#1568Medium~35 min

Minimum Number of Days to Disconnect Island

Time O(m * n) · Space O(m * n) · Official statement on LeetCode

graph dfs backtracking tarjans algorithm

Solutions

// Time:  O(m * n)
// Space: O(m * n)

// template: https://github.com/kamyu104/GoogleCodeJam-Farewell-Rounds/blob/main/Round%20B/railroad_maintenance.py3
// Reference: https://en.wikipedia.org/wiki/Biconnected_component#Algorithms
vector<int> iter_get_articulation_points(const auto& graph, int v) {  // modified
    int index_counter = 0;
    vector<int> index(size(graph), -1);
    vector<int> lowlinks(size(graph));
    vector<int> children_count(size(graph));
    vector<bool> is_cut(size(graph));
    vector<int> cutpoints;
    const auto& iter_dfs = [&](int v, int p) {
        vector<vector<int>> stk = {{1, v, p}};
        while (!empty(stk)) {
            const auto args = stk.back(); stk.pop_back();
            if (args[0] == 1) {
                const int v = args[1], p = args[2];
                index[v] = index_counter;
                lowlinks[v] = index_counter++;
                stk.push_back({4, v, p});
                for (const auto& w : graph[v]) {
                    if (w == p) {
                        continue;
                    }
                    stk.push_back({2, w, v});
                }
            } else if (args[0] == 2) {
                const int w = args[1], v = args[2];
                if (index[w] == -1) {
                    ++children_count[v];
                    stk.push_back({3, w, v});
                    stk.push_back({1, w, v});
                } else {
                    lowlinks[v] = min(lowlinks[v], index[w]);
                }
            } else if (args[0] == 3) {
                const int w = args[1], v = args[2];
                if (lowlinks[w] >= index[v]) {
                    is_cut[v] = true;
                }
                lowlinks[v] = min(lowlinks[v], lowlinks[w]);
            } else if (args[0] == 4) {
                const int v = args[1], p = args[2];
                if ((p != -1 && is_cut[v]) || (p == -1 && children_count[v] >= 2)) {
                    cutpoints.emplace_back(v);
                }
            }
        }
    };
    
    iter_dfs(v, -1);  // modified
    return cutpoints;
}

// flood fill, tarjan's algorithm, articulation points
class Solution {
public:
    int minDays(vector<vector<int>>& grid) {
        static const vector<pair<int, int>> DIRECTIONS{{0, 1}, {1, 0}, {0, -1}, {-1, 0}};

        const int R = size(grid), C = size(grid[0]);
        if (count_islands(grid) != 1) {
            return 0;
        }
        vector<vector<int>> adj(R * C);
        int cnt = 0, idx = -1;
        for (int i = 0; i < R; ++i) {
            for (int j = 0; j < C; ++j) {
                if (grid[i][j] == 0) {
                    continue;
                }
                ++cnt;
                if (idx == -1) {
                    idx = i * C + j;
                }
                for (const auto& [di, dj] : DIRECTIONS) {
                    const auto& [ni, nj] = pair(i + di, j + dj);
                    if (0 <= ni && ni < R && 0 <= nj && nj < C && grid[ni][nj] == grid[i][j]) {
                        adj[i * C + j].emplace_back(ni * C + nj);
                    }
                }
                
            }
        }
        return cnt == 1 || !empty(iter_get_articulation_points(adj, idx)) ? 1 : 2;
    }

private:
    int count_islands(const vector<vector<int>>& grid) {
        vector<vector<int>> lookup(grid.size(), vector<int>(grid[0].size()));
        int islands_cnt = 0;
        for (int i = 0; i < grid.size(); ++i) {
            for (int j = 0; j < grid[0].size(); ++j) {
                if (grid[i][j] == 0 || lookup[i][j]) {
                    continue;
                }
                ++islands_cnt;
                floodfill(grid, i, j, &lookup);
            }
        }
        return islands_cnt;
    }

    void floodfill(const vector<vector<int>>& grid, int i, int j, vector<vector<int>> *lookup) {
        static const vector<pair<int, int>> DIRECTIONS{{0, 1}, {1, 0}, {0, -1}, {-1, 0}};

        vector<pair<int, int>> stk = {{i, j}};
        (*lookup)[i][j] = 1;
        while (!stk.empty()) {
            auto [i, j] = stk.back(); stk.pop_back();
            for (const auto& [di, dj] : DIRECTIONS) {
                const auto& [ni, nj] = pair(i + di, j + dj);
                if (!(0 <= ni && ni < grid.size() &&
                      0 <= nj && nj < grid[0].size() &&
                      grid[ni][nj] &&
                      !(*lookup)[ni][nj])) {
                    continue;
                }
                (*lookup)[ni][nj] = 1;
                stk.emplace_back(ni, nj);
            }
        }
    }
};

