Given anm x nmatrix of non-negative integers representing the height of each unit cell in a continent, the "Pacific ocean" touches the left and top edges of the matrix and the "Atlantic ocean" touches the right and bottom edges.
Water can only flow in four directions (up, down, left, or right) from a cell to another one with height equal or lower.
Find the list of grid coordinates where water can flow to both the Pacific and Atlantic ocean.
Note:
- The order of returned grid coordinates does not matter.
- Both m and n are less than 150.
Example:
Given the following 5x5 matrix:
Pacific ~ ~ ~ ~ ~
~ 1 2 2 3 (5) *
~ 3 2 3 (4) (4) *
~ 2 4 (5) 3 1 *
~ (6) (7) 1 4 5 *
~ (5) 1 1 2 4 *
* * * * * Atlantic
Return:
[[0, 4], [1, 3], [1, 4], [2, 2], [3, 0], [3, 1], [4, 0]] (positions with parentheses in above matrix).
Solution 1: DFS，用2个boolean矩阵分别表示是否能到2个ocean
static int[] dx = {-1,0,0,1};
static int[] dy = {0,1,-1,0};
public List<int[]> pacificAtlantic(int[][] matrix) {
List<int[]> res = new ArrayList<>();
if (matrix == null || matrix.length == 0 || matrix[0].length == 0) return res;
boolean[][] pacific = new boolean[matrix.length][matrix[0].length];
boolean[][] atlantic = new boolean[matrix.length][matrix[0].length];
//其实不需要初始化
for (int i = 0; i < matrix.length; i++){
pacific[i][0] = true;
atlantic[i][matrix[0].length-1] = true;
}
//其实不需要初始化
for (int j = 0; j < matrix[0].length; j++){
pacific[0][j] = true;
atlantic[matrix.length-1][j] = true;
}
for (int i = 0; i < matrix.length; i++){
explore(pacific, matrix, i, 0);
explore(atlantic, matrix, i, matrix[0].length-1);
}
for (int j = 0; j < matrix[0].length; j++){
explore(pacific, matrix, 0, j);
explore(atlantic, matrix, matrix.length-1, j);
}
for (int i = 0; i < matrix.length; i++){
for (int j = 0; j < matrix[0].length; j++){
if (pacific[i][j] && atlantic[i][j] == true)
res.add(new int[]{i,j});
}
}
return res;
}
private void explore(boolean[][] grid, int[][] matrix, int i, int j){
grid[i][j] = true;
for (int d = 0; d < dx.length; d++){
if (i+dy[d] < grid.length && i+dy[d] >= 0 &&
j + dx[d] < grid[0].length && j + dx[d] >= 0 &&
grid[i+dy[d]][j+dx[d]] == false && matrix[i+dy[d]][j+dx[d]] >= matrix[i][j])
explore(grid, matrix, i+dy[d], j+dx[d]);
}
}
wrong:
int[] dx = {1, 0, -1, 0};
int[] dy = {0, 1, 0, -1};
public List<int[]> pacificAtlantic(int[][] matrix) {
List<int[]> ans = new ArrayList<>();
if(matrix == null || matrix.length == 0 || matrix[0] == null || matrix[0].length == 0){
return ans;
}
int m = matrix.length, n = matrix[0].length;
int[][] reach_P = new int[m][n], reach_A = new int[m][n];
for (int i = 0; i < m; i++) {
reach_P[i][0] = 1;
reach_A[i][n - 1] = 1;
}
for (int i = 0; i < n; i++) {
reach_P[0][i] = 1;
reach_A[m - 1][i] = 1;
}
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
int p = helper(matrix, i, j, reach_P), a = helper(matrix, i, j, reach_A);
if (p == 1 && a == 1) {
int[] temp = {i, j};
ans.add(temp);
}
}
}
return ans;
}
private int helper(int[][] matrix, int x, int y, int[][] dp) {
if (dp[x][y] != 0) {
return dp[x][y];
}
//-1:不能，1:能，0：还没看
dp[x][y] = -1;
for (int i = 0; i < 4; i++) {
int nx = x + dx[i], ny = y + dy[i];
if (nx < 0 || nx >= matrix.length || ny < 0 || ny >= matrix[0].length) {
continue;
}
if (matrix[x][y] >= matrix[nx][ny]) {
if (helper(matrix, nx, ny, dp) == 1) {
dp[x][y] = 1;
break;
}
}
}
return dp[x][y];
}
Solution 2: BFS. Two Queue and add all the Pacific border to one queue; Atlantic border to another queue. Keep a visited matrix for each queue. In the end, add the cell visited by two queue to the result. BFS: Water flood from ocean to the cell. Since water can only flow from high/equal cell to low cell, add the neighboor cell with height larger or equal to current cell to the queue and mark as visited.
