# Introduction to BFS (breadth first search)

As always we want to start of by talking about why we would actually be learning this material. BFS is a algorithm used to traverse a set of paths to find the “shortest path” to something in a graph problem; solving problems such as minimum steps of solving the Rubik’s Cube. Do not get discouraged because of solving it as a graph; since most problems can be converted into a graph problem and sometimes in a few steps. Why are we getting the “shortest” path tough; we are searching on a broad approach and exploring the field within that level fully before actually traversing to the next level of depth of a problem. For example analyzing networks, mapping routes, and scheduling are graph problems.

BFS is a algorithm used to traverse a set of paths to find the “shortest path” to something in a graph problem; solving problems such as minimum steps of solving the Rubik’s Cube

## BFS vs DFS

The act of searching or traversing through a graph data structure is fairly simple: it just means that we’re probably visiting every single vertex (and by proxy, every single edge) in the graph. At it’s very core, the only difference between traversing a graph by breadth or by depth is the order in which we visit the vertices in a graph. In other words, the order in which the vertices of a graph are visited is actually how we can classify different graph traversal algorithms. In both tree and graph traversal, the DFS algorithm uses a stack data structure. By comparison, the breadth-first search algorithm traverses broadly into a structure, by visiting neighboring sibling nodes before visiting children nodes. In both tree and graph traversal, the BFS algorithm implements a queue data structure.

In both tree and graph traversal, the DFS algorithm uses a stack data structure. By comparison, the breadth-first search algorithm traverses broadly into a structure, by visiting neighboring sibling nodes before visiting children nodes. In both tree and graph traversal, the BFS algorithm implements a queue data structure.

## Algorithm

The backbone of a breadth-first search consists of these basic steps:

• Add a node/vertex from the graph to a queue of nodes to be “visited”.
• Visit the topmost node in the queue, and mark it as such.
• If that node has any neighbors, check to see if they have been “visited” or not.
• Add any neighboring nodes that still need to be “visited” to the queue.
• Remove the node we’ve visited from the queue.

## Real world example

#  Number of Islands
#
# https://leetcode.com/problems/number-of-islands/description/
#
# algorithms
# Medium
#
# Given a 2d grid map of '1's (land) and '0's (water), count the number of
# islands. An island is surrounded by water and is formed by connecting
# adjacent lands horizontally or vertically. You may assume all four edges of
# the grid are all surrounded by water.
#
# Example 1:
#
#
# Input:
# 11110
# 11010
# 11000
# 00000
#
# Output: 1
class Solution:
def numIslands(self, grid: List[List[str]]) -> int:
"""
:type grid: List[List[str]]
:rtype: int
"""
if not grid or not grid:
return 0

num_count = 0
for i in range(len(grid)):  # <-- x
for j in range(len(grid[i])):  # <-- y
if grid[i][j] == "1":
self.bfs(grid, i, j)
num_count += 1

return num_count

def bfs(self, grid, r, c):
queue = collections.deque()
queue.append((r, c))
grid[r][c] = "0"
while queue:
directions = [(0, 1), (0, -1), (-1, 0), (1, 0)]
r, c = queue.popleft()
for d in directions:
nr, nc = r + d, c + d
if self.is_valid(grid, nr, nc) and grid[nr][nc] == "1":
queue.append((nr, nc))
grid[nr][nc] = "0"

def is_valid(self, grid, r, c):
m, n = len(grid), len(grid)
if r < 0 or c < 0 or r >= m or c >= n:
return False
return True