Two-pointer from both ends, and when you hit a mismatch you try skipping one side or the other.
Use a two-pointer technique to compare characters from both ends, and when a mismatch occurs, check if skipping either the left or right character results in a palindrome. This yields an O(n) time and O(1) space solution, which is optimal.
Pro tip: Clarify that 'at most one deletion' includes zero deletions, and mention that the two-pointer approach is optimal because it avoids unnecessary checks and handles all cases efficiently.
Confirm that the string contains only lowercase letters and that we can delete at most one character. Also, note that an empty string or a single-character string is already a palindrome.
Set left pointer at the start and right pointer at the end of the string. Compare characters while left < right.
When characters at left and right don't match, check if the substring skipping the left character or skipping the right character is a palindrome. If either is, return true.
If no mismatches are found, the string is already a palindrome, so return true. If a mismatch occurs and neither skip works, return false.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Model the courses and prerequisites as a directed graph where edges represent prerequisite relationships. Then determine if the graph is acyclic by performing a topological sort (using Kahn's algorithm or DFS) and checking if all courses can be processed. If a cycle exists, it's impossible to complete all courses.
Pro tip: Clarify edge direction upfront (e.g., prerequisite -> course) to avoid confusion, and mention that Kahn's algorithm naturally detects cycles by counting processed nodes. Also, discuss how you'd handle large inputs by using iterative DFS to avoid recursion depth issues.
Confirm the input format and edge direction (e.g., pair [a, b] means b depends on a). Represent courses as nodes and prerequisites as directed edges in a graph.
Decide between Kahn's algorithm (BFS-based topological sort) or DFS with recursion stack. Explain why one might be preferred (e.g., Kahn's is iterative and easy to reason about).
For Kahn's: compute in-degrees, enqueue nodes with in-degree 0, process and decrement neighbors, count processed nodes. For DFS: track visited and recursion stack to detect back edges.
State time and space complexity (O(V+E)). Discuss edge cases: no prerequisites, disconnected components, self-loops, and duplicate edges.
If all nodes are processed (Kahn's) or no cycle found (DFS), return true; otherwise false. Optionally, walk through a small example to validate.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.