# Swim in Rising Water LeetCode Solution

Here, We see Swim in Rising Water LeetCode Solution. This Leetcode problem is done in many programming languages like C++, Java, JavaScript, Python, etc. with different approaches.

# List of all LeetCode Solution

## 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.

## Swim in Rising Water LeetCode Solution C++

``````class Solution {
public:
int swimInWater(vector<vector<int>>& grid) {
int n = grid.size();
if (n == 1) {
return 0;
}
vector<vector<bool>> visited(n, vector<bool>(n));
visited[0][0] = true;
int result = max(grid[0][0], grid[n - 1][n - 1]);
priority_queue<vector<int>, vector<vector<int>>, greater<vector<int>>> pq;
pq.push({result, 0, 0});
while (!pq.empty()) {
vector<int> curr = pq.top();
pq.pop();
result = max(result, curr[0]);
for (int i = 0; i < 4; i++) {
int x = curr[1] + dirs[i][0];
int y = curr[2] + dirs[i][1];
if (x < 0 || x >= n || y < 0 || y >= n || visited[x][y]) {
continue;
}
if (x == n - 1 && y == n - 1) {
return result;
}
pq.push({grid[x][y], x, y});
visited[x][y] = true;
}
}
return -1;
}
private:
vector<vector<int>> dirs = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}};
};```Code language: PHP (php)```

## Swim in Rising Water LeetCode Solution Java

``````class Solution {
public int swimInWater(int[][] grid) {
PriorityQueue<Integer> pq = new PriorityQueue<>();
int N = grid.length - 1, ans = grid[0][0], i = 0, j = 0;
while (i < N || j < N) {
for (int[] m : moves) {
int ia = i + m[0], jb = j + m[1];
if (ia < 0 || ia > N || jb < 0 || jb > N || grid[ia][jb] > 2500) continue;
pq.add((grid[ia][jb] << 12) + (ia << 6) + jb);
grid[ia][jb] = 3000;
}
int next = pq.poll();
ans = Math.max(ans, next >> 12);
i = (next >> 6) & ((1 << 6) - 1);
j = next & ((1 << 6) - 1);
}
return ans;
}
private int[][] moves = {{1,0},{0,1},{-1,0},{0,-1}};
}```Code language: PHP (php)```

## Swim in Rising Water LeetCode Solution JavaScript

``````const moves = [[1,0],[0,1],[-1,0],[0,-1]]

var swimInWater = function(grid) {
let pq = new MinPriorityQueue(),
N = grid.length - 1, ans = grid[0][0], i = 0, j = 0
while (i < N || j < N) {
for (let [a,b] of moves) {
let ia = i + a, jb = j + b
if (ia < 0 || ia > N || jb < 0 || jb > N || grid[ia][jb] > 2500) continue
pq.enqueue((grid[ia][jb] << 12) + (ia << 6) + jb)
grid[ia][jb] = 3000
}
let next = pq.dequeue().element
ans = Math.max(ans, next >> 12)
i = (next >> 6) & ((1 << 6) - 1)
j = next & ((1 << 6) - 1)
}
return ans
};```Code language: JavaScript (javascript)```

## Swim in Rising Water LeetCode Solution Python

``````class Solution(object):
def swimInWater(self, grid):
N = len(grid)
visit = set()
minH = [[grid[0][0], 0, 0]]
directions = [[0, 1], [0, -1], [1, 0], [-1, 0]]