LeetCode #1568 — HARD

Minimum Number of Days to Disconnect Island

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

The Problem

Problem Statement

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.

Example 1:

Input: grid = [[0,1,1,0],[0,1,1,0],[0,0,0,0]]

Output: 2
Explanation: We need at least 2 days to get a disconnected grid.
Change land grid[1][1] and grid[0][2] to water and get 2 disconnected island.

Example 2:

Input: grid = [[1,1]]
Output: 2
Explanation: Grid of full water is also disconnected ([[1,1]] -> [[0,0]]), 0 islands.

Constraints:

m == grid.length
n == grid[i].length
1 <= m, n <= 30
grid[i][j] is either 0 or 1.

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array

Example 1

[[0,1,1,0],[0,1,1,0],[0,0,0,0]]

Example 2

[[1,1]]

Core Insight

What unlocks the optimal approach

Return 0 if the grid is already disconnected.
Return 1 if changing a single land to water disconnect the island.
Otherwise return 2.
We can disconnect the grid within at most 2 days.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Largest constraint values

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1568: Minimum Number of Days to Disconnect Island
class Solution {
    private static final int[] DIRS = new int[] {-1, 0, 1, 0, -1};
    private int[][] grid;
    private int m;
    private int n;

    public int minDays(int[][] grid) {
        this.grid = grid;
        m = grid.length;
        n = grid[0].length;
        if (count() != 1) {
            return 0;
        }
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                if (grid[i][j] == 1) {
                    grid[i][j] = 0;
                    if (count() != 1) {
                        return 1;
                    }
                    grid[i][j] = 1;
                }
            }
        }
        return 2;
    }

    private int count() {
        int cnt = 0;
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                if (grid[i][j] == 1) {
                    dfs(i, j);
                    ++cnt;
                }
            }
        }
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                if (grid[i][j] == 2) {
                    grid[i][j] = 1;
                }
            }
        }
        return cnt;
    }

    private void dfs(int i, int j) {
        grid[i][j] = 2;
        for (int k = 0; k < 4; ++k) {
            int x = i + DIRS[k], y = j + DIRS[k + 1];
            if (x >= 0 && x < m && y >= 0 && y < n && grid[x][y] == 1) {
                dfs(x, y);
            }
        }
    }
}

// Accepted solution for LeetCode #1568: Minimum Number of Days to Disconnect Island
func minDays(grid [][]int) int {
	m, n := len(grid), len(grid[0])
	dirs := []int{-1, 0, 1, 0, -1}

	var dfs func(i, j int)
	dfs = func(i, j int) {
		grid[i][j] = 2
		for k := 0; k < 4; k++ {
			x, y := i+dirs[k], j+dirs[k+1]
			if x >= 0 && x < m && y >= 0 && y < n && grid[x][y] == 1 {
				dfs(x, y)
			}
		}
	}

	count := func() int {
		cnt := 0
		for i := 0; i < m; i++ {
			for j := 0; j < n; j++ {
				if grid[i][j] == 1 {
					dfs(i, j)
					cnt++
				}
			}
		}
		for i := 0; i < m; i++ {
			for j := 0; j < n; j++ {
				if grid[i][j] == 2 {
					grid[i][j] = 1
				}
			}
		}
		return cnt
	}

	if count() != 1 {
		return 0
	}
	for i := 0; i < m; i++ {
		for j := 0; j < n; j++ {
			if grid[i][j] == 1 {
				grid[i][j] = 0
				if count() != 1 {
					return 1
				}
				grid[i][j] = 1
			}
		}
	}
	return 2
}

# Accepted solution for LeetCode #1568: Minimum Number of Days to Disconnect Island
class Solution:
    def minDays(self, grid: List[List[int]]) -> int:
        if self.count(grid) != 1:
            return 0
        m, n = len(grid), len(grid[0])
        for i in range(m):
            for j in range(n):
                if grid[i][j] == 1:
                    grid[i][j] = 0
                    if self.count(grid) != 1:
                        return 1
                    grid[i][j] = 1
        return 2

    def count(self, grid):
        def dfs(i, j):
            grid[i][j] = 2
            for a, b in [[0, -1], [0, 1], [1, 0], [-1, 0]]:
                x, y = i + a, j + b
                if 0 <= x < m and 0 <= y < n and grid[x][y] == 1:
                    dfs(x, y)

        m, n = len(grid), len(grid[0])
        cnt = 0
        for i in range(m):
            for j in range(n):
                if grid[i][j] == 1:
                    dfs(i, j)
                    cnt += 1
        for i in range(m):
            for j in range(n):
                if grid[i][j] == 2:
                    grid[i][j] = 1
        return cnt

