horizontal and vertical) to another 1.

Return the minimum number of operations to make grid well-isolated.

Example 1:

Input: grid = [[1,1,0],[0,1,1],[1,1,1]]
Output: 3
Explanation: Use 3 operations to change grid[0][1], grid[1][2], and grid[2][1] to 0.
After, no more 1's are 4-directionally connected and grid is well-isolated.


Example 2:

Input: grid = [[0,0,0],[0,0,0],[0,0,0]]
Output: 0
Explanation: There are no 1's in grid and it is well-isolated.
No operations were done so return 0.


Example 3:

Input: grid = [[0,1],[1,0]]
Output: 0
Explanation: None of the 1's are 4-directionally connected and grid is well-isolated.
No operations were done so return 0.


Constraints:

• m == grid.length
• n == grid[i].length
• 1 <= m, n <= 300
• grid[i][j] is either 0 or 1.

## Solutions

• class Solution {
private Map<Integer, List<Integer>> g = new HashMap<>();
private Set<Integer> vis = new HashSet<>();
private int[] match;

public int minimumOperations(int[][] grid) {
int m = grid.length, n = grid[0].length;
for (int i = 0; i < m; ++i) {
for (int j = 0; j < n; ++j) {
if ((i + j) % 2 == 1 && grid[i][j] == 1) {
int x = i * n + j;
if (i < m - 1 && grid[i + 1][j] == 1) {
g.computeIfAbsent(x, z -> new ArrayList<>()).add(x + n);
}
if (i > 0 && grid[i - 1][j] == 1) {
g.computeIfAbsent(x, z -> new ArrayList<>()).add(x - n);
}
if (j < n - 1 && grid[i][j + 1] == 1) {
g.computeIfAbsent(x, z -> new ArrayList<>()).add(x + 1);
}
if (j > 0 && grid[i][j - 1] == 1) {
g.computeIfAbsent(x, z -> new ArrayList<>()).add(x - 1);
}
}
}
}
match = new int[m * n];
Arrays.fill(match, -1);
int ans = 0;
for (int i : g.keySet()) {
ans += find(i);
vis.clear();
}
return ans;
}

private int find(int i) {
for (int j : g.get(i)) {
if (match[j] == -1 || find(match[j]) == 1) {
match[j] = i;
return 1;
}
}
}
return 0;
}
}

• class Solution {
public:
int minimumOperations(vector<vector<int>>& grid) {
int m = grid.size(), n = grid[0].size();
vector<int> match(m * n, -1);
unordered_set<int> vis;
unordered_map<int, vector<int>> g;
for (int i = 0; i < m; ++i) {
for (int j = 0; j < n; ++j) {
if ((i + j) % 2 && grid[i][j]) {
int x = i * n + j;
if (i < m - 1 && grid[i + 1][j]) {
g[x].push_back(x + n);
}
if (i && grid[i - 1][j]) {
g[x].push_back(x - n);
}
if (j < n - 1 && grid[i][j + 1]) {
g[x].push_back(x + 1);
}
if (j && grid[i][j - 1]) {
g[x].push_back(x - 1);
}
}
}
}
int ans = 0;
function<int(int)> find = [&](int i) -> int {
for (int& j : g[i]) {
if (!vis.count(j)) {
vis.insert(j);
if (match[j] == -1 || find(match[j])) {
match[j] = i;
return 1;
}
}
}
return 0;
};
for (auto& [i, _] : g) {
ans += find(i);
vis.clear();
}
return ans;
}
};

• class Solution:
def minimumOperations(self, grid: List[List[int]]) -> int:
def find(i: int) -> int:
for j in g[i]:
if j not in vis:
if match[j] == -1 or find(match[j]):
match[j] = i
return 1
return 0

g = defaultdict(list)
m, n = len(grid), len(grid[0])
for i, row in enumerate(grid):
for j, v in enumerate(row):
if (i + j) % 2 and v:
x = i * n + j
if i < m - 1 and grid[i + 1][j]:
g[x].append(x + n)
if i and grid[i - 1][j]:
g[x].append(x - n)
if j < n - 1 and grid[i][j + 1]:
g[x].append(x + 1)
if j and grid[i][j - 1]:
g[x].append(x - 1)

match = [-1] * (m * n)
ans = 0
for i in g.keys():
vis = set()
ans += find(i)
return ans


• func minimumOperations(grid [][]int) (ans int) {
m, n := len(grid), len(grid[0])
vis := map[int]bool{}
match := make([]int, m*n)
for i := range match {
match[i] = -1
}
g := map[int][]int{}
for i, row := range grid {
for j, v := range row {
if (i+j)&1 == 1 && v == 1 {
x := i*n + j
if i < m-1 && grid[i+1][j] == 1 {
g[x] = append(g[x], x+n)
}
if i > 0 && grid[i-1][j] == 1 {
g[x] = append(g[x], x-n)
}
if j < n-1 && grid[i][j+1] == 1 {
g[x] = append(g[x], x+1)
}
if j > 0 && grid[i][j-1] == 1 {
g[x] = append(g[x], x-1)
}
}
}
}
var find func(int) int
find = func(i int) int {
for _, j := range g[i] {
if !vis[j] {
vis[j] = true
if match[j] == -1 || find(match[j]) == 1 {
match[j] = i
return 1
}
}
}
return 0
}
for i := range g {
ans += find(i)
vis = map[int]bool{}
}
return
}

• function minimumOperations(grid: number[][]): number {
const m = grid.length;
const n = grid[0].length;
const match: number[] = Array(m * n).fill(-1);
const vis: Set<number> = new Set();
const g: Map<number, number[]> = new Map();
for (let i = 0; i < m; ++i) {
for (let j = 0; j < n; ++j) {
if ((i + j) % 2 && grid[i][j]) {
const x = i * n + j;
g.set(x, []);
if (i < m - 1 && grid[i + 1][j]) {
g.get(x)!.push(x + n);
}
if (i && grid[i - 1][j]) {
g.get(x)!.push(x - n);
}
if (j < n - 1 && grid[i][j + 1]) {
g.get(x)!.push(x + 1);
}
if (j && grid[i][j - 1]) {
g.get(x)!.push(x - 1);
}
}
}
}
const find = (i: number): number => {
for (const j of g.get(i)!) {
if (!vis.has(j)) {
if (match[j] === -1 || find(match[j])) {
match[j] = i;
return 1;
}
}
}
return 0;
};
let ans = 0;
for (const i of g.keys()) {
ans += find(i);
vis.clear();
}
return ans;
}