Use a hash map to group strings by a canonical key that is identical for all anagrams. For each string, compute the key (e.g., sorted characters or character count signature) and append the string to the corresponding group. Finally, return the groups as a list of lists.
Pro tip: Discuss the trade-offs between sorting each string (O(n * k log k)) and using a character count key (O(n * k)), where n is the number of strings and k is the maximum length. Mention that the count-based approach can be more efficient for long strings with small alphabets, but sorting is simpler and often fast enough.
Ask clarifying questions: Are all strings lowercase? Can the input be empty? Should the output groups be sorted? Confirm that anagrams are case-sensitive and that the order of groups and within groups does not matter.
Decide on a key that uniquely identifies anagrams. Common choices: sorted string (e.g., 'eat' -> 'aet') or a character count signature (e.g., 'a1e1t1'). Explain the trade-offs.
Iterate through the array, compute the key for each string, and use a hash map to map the key to a list of strings. Append the current string to the list for its key.
After processing all strings, return the values of the hash map as a list of lists. The order of groups is arbitrary.
State the time and space complexity. For sorting approach: O(n * k log k) time, O(n * k) space. For count approach: O(n * k) time, O(n * k) space. Mention that k is the max string length.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
BFS level-order traversal and grab the last node at each level.
Clarify that the right-side view consists of the rightmost node at each depth. Then perform a level-order traversal (BFS) and record the last node of each level, or a DFS that prioritizes the right child and records the first node seen at each depth.
Pro tip: Mention that BFS is more intuitive for level-based problems, but DFS with right-first traversal uses O(h) space instead of O(w), which can be a significant advantage for very wide trees.
Confirm that the right-side view includes the rightmost node at each level, and that the output should be ordered from top to bottom. Discuss edge cases like an empty tree or a tree with only left children.
Decide between BFS (level-order traversal) and DFS (right-first traversal). BFS is straightforward: process each level and take the last node. DFS is more space-efficient: traverse right subtree before left, and record the first node encountered at each depth.
For BFS, use a queue to process nodes level by level, tracking the last node of each level. For DFS, use recursion with a depth parameter, and if the depth is being visited for the first time, add the node's value to the result.
Check for null root and return an empty list. Ensure the algorithm works for skewed trees (all left or all right) and for trees with varying depths.
State that both approaches visit each node once, so time complexity is O(n). Space complexity is O(w) for BFS (where w is the maximum width) and O(h) for DFS (where h is the height).
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.