Algorithm
Problem Name: 778. Swim in Rising Water
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.lengthn == grid[i].length1 <= n <= 500 <= grid[i][j] < n2- Each value
grid[i][j]is unique.
Code Examples
#1 Code Example with Javascript Programming
Code -
Javascript Programming
const swimInWater = function(grid) {
let n = grid.length;
let low = grid[0][0],
hi = n * n - 1;
while (low < hi) {
let mid = low + Math.floor((hi - low) / 2);
if (valid(grid, mid)) hi = mid;
else low = mid + 1;
}
return low;
};
function valid(grid, waterHeight) {
let n = grid.length;
const visited = Array.from(new Array(n), el => new Array(n).fill(0));
const dir = [-1, 0, 1, 0, -1];
return dfs(grid, visited, dir, waterHeight, 0, 0, n);
}
function dfs(grid, visited, dir, waterHeight, row, col, n) {
visited[row][col] = 1;
for (let i = 0; i < 4; ++i) {
let r = row + dir[i],
c = col + dir[i + 1];
if (
r >= 0 &&
r < n &&
c >= 0 &&
c < n &&
visited[r][c] == 0 &&
grid[r][c] <= waterHeight
) {
if (r == n - 1 && c == n - 1) return true;
if (dfs(grid, visited, dir, waterHeight, r, c, n)) return true;
}
}
return false;
}
Input
Output
#2 Code Example with Python Programming
Code -
Python Programming
class Solution:
def swimInWater(self, grid):
heap, res, n, visited = [(grid[0][0], 0, 0)], 0, len(grid), set()
while True:
d, i, j = heapq.heappop(heap)
if d > res: res = d
if i == j == n - 1: return res
for x, y in ((i - 1, j), (i + 1, j), (i, j - 1), (i, j + 1)):
if 0 <= x < n and 0 <= y < n and (x, y) not in visited:
visited.add((x, y))
heapq.heappush(heap, (grid[x][y], x, y))
Input
Output