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 time0, 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 time3. When the depth of water is3, 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:
2 <= N <= 50.- grid[i][j] is a permutation of [0, ..., N*N - 1].
class Solution:
def swimInWater(self, grid: List[List[int]]) -> int:
n = len(grid)
p = list(range(n * n))
def find(x):
if p[x] != x:
p[x] = find(p[x])
return p[x]
def index(i, j):
return i * n + j
def check(i, j):
return 0 <= i < n and 0 <= j < n
hi = [0] * (n * n)
for i in range(n):
for j in range(n):
hi[grid[i][j]] = index(i, j)
for h in range(n * n):
x, y = hi[h] // n, hi[h] % n
for a, b in [(0, -1), (0, 1), (1, 0), (-1, 0)]:
x1, y1 = x + a, y + b
if check(x1, y1) and grid[x1][y1] <= h:
p[find(index(x1, y1))] = find(hi[h])
if find(0) == find(n * n - 1):
return h
return -1class Solution {
private int[] p;
private int n;
private int[][] dirs = new int[][]{{0, -1}, {0, 1}, {1, 0}, {-1, 0}};
public int swimInWater(int[][] grid) {
n = grid.length;
p = new int[n * n];
for (int i = 0; i < p.length; ++i) {
p[i] = i;
}
int[] hi = new int[n * n];
for (int i = 0; i < n; ++i) {
for (int j = 0; j < n; ++j) {
hi[grid[i][j]] = index(i, j);
}
}
for (int h = 0; h < n * n; ++h) {
int x = hi[h] / n, y = hi[h] % n;
for (int[] dir : dirs) {
int x1 = x + dir[0], y1 = y + dir[1];
if (check(x1, y1) && grid[x1][y1] <= h) {
p[find(index(x1, y1))] = find(hi[h]);
}
if (find(0) == find(n * n - 1)) {
return h;
}
}
}
return -1;
}
private int find(int x) {
if (p[x] != x) {
p[x] = find(p[x]);
}
return p[x];
}
private int index(int i, int j) {
return i * n + j;
}
private boolean check(int i, int j) {
return i >= 0 && i < n && j >= 0 && j < n;
}
}function swimInWater(grid: number[][]): number {
const m = grid.length, n = grid[0].length;
let visited = Array.from({ length: m }, () => new Array(n).fill(false));
let ans = 0;
let stack = [[0, 0, grid[0][0]]];
const dir = [[0, 1], [0, -1], [1, 0], [-1, 0]];
while (stack.length) {
let [i, j] = stack.shift();
ans = Math.max(grid[i][j], ans);
if (i == m - 1 && j == n - 1) break;
for (let [dx, dy] of dir) {
let x = i + dx, y = j + dy;
if (x < m && x > -1 && y < n && y > -1 && !visited[x][y]) {
visited[x][y] = true;
stack.push([x, y, grid[x][y]]);
}
}
stack.sort((a, b) => a[2] - b[2]);
}
return ans;
};class Solution {
public:
vector<int> p;
int n;
int dirs[4][2] = {{0, -1}, {0, 1}, {1, 0}, {-1, 0}};
int swimInWater(vector<vector<int>> &grid) {
n = grid.size();
for (int i = 0; i < n * n; ++i)
p.push_back(i);
vector<int> hi(n * n, 0);
for (int i = 0; i < n; ++i)
for (int j = 0; j < n; ++j)
hi[grid[i][j]] = index(i, j);
for (int h = 0; h < n * n; ++h)
{
int x = hi[h] / n, y = hi[h] % n;
for (auto dir : dirs)
{
int x1 = x + dir[0], y1 = y + dir[1];
if (check(x1, y1) && grid[x1][y1] <= h)
p[find(index(x1, y1))] = find(hi[h]);
if (find(0) == find(n * n - 1))
return h;
}
}
return -1;
}
int find(int x) {
if (p[x] != x)
p[x] = find(p[x]);
return p[x];
}
int index(int i, int j) {
return i * n + j;
}
bool check(int i, int j) {
return i >= 0 && i < n && j >= 0 && j < n;
}
};var p []int
var n int
func swimInWater(grid [][]int) int {
n = len(grid)
p = make([]int, n*n)
hi := make([]int, n*n)
for i := 0; i < len(p); i++ {
p[i] = i
}
for i := 0; i < n; i++ {
for j := 0; j < n; j++ {
hi[grid[i][j]] = index(i, j)
}
}
dirs := [4][2]int{{0, -1}, {0, 1}, {1, 0}, {-1, 0}}
for h := 0; h < n*n; h++ {
x, y := hi[h]/n, hi[h]%n
for _, dir := range dirs {
x1, y1 := x+dir[0], y+dir[1]
if check(x1, y1) && grid[x1][y1] <= h {
p[find(index(x1, y1))] = find(hi[h])
}
if find(0) == find(n*n-1) {
return h
}
}
}
return -1
}
func find(x int) int {
if p[x] != x {
p[x] = find(p[x])
}
return p[x]
}
func index(i, j int) int {
return i*n + j
}
func check(i, j int) bool {
return i >= 0 && i < n && j >= 0 && j < n
}