Swim In Rising Water Problem


Description

LeetCode Problem 778.

You are given an n x n integer matrix grid where each value grid[i][j] represents the elevation at that point (i, j).

The 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 distances in zero time. Of course, you must stay within the boundaries of the grid during your swim.

Return the least time until you can reach the bottom right square (n - 1, n - 1) if you start at the top left square (0, 0).

Example 1:

1
2
3
4
5
6
7
Input: grid = [[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:

1
2
3
4
Input: grid = [[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: The final route is shown.
We need to wait until time 16 so that (0, 0) and (4, 4) are connected.

Constraints:

  • n == grid.length
  • n == grid[i].length
  • 1 <= n <= 50
  • 0 <= grid[i][j] < n^2
  • Each value grid[i][j] is unique.


Sample C++ Code using Priority Queue

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
class Solution {
public:
    struct T {
        int t, x, y;
        T(int a, int b, int c) : t (a), x (b), y (c) {}
        bool operator< (const T &d) const {return t > d.t;}
    };

    int swimInWater(vector<vector<int>>& grid) {
        int N = grid.size (), res = 0;
        priority_queue<T> pq;
        pq.push(T(grid[0][0], 0, 0));
        
        vector<vector<int>> seen(N, vector<int>(N, 0));
        seen[0][0] = 1;
        static int dir[4][2] = { {0, 1}, {0, -1}, {1, 0}, { -1, 0} };

        while (true) {
            auto p = pq.top ();
            pq.pop ();
            res = max(res, p.t);
            if (p.x == N - 1 && p.y == N - 1) 
                return res;
            for (auto& d : dir) {
                int i = p.x + d[0], j = p.y + d[1];
                if (i >= 0 && i < N && j >= 0 && j < N && !seen[i][j]) {
                    seen[i][j] = 1;
                    pq.push (T(grid[i][j], i, j));
                }
            }
        }
        
    }
};




Related Posts

Shortest Subarray With Sum At Least K Problem

LeetCode 862. Given an integer array nums and an integer...

Reachable Nodes In Subdivided Graph Problem

LeetCode 882. You are given an undirected graph (the “original...

Minimum Number Of Refueling Stops Problem

LeetCode 871. A car travels from a starting position to...

Minimum Cost To Hire K Workers Problem

LeetCode 857. There are n workers. You are given two...

K Closest Points To Origin Problem

LeetCode 973. Given an array of points where points[i] =...

Swim In Rising Water Problem

LeetCode 778. You are given an n x n integer...