# 827. Making A Large Island

## Description

You are given an n x n binary matrix grid. You are allowed to change at most one 0 to be 1.

Return the size of the largest island in grid after applying this operation.

An island is a 4-directionally connected group of 1s.

Example 1:

Input: grid = [[1,0],[0,1]]
Output: 3
Explanation: Change one 0 to 1 and connect two 1s, then we get an island with area = 3.


Example 2:

Input: grid = [[1,1],[1,0]]
Output: 4
Explanation: Change the 0 to 1 and make the island bigger, only one island with area = 4.

Example 3:

Input: grid = [[1,1],[1,1]]
Output: 4
Explanation: Can't change any 0 to 1, only one island with area = 4.


Constraints:

• n == grid.length
• n == grid[i].length
• 1 <= n <= 500
• grid[i][j] is either 0 or 1.

## Solutions

Union find.

• class Solution {
private int n;
private int[] p;
private int[] size;
private int ans = 1;
private int[] dirs = new int[] {-1, 0, 1, 0, -1};

public int largestIsland(int[][] grid) {
n = grid.length;
p = new int[n * n];
size = new int[n * n];
for (int i = 0; i < p.length; ++i) {
p[i] = i;
size[i] = 1;
}
for (int i = 0; i < n; ++i) {
for (int j = 0; j < n; ++j) {
if (grid[i][j] == 1) {
for (int k = 0; k < 4; ++k) {
int x = i + dirs[k], y = j + dirs[k + 1];
if (x >= 0 && x < n && y >= 0 && y < n && grid[x][y] == 1) {
int pa = find(x * n + y), pb = find(i * n + j);
if (pa == pb) {
continue;
}
p[pa] = pb;
size[pb] += size[pa];
ans = Math.max(ans, size[pb]);
}
}
}
}
}
for (int i = 0; i < n; ++i) {
for (int j = 0; j < n; ++j) {
if (grid[i][j] == 0) {
int t = 1;
Set<Integer> vis = new HashSet<>();
for (int k = 0; k < 4; ++k) {
int x = i + dirs[k], y = j + dirs[k + 1];
if (x >= 0 && x < n && y >= 0 && y < n && grid[x][y] == 1) {
int root = find(x * n + y);
if (!vis.contains(root)) {
t += size[root];
}
}
}
ans = Math.max(ans, t);
}
}
}
return ans;
}

private int find(int x) {
if (p[x] != x) {
p[x] = find(p[x]);
}
return p[x];
}
}

• class Solution {
public:
const static inline vector<int> dirs = {-1, 0, 1, 0, -1};

int largestIsland(vector<vector<int>>& grid) {
int n = grid.size();
vector<int> p(n * n);
vector<int> size(n * n, 1);
iota(p.begin(), p.end(), 0);

function<int(int)> find;
find = [&](int x) {
if (p[x] != x) {
p[x] = find(p[x]);
}
return p[x];
};

int ans = 1;
for (int i = 0; i < n; ++i) {
for (int j = 0; j < n; ++j) {
if (grid[i][j]) {
for (int k = 0; k < 4; ++k) {
int x = i + dirs[k], y = j + dirs[k + 1];
if (x >= 0 && x < n && y >= 0 && y < n && grid[x][y]) {
int pa = find(x * n + y), pb = find(i * n + j);
if (pa == pb) continue;
p[pa] = pb;
size[pb] += size[pa];
ans = max(ans, size[pb]);
}
}
}
}
}
for (int i = 0; i < n; ++i) {
for (int j = 0; j < n; ++j) {
if (!grid[i][j]) {
int t = 1;
unordered_set<int> vis;
for (int k = 0; k < 4; ++k) {
int x = i + dirs[k], y = j + dirs[k + 1];
if (x >= 0 && x < n && y >= 0 && y < n && grid[x][y]) {
int root = find(x * n + y);
if (!vis.count(root)) {
vis.insert(root);
t += size[root];
}
}
}
ans = max(ans, t);
}
}
}
return ans;
}
};

• class Solution:
def largestIsland(self, grid: List[List[int]]) -> int:
def find(x):
if p[x] != x:
p[x] = find(p[x])
return p[x]

def union(a, b):
pa, pb = find(a), find(b)
if pa == pb:
return
p[pa] = pb
size[pb] += size[pa]

n = len(grid)
p = list(range(n * n))
size = [1] * (n * n)
for i, row in enumerate(grid):
for j, v in enumerate(row):
if v:
for a, b in [[0, -1], [-1, 0]]:
x, y = i + a, j + b
if 0 <= x < n and 0 <= y < n and grid[x][y]:
union(x * n + y, i * n + j)
ans = max(size)
for i, row in enumerate(grid):
for j, v in enumerate(row):
if v == 0:
vis = set()
t = 1
for a, b in [[0, -1], [0, 1], [1, 0], [-1, 0]]:
x, y = i + a, j + b
if 0 <= x < n and 0 <= y < n and grid[x][y]:
root = find(x * n + y)
if root not in vis:
t += size[root]
ans = max(ans, t)
return ans