// Accepted solution for LeetCode #1568: Minimum Number of Days to Disconnect Island
struct Solution;

impl Solution {
    fn min_days(mut grid: Vec<Vec<i32>>) -> i32 {
        let n = grid.len();
        let m = grid[0].len();

        if Self::islands(&grid, n, m) != 1 {
            return 0;
        }

        for i in 0..n {
            for j in 0..m {
                if grid[i][j] == 1 {
                    grid[i][j] = 0;
                    if Self::islands(&grid, n, m) > 1 {
                        return 1;
                    }
                    grid[i][j] = 1;
                }
            }
        }

        2
    }

    fn islands(grid: &[Vec<i32>], n: usize, m: usize) -> i32 {
        let mut visited = vec![vec![false; m]; n];
        let mut res = 0;
        for i in 0..n {
            for j in 0..m {
                if !visited[i][j] && grid[i][j] == 1 {
                    res += 1;
                    visited[i][j] = true;
                    Self::dfs(i, j, &mut visited, grid, n, m);
                }
            }
        }
        res
    }

    fn dfs(
        i: usize,
        j: usize,
        visited: &mut Vec<Vec<bool>>,
        grid: &[Vec<i32>],
        n: usize,
        m: usize,
    ) {
        if i > 0 && grid[i - 1][j] == 1 && !visited[i - 1][j] {
            visited[i - 1][j] = true;
            Self::dfs(i - 1, j, visited, grid, n, m);
        }
        if j > 0 && grid[i][j - 1] == 1 && !visited[i][j - 1] {
            visited[i][j - 1] = true;
            Self::dfs(i, j - 1, visited, grid, n, m);
        }
        if i + 1 < n && grid[i + 1][j] == 1 && !visited[i + 1][j] {
            visited[i + 1][j] = true;
            Self::dfs(i + 1, j, visited, grid, n, m);
        }
        if j + 1 < m && grid[i][j + 1] == 1 && !visited[i][j + 1] {
            visited[i][j + 1] = true;
            Self::dfs(i, j + 1, visited, grid, n, m);
        }
    }
}

#[test]
fn test() {
    let grid = vec_vec_i32![[0, 1, 1, 0], [0, 1, 1, 0], [0, 0, 0, 0]];
    let res = 2;
    assert_eq!(Solution::min_days(grid), res);
    let grid = vec_vec_i32![[1, 1]];
    let res = 2;
    assert_eq!(Solution::min_days(grid), res);
    let grid = vec_vec_i32![[1, 0, 1, 0]];
    let res = 0;
    assert_eq!(Solution::min_days(grid), res);
    let grid = vec_vec_i32![
        [1, 1, 0, 1, 1],
        [1, 1, 1, 1, 1],
        [1, 1, 0, 1, 1],
        [1, 1, 0, 1, 1]
    ];
    let res = 1;
    assert_eq!(Solution::min_days(grid), res);
    let grid = vec_vec_i32![
        [1, 1, 0, 1, 1],
        [1, 1, 1, 1, 1],
        [1, 1, 0, 1, 1],
        [1, 1, 1, 1, 1]
    ];
    let res = 2;
    assert_eq!(Solution::min_days(grid), res);
}

// Accepted solution for LeetCode #1568: Minimum Number of Days to Disconnect Island
function minDays(grid: number[][]): number {
    const [m, n] = [grid.length, grid[0].length];

    const dfs = (i: number, j: number) => {
        if (i < 0 || m <= i || j < 0 || n <= j || [0, 2].includes(grid[i][j])) return;

        grid[i][j] = 2;
        const dir = [-1, 0, 1, 0, -1];
        for (let k = 0; k < 4; k++) {
            const [y, x] = [i + dir[k], j + dir[k + 1]];
            dfs(y, x);
        }
    };

    const count = () => {
        let c = 0;

        for (let i = 0; i < m; i++) {
            for (let j = 0; j < n; j++) {
                if (grid[i][j] === 1) {
                    dfs(i, j);
                    c++;
                }
            }
        }

        for (let i = 0; i < m; i++) {
            for (let j = 0; j < n; j++) {
                if (grid[i][j] === 2) {
                    grid[i][j] = 1;
                }
            }
        }

        return c;
    };

    if (count() !== 1) return 0;

    for (let i = 0; i < m; i++) {
        for (let j = 0; j < n; j++) {
            if (grid[i][j] === 1) {
                grid[i][j] = 0;

                if (count() !== 1) return 1;

                grid[i][j] = 1;
            }
        }
    }

    return 2;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(1)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

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.