778. Swim in Rising Water

On an N x N grid, each square grid[i][j] represents the elevation at that point (i,j).

Now rain starts to fall. At time t, the depth of the water everywhere is t. You can swim from a square to another 4-directionally adjacent square if and only if the elevation of both squares individually are at most t. You can swim infinite distance in zero time. Of course, you must stay within the boundaries of the grid during your swim.

You start at the top left square (0, 0). What is the least time until you can reach the bottom right square (N-1, N-1)?

Example 1:

Input: [[0,2],[1,3]]
Output: 3
Explanation:
At time 0, you are in grid location (0, 0).
You cannot go anywhere else because 4-directionally adjacent neighbors have a higher elevation than t = 0.

You cannot reach point (1, 1) until time 3.
When the depth of water is 3, we can swim anywhere inside the grid.

Example 2:

Input: [[0,1,2,3,4],[24,23,22,21,5],[12,13,14,15,16],[11,17,18,19,20],[10,9,8,7,6]]
Output: 16
Explanation:
 0  1  2  3  4
24 23 22 21  5
12 13 14 15 16
11 17 18 19 20
10  9  8  7  6

The final route is marked in bold.
We need to wait until time 16 so that (0, 0) and (4, 4) are connected.

Note:

1. 2 <= N <= 50.

2. grid[i][j] is a permutation of [0, ..., N*N - 1].

// BFS Dijkstra
int swimInWater(vector<vector<int>>& grid) { // time: (n^2 * log(n)); space: O(n^2)
    const vector<pair<int, int> > dirs({ {-1, 0}, {1, 0}, {0, -1}, {0, 1} });
    const int n = grid.size();
    priority_queue<tuple<int, int, int>, vector<tuple<int, int, int> >, greater<tuple<int, int, int> > > pq;
    pq.emplace(grid[0][0], 0, 0);
    int res = 0;
    grid[0][0] = -1; // mark as visited
    while (!pq.empty()) {
        auto [elevation, i, j] = pq.top(); 
        pq.pop();
        res = max(res, elevation);
        if (i == n - 1 && j == n - 1) return res;
        for (const pair<int, int>& dir : dirs) {
            int new_i = i + dir.first, new_j = j + dir.second;
            if (new_i < 0 || new_i >= n || new_j < 0 || new_j >= n || grid[new_i][new_j] == -1) continue;
            pq.emplace(grid[new_i][new_j], new_i, new_j);
            grid[new_i][new_j] = -1; // mark as visited
        }
    }
    return -1;
}

// Binary Search + DFS
bool dfs(const vector<vector<int>>& grid, vector<vector<bool> >& visited, const vector<int>& dir, int waterLevel, int row, int col) {
    const int n = grid.size();
    visited[row][col] = true;
    for (int i = 0; i < 4; ++i) {
        int r = row + dir[i], c = col + dir[i+1];
        if (r < 0 || r >= n || c < 0 || c >= n || visited[r][c] || grid[r][c] > waterLevel) continue;
        if (r == n-1 && c == n-1) return true;
        if (dfs(grid, visited, dir, waterLevel, r, c)) return true;
    }
    return false;
}
bool valid(const vector<vector<int> >& grid, int waterLevel) {
    const int n = grid.size();
    vector<vector<bool> > visited(n, vector<bool>(n, false));
    vector<int> dir({-1, 0, 1, 0, -1});
    return dfs(grid, visited, dir, waterLevel, 0, 0);
}
int swimInWater(vector<vector<int>>& grid) { // time: O(n^2 * log(n)); space: O(n^2)
    const int n = grid.size();
    int low = grid[0][0], high = n * n - 1;
    while (low < high) {
        int mid = low + (high - low) / 2;
        if (valid(grid, mid) ) high = mid;
        else low = mid+1;
    }
    return low;
}

// Union-Find
int find(vector<int>& roots, int i) {
    return roots[i] == i ? i : roots[i] = find(roots, roots[i]);
}
void uni(vector<int>& roots, int x, int y) {
    x = find(roots, x), y = find(roots, y);
    if (x != y) roots[y] = x;
}
int swimInWater(vector<vector<int>>& grid) { // time: (n^2 * log(n)); space: O(n^2)
    const vector<pair<int, int> > dirs({ {-1, 0}, {1, 0}, {0, -1}, {0, 1} });
    const int n = grid.size();
    vector<bool> visited(n * n, false);
    vector<int> roots(n * n);
    vector<pair<int, int> > ids(n * n);
    // Initialization of roots vector
    for (int i = 0; i < n; ++i) {
        for (int j = 0; j < n; ++j) {
            int k = i * n + j;
            roots[k] = k;
            ids[k] = {i, j};
        }
    }
    // sort in ascending order
    auto cmp = [&grid](const pair<int, int>& id1, const pair<int, int>& id2) {
        return grid[id1.first][id1.second] < grid[id2.first][id2.second];  
    };
    sort(ids.begin(), ids.end(), cmp);
    for (const pair<int, int>& id : ids) {
        int i = id.first, j = id.second, k = i * n + j;
        visited[k] = true;
        for (const pair<int, int>& dir : dirs) {
            int new_i = i + dir.first, new_j = j + dir.second, new_k = new_i * n + new_j;
            if (new_i < 0 || new_i >= n || new_j < 0 || new_j >= n || !visited[new_k]) continue;
            uni(roots, k, new_k);
            if (find(roots, 0) == find(roots, n * n - 1)) return grid[i][j];
        }
    }
    return -1;
}