Question: I am working on Leetcode 417 Pacific Atlantic Water Flow:  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.  My solution is below. I am having Time Limit Exceeded error for a very large test case like [[1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19],[72,73,74,75,76,77,78,79,80,81,82,83,84,85,86,87,88,89,20],[71,136,137,138,139,140,141,142,143,144,145,146,147,148,149,150,151,90,21],[70,135,192,193,194,195,196,197,198,199,200,201,202,203,204,205,152,91,22], ...]  I cannot figure out why my BFS did not work within a reasonable time. What am I missing? class Solution: def pacificAtlantic(self, heights): rows, cols = len(heights), len(heights[0]) # define directions directions = [[0, 1], [0, -1], [1, 0], [-1, 0]] def bfs(node, visited): Q = [node] while Q: x, y = Q.pop(0) visited[x][y] = True for dx, dy in directions: next_x, next_y = x + dx, y + dy if next_x < 0 or next_x >= rows: continue if next_y < 0 or next_y >= cols: continue if visited[next_x][next_y]: continue if heights[x][y] > heights[next_x][next_y]: continue Q.append((next_x, next_y)) # pacific pacific_start = [[0, i] for i in range(cols)] + [[i, 0] for i in range(1, rows)] pacific_visited = [[False for _ in range(cols)] for _ in range(rows)] for row, col in pacific_start: if not pacific_visited[row][col]: bfs((row, col), pacific_visited, 0) # atlantic atlantic_start = [[rows - 1, i] for i in range(cols)] + [[i, cols - 1] for i in range(0, rows - 1)] atlantic_visited = [[False for _ in range(cols)] for _ in range(rows)] for row, col in atlantic_start: if not atlantic_visited[row][col]: bfs((row, col), atlantic_visited, 0) # find the common land ans = [] for i in range(rows): for j in range(cols): if pacific_visited[i][j] and atlantic_visited[i][j]: ans.append((i, j)) return ans 

Answer: There are two issues in your bfs function that negatively affect performance:  pop(0) is not efficient on a list. Instead use a deque You mark a node as visited, after having taken it from the queue, but that means you can have several copies of the same cell in the queue, increasing the number of iterations the BFS loop will make. Instead mark a node as visited at the time you push it on the queue.  Here is the code of your bfs function, with minimal changes needed to resolve those two issues: def bfs(node, visited, test): visited[node[0]][node[1]] = True # mark node when it enters the queue Q = deque([node]) # Use a deque, not a list while Q: x, y = Q.popleft() # now it's an efficient operation for dx, dy in directions: next_x, next_y = x + dx, y + dy if next_x < 0 or next_x >= rows: continue if next_y < 0 or next_y >= cols: continue if visited[next_x][next_y]: continue if heights[x][y] > heights[next_x][next_y]: continue visited[next_x][next_y] = True # mark node when it enters the queue Q.append((next_x, next_y))