778. Swim in Rising Water

Problem Statement

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:

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:

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] < n2
Each value grid[i][j] is unique.

Intuition

Approach:

We Try to traverse all paths and maintain , 
Prioirity Queue like Dikjstra type

{max_distance(peak), {row,col}} info

Max peak will be the peak observed till now
then if we hot condition n-1, n-1 we stop as min heap will give us min at top

Mainatin visited to avoid multiple visits

Links

https://leetcode.com/problems/swim-in-rising-water/description/

Video Links

https://www.youtube.com/watch?v=amvrKlMLuGY&ab_channel=NeetCode

Approach 1:

C++

typedef pair<int, pair<int,int>> pii;

class Solution {
public:
    bool check_val(int x, int y, int n){
        return x>=0 and y>=0 and x<n and y<n;
    }

    int swimInWater(vector<vector<int>>& grid) {
        int n = grid.size();
        vector<vector<bool>> visited(n, vector<bool>(n,false));
        priority_queue< pii, vector<pii>, greater<pii> > pq;
        pq.push({grid[0][0], {0,0}});
        int dx[] = {-1, 1, 0, 0};
        int dy[] = {0, 0, -1, 1};

        while(!pq.empty()){
            int peak = pq.top().first;
            int row = pq.top().second.first, col = pq.top().second.second;
            pq.pop();

            if(row == n-1 and col == n-1)
                return peak;

            for(int i=0; i<4; i++){
                int nrow = row + dx[i];
                int ncol = col + dy[i];

                if(check_val(nrow, ncol, n) and !visited[nrow][ncol]){
                    int updated_peak = max(grid[nrow][ncol], peak);
                    pq.push({updated_peak, {nrow, ncol}});
                    visited[nrow][ncol] = true;
                }
            }
        }

        return -1;
    }
};

Approach 2:

C++

Approach 3:

C++

Approach 4:

C++