// Time:  O((m * n) * log(m * n))
// Space: O((m * n) * log(m * n))
// flood fill, persistent union find
class Solution2 {
public:
    int minDays(vector<vector<int>>& grid) {
        static const vector<pair<int, int>> DIRECTIONS{{0, 1}, {1, 0}, {0, -1}, {-1, 0}};

        const int R = size(grid), C = size(grid[0]);
        if (count_islands(grid) != 1) {
            return 0;
        }
    
        PersistentUnionFind uf(R * C);
        vector<int> lookup(R * C);
        const auto& merge = [&](int i) {
            const int r = i / C, c = i % C;
            for (const auto& [dr, dc] : DIRECTIONS) {
                const int nr = r + dr, nc = c + dc;
                const int ni = nr * C + nc;
                if (0 <= nr && nr < R && 0 <= nc && nc < C && grid[nr][nc] == grid[r][c] && lookup[ni]) {
                    uf.union_set(i, ni);
                }
            }
        };

        const auto& check = [&](int i) {
            const int r = i / C, c = i % C;
            if (grid[r][c] == 0) {
                return false;
            }
            unordered_set<int> lookup;
            for (const auto& [dr, dc] : DIRECTIONS) {
                const int nr = r + dr, nc = c + dc;
                if (0 <= nr && nr < R && 0 <= nc && nc < C && grid[nr][nc] == grid[r][c]) {
                    lookup.emplace(uf.find_set(nr * C + nc));
                }
            }
            return size(lookup) != 1;
        };

        const function<bool (int, int)> dfs = [&](int left, int right) {
            if (left == right) {
                return check(left);
            }
            const int mid = left + (right - left) / 2;
            int l1 = left, r1 = mid, l2 = mid + 1, r2 = right;
            for (int _ = 0; _ < 2; ++_) {
                uf.snapshot();
                for (int i = l1; i <= r1; ++i) {
                    lookup[i] = true;
                    merge(i);
                }
                if (dfs(l2, r2)) {
                    return true;
                }
                for (int i = l1; i <= r1; ++i) {
                    lookup[i] = false;
                }
                uf.rollback();
                swap(l1, l2), swap(r1, r2);
            };
            return false;
        };

        return dfs(0, R * C - 1) ? 1 : 2;
    }

private:
    int count_islands(const vector<vector<int>>& grid) {
        vector<vector<int>> lookup(grid.size(), vector<int>(grid[0].size()));
        int islands_cnt = 0;
        for (int i = 0; i < grid.size(); ++i) {
            for (int j = 0; j < grid[0].size(); ++j) {
                if (grid[i][j] == 0 || lookup[i][j]) {
                    continue;
                }
                ++islands_cnt;
                floodfill(grid, i, j, &lookup);
            }
        }
        return islands_cnt;
    }

    void floodfill(const vector<vector<int>>& grid, int i, int j, vector<vector<int>> *lookup) {
        static const vector<pair<int, int>> DIRECTIONS{{0, 1}, {1, 0}, {0, -1}, {-1, 0}};

        vector<pair<int, int>> stk = {{i, j}};
        (*lookup)[i][j] = 1;
        while (!stk.empty()) {
            auto [i, j] = stk.back(); stk.pop_back();
            for (const auto& [di, dj] : DIRECTIONS) {
                const auto& [ni, nj] = pair(i + di, j + dj);
                if (!(0 <= ni && ni < grid.size() &&
                      0 <= nj && nj < grid[0].size() &&
                      grid[ni][nj] &&
                      !(*lookup)[ni][nj])) {
                    continue;
                }
                (*lookup)[ni][nj] = 1;
                stk.emplace_back(ni, nj);
            }
        }
    }

    class PersistentUnionFind {
    public:
        PersistentUnionFind(int n)
         : set_(n)
         , size_(n) {
            iota(begin(set_), end(set_), 0);
        }

        int find_set(int x) {
            vector<int> stk;
            while (set_[x] != x) {  // path compression
                stk.emplace_back(x);
                x = set_[x];
            }
            while (!empty(stk)) {
                const int y = stk.back(); stk.pop_back();
                undos_.emplace_back(~y, set_[y]);  // added
                set_[y] = x;
            }
            return x;
        }

        bool union_set(int x, int y) {
            x = find_set(x), y = find_set(y);
            if (x == y) {
                return false;
            }
            if (size_[x] > size_[y]) {
                swap(x, y);
            }
            undos_.emplace_back(x, y);  // added
            set_[x] = y;  // Union by size.
            size_[y] += size_[x];
            return true;
        }

