In this Leetcode Pacific Atlantic Water Flow problem solution, There is an m x n rectangular island that borders both the Pacific Ocean and Atlantic Ocean. The Pacific Ocean touches the island's left and top edges, and the Atlantic Ocean touches the island's right and bottom edges.

The island is partitioned into a grid of square cells. You are given an m x n integer matrix heights where heights[r][c] represents the height above sea level of the cell at coordinate (r, c).

The island receives a lot of rain, and the rain water can flow to neighboring cells directly north, south, east, and west if the neighboring cell's height is less than or equal to the current cell's height. Water can flow from any cell adjacent to an ocean into the ocean.

Return a 2D list of grid coordinates result where result[i] = [ri, ci] denotes that rain water can flow from cell (ri, ci) to both the Pacific and Atlantic oceans.

leetcode pacific atlantic water flow problem solution


Problem solution in Python.

class Solution:
    def pacificAtlantic(self, matrix: List[List[int]]) -> List[List[int]]:
        if not matrix: return []
        row = len(matrix)
        col = len(matrix[0])
        pacific = [["0" for _ in range(col)] for _ in  range(row)]
        atlantic = [["0" for _ in range(col)] for _ in  range(row)]
        def function(r,c,sea,val,M) :
            if 0<=r<row and 0<=c<col and M[r][c]  == "0" :
                if matrix[r][c] >= val : 
                    #print(r,c)
                    M[r][c] = sea
                    function(r-1,c,sea,matrix[r][c],M) 
                    function(r+1,c,sea,matrix[r][c],M) 
                    function(r,c-1,sea,matrix[r][c],M) 
                    function(r,c+1,sea,matrix[r][c],M) 
                
        for i in range(col) :
            function(0,i,"P",-1,pacific)
        for i in range(row) :
            function(i,0,"P",-1,pacific)
        for i in range(col-1,-1,-1) :
            function(row-1,i,"A",-1,atlantic)
        for i in range(row-1,-1,-1) :
            function(i,col-1,"A",-1,atlantic)
        answer = []
        
        for i in range(row) :
            for j in range(col) :
                if atlantic[i][j] == "A" and pacific[i][j] == "P" :
                    answer.append([i,j])
                    
        return answer



Problem solution in Java.

class Solution {
    List<List<Integer>> res = new ArrayList<>();
    public int[][] direction = {{1,0},{-1,0},{0,1},{0,-1}};
    public List<List<Integer>> pacificAtlantic(int[][] matrix) {
        //corner case
        if(matrix==null || matrix.length==0 || matrix[0].length==0) return res;
        int m=matrix.length;
        int n=matrix[0].length;
        
        
        boolean[][] pacific = new boolean[m][n];
        boolean[][] atlantic = new boolean[m][n];
        for(int i=0;i<m;i++){
            dfs(matrix,i,0,pacific,matrix[i][0]);
            dfs(matrix,i,n-1,atlantic,matrix[i][n-1]);
        }
        for(int j=0;j<n;j++){
            dfs(matrix,0,j,pacific,matrix[0][j]);
            dfs(matrix,m-1,j,atlantic,matrix[m-1][j]);
        }
        
        for(int i=0;i<m;i++){
            for(int j=0;j<n;j++){
                if(pacific[i][j]==true && atlantic[i][j]==true){
                    List<Integer> temp=new ArrayList<>();
                    temp.add(i);
                    temp.add(j);
                    res.add(temp);
                }
            }
        }
        return res;
    }
    
    private void dfs(int[][] matrix,int r,int c,boolean[][] isTrue,int val){
        int m=matrix.length;
        int n=matrix[0].length;
        
        if(r<0 || r>=m || c<0 || c>=n || matrix[r][c]<val) return;
        if(isTrue[r][c]) return;
        
        isTrue[r][c]=true;
        
        for(int i=0;i<direction.length;i++){
            dfs(matrix,r+direction[i][0],c+direction[i][1],isTrue,matrix[r][c]);
        }
        
    }
}


Problem solution in C++.