• func largestIsland(grid [][]int) int {
n := len(grid)
p := make([]int, n*n)
size := make([]int, n*n)
for i := range p {
p[i] = i
size[i] = 1
}
var find func(int) int
find = func(x int) int {
if p[x] != x {
p[x] = find(p[x])
}
return p[x]
}
dirs := []int{-1, 0, 1, 0, -1}
ans := 1
for i, row := range grid {
for j, v := range row {
if v == 1 {
for k := 0; k < 4; k++ {
x, y := i+dirs[k], j+dirs[k+1]
if x >= 0 && x < n && y >= 0 && y < n && grid[x][y] == 1 {
pa, pb := find(x*n+y), find(i*n+j)
if pa != pb {
p[pa] = pb
size[pb] += size[pa]
ans = max(ans, size[pb])
}
}
}
}
}
}
for i, row := range grid {
for j, v := range row {
if v == 0 {
t := 1
vis := map[int]struct{}{}
for k := 0; k < 4; k++ {
x, y := i+dirs[k], j+dirs[k+1]
if x >= 0 && x < n && y >= 0 && y < n && grid[x][y] == 1 {
root := find(x*n + y)
if _, ok := vis[root]; !ok {
vis[root] = struct{}{}
t += size[root]
}
}
}
ans = max(ans, t)
}
}
}
return ans
}

• function largestIsland(grid: number[][]): number {
const n = grid.length;
const vis = Array.from({ length: n }, () => new Array(n).fill(false));
const group = Array.from({ length: n }, () => new Array(n).fill(0));
const dfs = (i: number, j: number, paths: [number, number][]) => {
if (i < 0 || j < 0 || i === n || j === n || vis[i][j] || grid[i][j] !== 1) {
return;
}
vis[i][j] = true;
paths.push([i, j]);
dfs(i + 1, j, paths);
dfs(i, j + 1, paths);
dfs(i - 1, j, paths);
dfs(i, j - 1, paths);
};
let count = 1;
for (let i = 0; i < n; i++) {
for (let j = 0; j < n; j++) {
const paths: [number, number][] = [];
dfs(i, j, paths);
if (paths.length !== 0) {
for (const [x, y] of paths) {
group[x][y] = count;
grid[x][y] = paths.length;
}
count++;
}
}
}

let res = 0;
for (let i = 0; i < n; i++) {
for (let j = 0; j < n; j++) {
let sum = grid[i][j];
if (grid[i][j] === 0) {
sum++;
const set = new Set();
if (i !== 0) {
sum += grid[i - 1][j];
}
if (i !== n - 1 && !set.has(group[i + 1][j])) {
sum += grid[i + 1][j];
}
if (j !== 0 && !set.has(group[i][j - 1])) {
sum += grid[i][j - 1];
}
if (j !== n - 1 && !set.has(group[i][j + 1])) {
sum += grid[i][j + 1];
}
}
res = Math.max(res, sum);
}
}
return res;
}


• use std::collections::HashSet;
impl Solution {
fn dfs(
i: usize,
j: usize,
grid: &Vec<Vec<i32>>,
paths: &mut Vec<(usize, usize)>,
vis: &mut Vec<Vec<bool>>
) {
let n = vis.len();
if vis[i][j] || grid[i][j] != 1 {
return;
}
paths.push((i, j));
vis[i][j] = true;
if i != 0 {
Self::dfs(i - 1, j, grid, paths, vis);
}
if j != 0 {
Self::dfs(i, j - 1, grid, paths, vis);
}
if i != n - 1 {
Self::dfs(i + 1, j, grid, paths, vis);
}
if j != n - 1 {
Self::dfs(i, j + 1, grid, paths, vis);
}
}

pub fn largest_island(mut grid: Vec<Vec<i32>>) -> i32 {
let n = grid.len();
let mut vis = vec![vec![false; n]; n];
let mut group = vec![vec![0; n]; n];
let mut count = 1;
for i in 0..n {
for j in 0..n {
let mut paths: Vec<(usize, usize)> = Vec::new();
Self::dfs(i, j, &grid, &mut paths, &mut vis);
let m = paths.len() as i32;
if m != 0 {
for (x, y) in paths {
grid[x][y] = m;
group[x][y] = count;
}
count += 1;
}
}
}
let mut res = 0;
for i in 0..n {
for j in 0..n {
let mut sum = grid[i][j];
if grid[i][j] == 0 {
sum += 1;
let mut set = HashSet::new();
if i != 0 {
sum += grid[i - 1][j];
set.insert(group[i - 1][j]);
}
if j != 0 && !set.contains(&group[i][j - 1]) {
sum += grid[i][j - 1];
set.insert(group[i][j - 1]);
}
if i != n - 1 && !set.contains(&group[i + 1][j]) {
sum += grid[i + 1][j];
set.insert(group[i + 1][j]);
}
if j != n - 1 && !set.contains(&group[i][j + 1]) {
sum += grid[i][j + 1];
set.insert(group[i][j + 1]);
}
}
res = res.max(sum);
}
}
res
}
}