Off-by-one on range boundaries
Wrong move: Loop endpoints miss first/last candidate.
Usually fails on: Fails on minimal arrays and exact-boundary answers.
Fix: Re-derive loops from inclusive/exclusive ranges before coding.
Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.
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.lengthn == grid[i].length1 <= n <= 500grid[i][j] is either 0 or 1.Problem summary: 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.
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Array · Union-Find
[[1,0],[0,1]]
[[1,1],[1,0]]
[[1,1],[1,1]]
Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #827: Making A Large Island
class Solution {
private int n;
private int root;
private int[] cnt;
private int[][] p;
private int[][] grid;
private final int[] dirs = {-1, 0, 1, 0, -1};
public int largestIsland(int[][] grid) {
n = grid.length;
p = new int[n][n];
this.grid = grid;
cnt = new int[n * n + 1];
int ans = 0;
for (int i = 0; i < n; ++i) {
for (int j = 0; j < n; ++j) {
if (grid[i][j] == 1 && p[i][j] == 0) {
++root;
ans = Math.max(ans, dfs(i, j));
}
}
}
for (int i = 0; i < n; ++i) {
for (int j = 0; j < n; ++j) {
if (grid[i][j] == 0) {
Set<Integer> s = new HashSet<>();
for (int k = 0; k < 4; ++k) {
int x = i + dirs[k];
int y = j + dirs[k + 1];
if (x >= 0 && x < n && y >= 0 && y < n) {
s.add(p[x][y]);
}
}
int t = 1;
for (int x : s) {
t += cnt[x];
}
ans = Math.max(ans, t);
}
}
}
return ans;
}
private int dfs(int i, int j) {
p[i][j] = root;
++cnt[root];
for (int k = 0; k < 4; ++k) {
int x = i + dirs[k];
int y = j + dirs[k + 1];
if (x >= 0 && x < n && y >= 0 && y < n && grid[x][y] == 1 && p[x][y] == 0) {
dfs(x, y);
}
}
return cnt[root];
}
}
// Accepted solution for LeetCode #827: Making A Large Island
func largestIsland(grid [][]int) int {
n := len(grid)
p := make([][]int, n)
for i := range p {
p[i] = make([]int, n)
}
cnt := make([]int, n*n+1)
dirs := []int{-1, 0, 1, 0, -1}
root := 0
ans := 0
var dfs func(int, int) int
dfs = func(i, j int) int {
p[i][j] = root
cnt[root]++
for k := 0; k < 4; k++ {
x := i + dirs[k]
y := j + dirs[k+1]
if x >= 0 && x < n && y >= 0 && y < n && grid[x][y] == 1 && p[x][y] == 0 {
dfs(x, y)
}
}
return cnt[root]
}
for i := 0; i < n; i++ {
for j := 0; j < n; j++ {
if grid[i][j] == 1 && p[i][j] == 0 {
root++
ans = max(ans, dfs(i, j))
}
}
}
for i := 0; i < n; i++ {
for j := 0; j < n; j++ {
if grid[i][j] == 0 {
s := make(map[int]struct{})
for k := 0; k < 4; k++ {
x := i + dirs[k]
y := j + dirs[k+1]
if x >= 0 && x < n && y >= 0 && y < n {
s[p[x][y]] = struct{}{}
}
}
t := 1
for x := range s {
t += cnt[x]
}
ans = max(ans, t)
}
}
}
return ans
}
# Accepted solution for LeetCode #827: Making A Large Island
class Solution:
def largestIsland(self, grid: List[List[int]]) -> int:
def dfs(i: int, j: int):
p[i][j] = root
cnt[root] += 1
for a, b in pairwise(dirs):
x, y = i + a, j + b
if 0 <= x < n and 0 <= y < n and grid[x][y] and p[x][y] == 0:
dfs(x, y)
n = len(grid)
cnt = Counter()
p = [[0] * n for _ in range(n)]
dirs = (-1, 0, 1, 0, -1)
root = 0
for i, row in enumerate(grid):
for j, x in enumerate(row):
if x and p[i][j] == 0:
root += 1
dfs(i, j)
ans = max(cnt.values() or [0])
for i, row in enumerate(grid):
for j, x in enumerate(row):
if x == 0:
s = set()
for a, b in pairwise(dirs):
x, y = i + a, j + b