int[][]dir = new int[][]{{1,0},{-1,0},{0,1},{0,-1}};
public List<int[]> pacificAtlantic(int[][] matrix) {
List<int[]> res = new LinkedList<>();
if(matrix == null || matrix.length == 0 || matrix[0].length == 0){
return res;
}
int n = matrix.length, m = matrix[0].length;
//One visited map for each ocean
boolean[][] pacific = new boolean[n][m];
boolean[][] atlantic = new boolean[n][m];
Queue<int[]> pQueue = new LinkedList<>();
Queue<int[]> aQueue = new LinkedList<>();
for(int i=0; i<n; i++){ //Vertical border
pQueue.offer(new int[]{i, 0});
aQueue.offer(new int[]{i, m-1});
pacific[i][0] = true;
atlantic[i][m-1] = true;
}
for(int i=0; i<m; i++){ //Horizontal border
pQueue.offer(new int[]{0, i});
aQueue.offer(new int[]{n-1, i});
pacific[0][i] = true;
atlantic[n-1][i] = true;
}
bfs(matrix, pQueue, pacific);
bfs(matrix, aQueue, atlantic);
for(int i=0; i<n; i++){
for(int j=0; j<m; j++){
if(pacific[i][j] && atlantic[i][j])
res.add(new int[]{i,j});
}
}
return res;
}
public void bfs(int[][]matrix, Queue<int[]> queue, boolean[][]visited){
int n = matrix.length, m = matrix[0].length;
while(!queue.isEmpty()){
int[] cur = queue.poll();
for(int[] d:dir){
int x = cur[0]+d[0];
int y = cur[1]+d[1];
if(x<0 || x>=n || y<0 || y>=m || visited[x][y] || matrix[x][y] < matrix[cur[0]][cur[1]]){
continue;
}
visited[x][y] = true;
queue.offer(new int[]{x, y});
}
}
}
one queue + bit manipulation:
00: cannot reach any ocean01: can reach pacific ocean10: can reach atlantic ocean11: can reach two oceans
Step 1: Update the status of border cells and put them into the queue
Step 2: Iterate the queue and explore the four directions. We only put a new cell into the queue if :
- row and col index are valid
- the height of the new cell is larger or equals to the height of the current cell
- the new cell can benifit from the current cell (check status)
public List<int[]> pacificAtlantic(int[][] matrix) {
List<int[]> res = new ArrayList<>();
int m = matrix.length;
if (m == 0) return res;
int n = matrix[0].length;
int[][] state = new int[m][n];
Queue<int[]> q = new LinkedList<>();
for (int i = 0; i < m; i++) {
state[i][0] |= 1;
if (i == m - 1 || n == 1) state[i][0] |= 2;
if (state[i][0] == 3) res.add(new int[]{i, 0});
q.add(new int[]{i, 0});
if (n > 1) {
state[i][n - 1] |= 2;
if (i == 0) state[i][n - 1] |= 1;
if (state[i][n - 1] == 3) res.add(new int[]{i, n - 1});
q.add(new int[]{i, n - 1});
}
}
for (int j = 1; j < n - 1; j++) {
state[0][j] |= 1;
if (m == 1) state[0][j] |= 2;
if (state[0][j] == 3) res.add(new int[]{0, j});
q.add(new int[]{0, j});
if (m > 1) {
state[m - 1][j] |= 2;
if (state[m - 1][j] == 3) res.add(new int[]{m - 1, j});
q.add(new int[]{m - 1, j});
}
}
int[][] dirs = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}};
while (!q.isEmpty()) {
int[] cell = q.poll();
for (int[] dir : dirs) {
int row = cell[0] + dir[0];
int col = cell[1] + dir[1];
if (row < 0 || col < 0 || row == m || col == n || matrix[row][col] < matrix[cell[0]][cell[1]] || ((state[cell[0]][cell[1]] | state[row][col]) == state[row][col])) continue;
state[row][col] |= state[cell[0]][cell[1]];
if (state[row][col] == 3) res.add(new int[]{row, col});
q.add(new int[]{row, col});
}
}
return res;
}