        void snapshot() {  // added
            snapshots_.emplace_back(size(undos_));
        }
    
        void rollback() {  // added
            const int cnt = snapshots_.back(); snapshots_.pop_back();
            while (size(undos_) != cnt) {
                const auto [x, y] = undos_.back(); undos_.pop_back();
                if (x >= 0) {
                    size_[y] -= size_[x];
                    set_[x] = x;
                } else {
                    set_[~x] = y;
                }
            }
        }

    private:
        vector<int> set_;
        vector<int> size_;
        vector<pair<int, int>> undos_;  // added
        vector<int> snapshots_;  // added
    };
};

// Time:  O(m^2 * n^2)
// Space: O(m * n)
// flood fill
class Solution3 {
public:
    int minDays(vector<vector<int>>& grid) {
        if (count_islands(grid) != 1) {
            return 0;
        }
        for (int i = 0; i < grid.size(); ++i) {
            for (int j = 0; j < grid[0].size(); ++j) {
                if (grid[i][j] == 0) {
                    continue;
                }
                grid[i][j] = 0;
                int islands_cnt = count_islands(grid);
                grid[i][j] = 1;
                if (islands_cnt != 1) {
                    return 1;
                }
            }
        }
        return 2;
    }

private:
    int count_islands(const vector<vector<int>>& grid) {
        vector<vector<int>> lookup(grid.size(), vector<int>(grid[0].size()));
        int islands_cnt = 0;
        for (int i = 0; i < grid.size(); ++i) {
            for (int j = 0; j < grid[0].size(); ++j) {
                if (grid[i][j] == 0 || lookup[i][j]) {
                    continue;
                }
                ++islands_cnt;
                floodfill(grid, i, j, &lookup);
            }
        }
        return islands_cnt;
    }

    void floodfill(const vector<vector<int>>& grid, int i, int j, vector<vector<int>> *lookup) {
        static const vector<pair<int, int>> DIRECTIONS{{0, 1}, {1, 0}, {0, -1}, {-1, 0}};

        vector<pair<int, int>> stk = {{i, j}};
        (*lookup)[i][j] = 1;
        while (!stk.empty()) {
            auto [i, j] = stk.back(); stk.pop_back();
            for (const auto& [di, dj] : DIRECTIONS) {
                const auto& [ni, nj] = pair(i + di, j + dj);
                if (!(0 <= ni && ni < grid.size() &&
                      0 <= nj && nj < grid[0].size() &&
                      grid[ni][nj] &&
                      !(*lookup)[ni][nj])) {
                    continue;
                }
                (*lookup)[ni][nj] = 1;
                stk.emplace_back(ni, nj);
            }
        }
    }
};

Beginner Explanation

What is Minimum Number of Days to Disconnect Island?

Minimum Number of Days to Disconnect Island (LeetCode #1568) is a Medium problem that primarily trains graph.

How to think about it

Restate the goal in your own words before coding.
Work a tiny example by hand so the invariant becomes obvious.
Identify the pattern — this problem aligns with dfs backtracking and tarjans algorithm.
Only then translate the idea into code.

Why this problem matters

It sits in the sweet spot of interview difficulty: multiple valid approaches, clear trade-offs. Official solution notes mention: DFS, Persistent Union Find, Tarjan's Algorithm, Articulation Points.

AlgoForge explanations are original teaching notes. Always open the official problem statement on LeetCode for constraints and examples.

Interview Walkthrough

Interview approach for Minimum Number of Days to Disconnect Island

Opening (30–60 seconds)

Clarify inputs/outputs and edge cases (empty input, single element, duplicates, overflow).
State a brute force so the interviewer knows you can solve it naively.
Propose the optimal direction tied to dfs backtracking and tarjans algorithm.

Core solution narrative

Define the state you track (pointers, DP cell, set membership, stack top, etc.).
Explain the transition when you process the next element.
Call out time (O(m * n)) and space (O(m * n)) before coding.
Code cleanly; narrate variable names.

What interviewers listen for

Correctness on edge cases
Complexity honesty
Ability to discuss trade-offs (e.g., hash map space vs. sort + two pointers)

Follow-up questions they may ask

Can you solve it with less memory?
What if the input stream is infinite / doesn't fit in RAM?
How would tests look for adversarial inputs?

Optimized Approach

Optimized solution notes

The reference solutions on AlgoForge target O(m * n) time and O(m * n) space.

