You are given an undirected graph represented as an adjacency list, a starting node start, and a target node end. Write a function to determine if there is a path between start and end in the graph. Return true if a path exists; otherwise, return false. The graph is represented as an adjacency list, where each key is a node and its value is a list of neighboring nodes. Select the optimal solution in terms of time complexity. If time complexities are same for multiple solutions, select the one with optimal space complexity. function path_exists(graph, start, end): stack = [start] visited = set() while stack: node = stack.pop() if node == end: return true if node not in visited: visited.add(node) for neighbor in graph[node]: stack.append(neighbor) return false function path_exists(graph, start, end, visited=set()): if start == end: return true visited.add(start) for neighbor in graph[start]: if neighbor not in visited: if path_exists(graph, neighbor, end, visited): return true return false function path_exists(graph, start, end): queue = [start] visited = set([start]) while queue: node = queue.pop(0) if node == end: return true for neighbor in graph[node]: if neighbor not in visited: visited.add(neighbor) queue.append(neighbor) return false
You are given an undirected graph represented as an adjacency list, a starting node start, and a target node end. Write a function to determine if there is a path between start and end in the graph. Return true if a path exists; otherwise, return false. The graph is represented as an adjacency list, where each key is a node and its value is a list of neighboring nodes. Select the optimal solution in terms of time complexity. If time complexities are same for multiple solutions, select the one with optimal space complexity. function path_exists(graph, start, end): stack = [start] visited = set() while stack: node = stack.pop() if node == end: return true if node not in visited: visited.add(node) for neighbor in graph[node]: stack.append(neighbor) return false function path_exists(graph, start, end, visited=set()): if start == end: return true visited.add(start) for neighbor in graph[start]: if neighbor not in visited: if path_exists(graph, neighbor, end, visited): return true return false function path_exists(graph, start, end): queue = [start] visited = set([start]) while queue: node = queue.pop(0) if node == end: return true for neighbor in graph[node]: if neighbor not in visited: visited.add(neighbor) queue.append(neighbor) return false
Related questions
Question
You are given an undirected graph represented as an adjacency list, a starting node start, and a target node end. Write a function to determine if there is a path between start and end in the graph. Return true if a path exists; otherwise, return false. The graph is represented as an adjacency list, where each key is a node and its value is a list of neighboring nodes. Select the optimal solution in terms of time complexity. If time complexities are same for multiple solutions, select the one with optimal space complexity.
function path_exists(graph, start, end):
stack = [start]
visited = set()
while stack:
node = stack.pop()
if node == end:
return true
if node not in visited:
visited.add(node)
for neighbor in graph[node]:
stack.append(neighbor)
return false
stack = [start]
visited = set()
while stack:
node = stack.pop()
if node == end:
return true
if node not in visited:
visited.add(node)
for neighbor in graph[node]:
stack.append(neighbor)
return false
function path_exists(graph, start, end, visited=set()):
if start == end:
return true
visited.add(start)
for neighbor in graph[start]:
if neighbor not in visited:
if path_exists(graph, neighbor, end, visited):
return true
return false
if start == end:
return true
visited.add(start)
for neighbor in graph[start]:
if neighbor not in visited:
if path_exists(graph, neighbor, end, visited):
return true
return false
function path_exists(graph, start, end):
queue = [start]
visited = set([start])
while queue:
node = queue.pop(0)
if node == end:
return true
for neighbor in graph[node]:
if neighbor not in visited:
visited.add(neighbor)
queue.append(neighbor)
return false
queue = [start]
visited = set([start])
while queue:
node = queue.pop(0)
if node == end:
return true
for neighbor in graph[node]:
if neighbor not in visited:
visited.add(neighbor)
queue.append(neighbor)
return false
AI-Generated Solution
AI-generated content may present inaccurate or offensive content that does not represent bartleby’s views.
Unlock instant AI solutions
Tap the button
to generate a solution