if 0 <= x < n and 0 <= y < n:
s.add(p[x][y])
ans = max(ans, sum(cnt[root] for root in s) + 1)
return ans
// Accepted solution for LeetCode #827: Making A Large Island
use std::collections::HashSet;
impl Solution {
pub fn largest_island(grid: Vec<Vec<i32>>) -> i32 {
let n = grid.len();
let mut p = vec![vec![0; n]; n];
let mut cnt = vec![0; n * n + 1];
let dirs = [-1, 0, 1, 0, -1];
let mut root = 0;
let mut ans = 0;
fn dfs(
grid: &Vec<Vec<i32>>,
p: &mut Vec<Vec<i32>>,
cnt: &mut Vec<i32>,
root: i32,
i: usize,
j: usize,
dirs: &[i32; 5],
) -> i32 {
p[i][j] = root;
cnt[root as usize] += 1;
for k in 0..4 {
let x = (i as i32) + dirs[k];
let y = (j as i32) + dirs[k + 1];
if x >= 0
&& (x as usize) < grid.len()
&& y >= 0
&& (y as usize) < grid.len()
&& grid[x as usize][y as usize] == 1
&& p[x as usize][y as usize] == 0
{
dfs(grid, p, cnt, root, x as usize, y as usize, dirs);
}
}
cnt[root as usize]
}
for i in 0..n {
for j in 0..n {
if grid[i][j] == 1 && p[i][j] == 0 {
root += 1;
ans = ans.max(dfs(&grid, &mut p, &mut cnt, root, i, j, &dirs));
}
}
}
for i in 0..n {
for j in 0..n {
if grid[i][j] == 0 {
let mut s = HashSet::new();
for k in 0..4 {
let x = (i as i32) + dirs[k];
let y = (j as i32) + dirs[k + 1];
if x >= 0 && (x as usize) < n && y >= 0 && (y as usize) < n {
s.insert(p[x as usize][y as usize]);
}
}
let mut t = 1;
for &x in &s {
t += cnt[x as usize];
}
ans = ans.max(t);
}
}
}
ans
}
}
// Accepted solution for LeetCode #827: Making A Large Island
function largestIsland(grid: number[][]): number {
const n = grid.length;
const p = Array.from({ length: n }, () => Array(n).fill(0));
const cnt = Array(n * n + 1).fill(0);
const dirs = [-1, 0, 1, 0, -1];
let root = 0;
let ans = 0;
const dfs = (i: number, j: number): number => {
p[i][j] = root;
cnt[root]++;
for (let k = 0; k < 4; ++k) {
const x = i + dirs[k];
const y = j + dirs[k + 1];
if (x >= 0 && x < n && y >= 0 && y < n && grid[x][y] === 1 && p[x][y] === 0) {
dfs(x, y);
}
}
return cnt[root];
};
for (let i = 0; i < n; ++i) {
for (let j = 0; j < n; ++j) {
if (grid[i][j] === 1 && p[i][j] === 0) {
root++;
ans = Math.max(ans, dfs(i, j));
}
}
}
for (let i = 0; i < n; ++i) {
for (let j = 0; j < n; ++j) {
if (grid[i][j] === 0) {
const s = new Set<number>();
for (let k = 0; k < 4; ++k) {
const x = i + dirs[k];
const y = j + dirs[k + 1];
if (x >= 0 && x < n && y >= 0 && y < n) {
s.add(p[x][y]);
}
}
let t = 1;
for (const x of s) {
t += cnt[x];
}
ans = Math.max(ans, t);
}
}
}
return ans;
}
Use this to step through a reusable interview workflow for this problem.
Track components with a list or adjacency matrix. Each union operation may need to update all n elements’ component labels, giving O(n) per union. For n union operations total: O(n²). Find is O(1) with direct lookup, but union dominates.
With path compression and union by rank, each find/union operation takes O(α(n)) amortized time, where α is the inverse Ackermann function — effectively constant. Space is O(n) for the parent and rank arrays. For m operations on n elements: O(m × α(n)) total.
Review these before coding to avoid predictable interview regressions.
Wrong move: Loop endpoints miss first/last candidate.
Usually fails on: Fails on minimal arrays and exact-boundary answers.
Fix: Re-derive loops from inclusive/exclusive ranges before coding.