Pattern focus: dfs backtracking and tarjans algorithm

Use the pattern as a checklist:

dfs backtracking — confirm the invariant holds after each step
tarjans algorithm — confirm the invariant holds after each step

Multiple methods appear in the source solutions — compare them and explain when each is preferable.

Implementation tips

Prefer readable names over micro-optimizations in interviews.
Extract helpers only when they clarify (e.g., expand-around-center, DFS visit).
After AC-level logic, re-scan for off-by-one and null checks.

Complexity Analysis

Complexity

Measure	Bound
Time	*O(m n)**
Space	*O(m n)**

How to justify this in an interview

Time: count loops, map/set operations, and recursive branching; state average vs worst case if relevant.
Space: include hash maps, recursion stack, and output allocation when the problem asks for it.

If your implementation differs from the reference, re-derive big-O from your code — never memorize a complexity you cannot defend.

Common Mistakes

Common mistakes on Minimum Number of Days to Disconnect Island

Skipping edge cases — empty collections, single-element inputs, max constraints.
Wrong invariant for dfs backtracking and tarjans algorithm — updating state too early or too late.
Mutating input unexpectedly when the problem forbids it.
Off-by-one in windows, ranges, or binary search bounds.
Ignoring overflow / precision for integer arithmetic problems.
Overengineering — jumping to an advanced structure when a simpler approach works.

Alternative Approaches

Alternatives

The source file includes more than one method. Compare:

Primary optimized path — best complexity for typical interviews.
Secondary approach — often brute force, sorting-based, or space-optimized variant.

Practice articulating when you would pick each (constraints, readability, follow-ups).

Edge Cases

Edge cases checklist

Minimum input size
Maximum input size / time limits
Duplicates and already-sorted input
Negative numbers / zeros (if applicable)
Disconnected structures (graphs/trees)
Single path vs branching recursion depth

Pattern Recognition

Spotting this pattern

Signal phrases that point to dfs backtracking and tarjans algorithm:

Sorted input or ability to sort without changing the answer class
Need for contiguous subarray / substring → consider sliding window
Need for O(1) membership → hash set/map
Optimal substructure + overlapping subproblems → DP
Connectivity / components → graph DFS/BFS or Union-Find

Primary topics: graph.

Follow-up Interview Questions

Follow-ups

How does the solution change if the input is a stream?
Can you solve it in-place?
What if duplicates must be handled differently?
How would you parallelize the approach?
Design tests that would break a buggy implementation.

Practice Recommendations

What to practice next

Re-solve Minimum Number of Days to Disconnect Island in a second language (cpp, python).
Drill 3–5 more problems tagged graph.
Teach the solution out loud in under 5 minutes.
Add this problem to your revision calendar in 3 days and 14 days.

Visualization

Conceptual diagram for Minimum Number of Days to Disconnect Island: show input structure (graph), highlight the moving parts of the dfs backtracking and tarjans algorithm approach, and annotate each step with the maintained invariant and complexity.

Study checklist

Read the official problem statement on LeetCode
Solve on paper / whiteboard first
Implement the dfs backtracking and tarjans algorithm approach
Verify edge cases from the checklist
State time and space complexity aloud
Compare with the AlgoForge reference solution
Schedule a revision session

Revision notes

Minimum Number of Days to Disconnect Island (#1568) — Medium. Pattern: dfs backtracking and tarjans algorithm. Complexity: O(m * n) time / O(m * n) space. Re-derive the invariant before coding.

FAQs

What is the time complexity of Minimum Number of Days to Disconnect Island?+

The reference solutions aim for O(m * n) time and O(m * n) space. Always re-derive complexity from the code you write in the interview.

What pattern does Minimum Number of Days to Disconnect Island use?+

It primarily maps to dfs backtracking and tarjans algorithm, within the broader topic of graph.

Is Minimum Number of Days to Disconnect Island good for interviews?+

Yes — as a Medium problem it is a solid practice target. Pair it with related problems in the same pattern family for spaced repetition.

Where can I read the official statement?+

Open the official LeetCode page for constraints and examples: https://leetcode.com/problems/minimum-number-of-days-to-disconnect-island/

#	Title	Difficulty	Topics
2850	Minimum Moves to Spread Stones Over Grid	Medium	graph
1601	Maximum Number of Achievable Transfer Requests	Hard	graph
1361	Validate Binary Tree Nodes	Medium	graph
1778	Shortest Path in a Hidden Grid	Medium	graph
1810	Minimum Path Cost in a Hidden Grid	Medium	graph
2307	Check for Contradictions in Equations	Hard	graph
3910	Count Connected Subgraphs with Even Node Sum	Hard	graph
2359	Find Closest Node to Given Two Nodes	Medium	graph