Leetcode Pacific Atlantic Water Flow problem solution YASH PAL, 31 July 2024 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. Topics we are covering Toggle Problem solution in Python.Problem solution in Java.Problem solution in C++.Problem solution in C. 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; } coding problems solutions