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Code Problem / Data Structures and Algorithms / Recursion (32)
Code Problem / Data Structures and Algorithms / Binary Search (30)
Code Problem / Data Structures and Algorithms / Linked Lists (26)
Code Problem / Data Structures and Algorithms / Backtracking (26)
Code Problem / Data Structures and Algorithms / Stacks and Queues (25)
Code Problem / Data Structures and Algorithms / Hashing (24)
Databases / SQL / Backup and Recovery (19)
Code Problem / Data Structures and Algorithms / Arrays and Strings (19)
Databases / SQL / Database Normalization (18)
Code Problem / Data Structures and Algorithms / Time Complexity (17)
Databases / SQL / Locking (16)
Databases / SQL / Replication (15)
Databases / SQL / SQL Queries (13)
Code Problem / Data Structures and Algorithms / Graphs (13)
Databases / SQL / Database Security (13)
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Recent solutions:
Check these solutions from our community:
This technical problem deals with finding a route between two nodes in a directed graph. An example input is given, as well as the desired output.
About this solution: The candidate's solution is correct and demonstrates a level of completeness. The candidate has correctly implemented a breadth-first search algorithm to find whether there is a route between two nodes in a directed graph.
The problem asks for a way to determine whether there is a route between two nodes in a directed graph. One example is given, where the nodes are connected in a linear fashion. The solution should work for more general cases as well.
About this solution: The candidate's solution is complete and solves the problem. The candidate has used a breadth-first search approach, which is a good choice for this problem.
This problem asks for the shortest route between two nodes in a graph. The input is a list of nodes and their connections. The output should be the shortest route between the two given nodes.
About this solution: The candidate's solution correctly implements a breadth-first search to find a route between two nodes in a graph. The solution is complete and correctly handles edge cases such as null input values.
This problem asks us to find the shortest path between any two words in a list. For example, given the list ["cat", "bat", "rat"], the shortest path would be " cat" -> "bat" -> "rat".
About this solution: The candidate's solution does not demonstrate a level of completeness and does not solve the problem. The general approach is not clear.
This technical problem involves determining whether a route exists between two nodes in a directed graph. An example input and output are given.
About this solution: The candidate's solution is correct and uses a breadth-first search to find the shortest path between two nodes. The time complexity is O(n) and the space complexity is O(n).
This problem is about finding the shortest path through a maze. The input is a 2D array representing the maze, and the output is the shortest path from the top-left corner to the bottom-right corner.
About this solution: The candidate's solution correctly solves the problem and demonstrates a level of completeness. The candidate uses a breadth-first search algorithm to find the shortest path through the maze. The time and space complexity of the algorithm is correctly analyzed.
The shortest path between two nodes in an undirected graph can be found by traversing the graph from one node to the other.
About this solution: The candidate's solution is complete and solves the problem. The candidate uses a breadth-first search approach to find the shortest path between two nodes.
This function takes in a graph represented as an adjacency list, and returns whether or not the graph contains a cycle.
About this solution: The candidate's solution is correct and demonstrates a good understanding of depth first search. However, the code could be more concise. For example, the has_cycle function could be written as follows:
def has_cycle(g):
visited = set()
for node in g:
if node not in visited:
if dfs(node, visited, None):
return 1
return 0
def dfs(node, visited, parent):
visited.add(node)
for neighbor in node:
if neighbor not in visited:
if dfs(neighbor, visited, node):
return True
elif neighbor != parent:
return True
return False