#1568
Hard Algorithms Minimum number of days to disconnect island
Array Depth-First Search Breadth-First Search Matrix Strongly Connected Component
58.8% acceptance
Feb 25, 2026
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You are given an m x n binary grid grid where 1 represents land and 0 represents water. An island is a maximal 4-directionally (horizontal or vertical) connected group of 1's.
The grid is said to be connected if we have exactly one island, otherwise is said disconnected.
In one day, we are allowed to change any single land cell (1) into a water cell (0).
Return the minimum number of days to disconnect the grid.
Solution
Rust
Time O(n * m)
Space O(n * m)
impl Solution {
pub fn min_days(grid: Vec<Vec<i32>>) -> i32 {
fn count_islands(g: &Vec<Vec<i32>>) -> i32 {
let m = g.len();
let n = g[0].len();
let mut visited = vec![vec![false; n]; m];
let mut count = 0;
for r in 0..m {
for c in 0..n {
if g[r][c] == 1 && !visited[r][c] {
count += 1;
// BFS
let mut queue = std::collections::VecDeque::new();
queue.push_back((r, c));
visited[r][c] = true;
while let Some((row, col)) = queue.pop_front() {
for (dr, dc) in [(-1i32, 0), (1, 0), (0, -1i32), (0, 1)] {
let nr = row as i32 + dr;
let nc = col as i32 + dc;
if nr >= 0 && nr < m as i32 && nc >= 0 && nc < n as i32 {
let nr = nr as usize;
let nc = nc as usize;
if g[nr][nc] == 1 && !visited[nr][nc] {
visited[nr][nc] = true;
queue.push_back((nr, nc));
}
}
}
}
}
}
}
count
}
// Already disconnected or no island?
let islands = count_islands(&grid);
if islands != 1 {
return 0;
}
// Try removing 1 cell
let m = grid.len();
let n = grid[0].len();
for r in 0..m {
for c in 0..n {
if grid[r][c] == 1 {
let mut g2 = grid.clone();
g2[r][c] = 0;
if count_islands(&g2) != 1 {
return 1;
}
}
}
}
// Any connected island can be disconnected in at most 2 moves
2
}
}