class Solution {
public:
    int m, n;
    void dfs(vector<vector<int>>& matrix, vector<vector<bool>>& vis, int i, int j)
    {
        vis[i][j] = true;
        int X[] = {1, 0, -1, 0};
        int Y[] = {0, 1, 0, -1};
        for(int k = 0; k < 4; k++)
        {
            int x = i + X[k];
            int y = j + Y[k];
            if(x >= 0 && x < m && y >= 0 && y < n && !vis[x][y] && matrix[x][y] >= matrix[i][j] )
            {
                vis[x][y] = true;
                dfs(matrix, vis, x, y);
            }
        }
    }
    vector<vector<int>> pacificAtlantic(vector<vector<int>>& matrix) {
        vector<vector<int>>ans;
        m = matrix.size();
        if(m == 0)return ans;
        n = matrix[0].size();
        vector<vector<bool>>p(m, vector<bool>(n, false)), a(m, vector<bool>(n, false));
        for(int i = 0; i < m; i++)
        {
            if(!p[i][0])
                dfs(matrix, p, i, 0);
            if(!a[i][n - 1])
                dfs(matrix, a, i, n - 1);
        }
        for(int i = 0; i < n; i++)
        {
            if(!p[0][i])
                dfs(matrix, p, 0, i);
            if(!a[m - 1][i])
                dfs(matrix, a, m - 1, i);
        }
        for(int i = 0; i < m; i++)
        {
            for(int j = 0; j < n; j++)
            {
                if(p[i][j] && a[i][j])
                    ans.push_back({i, j});
            }
        }
        return ans;
    }
};


Problem solution in C.

void dfs(int** matrix, int matrixRowSize, int *matrixColSizes,int** map,int row,int col,int value,int** ret,int* returnSize){
    if(map[row][col]==value){
        return;
    }
    int direction[4][2]={{1,0},{-1,0},{0,1},{0,-1}};
    if(map[row][col]==0){
        map[row][col]=value;
    }else{
        map[row][col]=value;
        ret[(*returnSize)++]=(int*)calloc(2,sizeof(int));
        ret[(*returnSize)-1][0]=row;
        ret[(*returnSize)-1][1]=col;
    }
    for(int i=0;i<4;i++){
        int tmp_row=row+direction[i][0];
        int tmp_col=col+direction[i][1];
        if(tmp_row>-1&&tmp_row<matrixRowSize&&tmp_col>-1
           &&tmp_col<matrixColSizes[tmp_row]&&matrix[row][col]<=matrix[tmp_row][tmp_col]){
            dfs(matrix,matrixRowSize,matrixColSizes,map,tmp_row,tmp_col,value,ret,returnSize);
        }
    }
}
int** pacificAtlantic(int** matrix, int matrixRowSize, int *matrixColSizes, int** columnSizes, int* returnSize) {
    int** map=(int**)malloc(matrixRowSize*sizeof(int*));
    *returnSize=0;
    if(matrixRowSize==0){
        return NULL;
    }
    columnSizes[0]=(int*)malloc(matrixRowSize*matrixColSizes[0]*sizeof(int));
    int** ret=(int**)malloc(matrixRowSize*matrixColSizes[0]*sizeof(int*));
    for(int i=0;i<matrixRowSize;i++){
        map[i]=(int*)calloc(matrixColSizes[i],sizeof(int));
    }
    for(int i=0;i<matrixColSizes[0];i++){
        dfs(matrix,matrixRowSize,matrixColSizes,map,0,i,1,ret,returnSize);
    }
    for(int i=0;i<matrixRowSize;i++){
        dfs(matrix,matrixRowSize,matrixColSizes,map,i,0,1,ret,returnSize);
    }
    for(int i=0;i<matrixColSizes[0];i++){
        dfs(matrix,matrixRowSize,matrixColSizes,map,matrixRowSize-1,i,-1,ret,returnSize);
    }
    for(int i=0;i<matrixRowSize;i++){
        dfs(matrix,matrixRowSize,matrixColSizes,map,i,matrixColSizes[0]-1,-1,ret,returnSize);
    }
    for(int i=0;i<*returnSize;i++){
        columnSizes[0][i]=2;
    }
    return ret;
}