Start by clarifying the problem constraints and edge cases, then explain the sorting-based two-pointer approach. Walk through the algorithm step-by-step, emphasizing how sorting enables efficient pair-finding and how duplicate handling ensures unique triplets. Conclude with complexity analysis and potential optimizations.
Pro tip: Mention that sorting the array first allows skipping duplicates efficiently, and that the two-pointer technique reduces the inner loop from O(n) to O(1) per element, achieving O(n^2) overall. Also, note that this approach is optimal for comparison-based sorting and is commonly used in industry interviews.
Confirm input constraints (e.g., array size, possible duplicates, integer range) and sort the array to enable two-pointer traversal and easy duplicate skipping.
Loop through the sorted array, fixing each element as the first of a triplet, and skip duplicates for this fixed element to avoid redundant triplets.
Use two pointers (left and right) to find pairs that sum to the remaining target. Move pointers based on the sum comparison, and skip duplicates for both pointers when a valid triplet is found.
Add each valid triplet to the result list, ensuring uniqueness by skipping duplicate values during pointer movements.
State that sorting takes O(n log n) and the two-pointer traversal takes O(n^2), resulting in overall O(n^2) time and O(1) extra space (excluding output).
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
First, count the frequency of each element using a hash map. Then, iterate through the map and maintain a min-heap of size k, where the heap is ordered by frequency and, for ties, by value (so the smallest frequency and smallest value are at the top). After processing all elements, extract the k elements from the heap to form the result, ensuring the order matches the problem's requirement (higher frequency first, then higher value).
Pro tip: Clarify the tie-breaking rule upfront and confirm whether the output should be sorted or just a set; this shows attention to detail and avoids incorrect assumptions. Also, mention that the heap comparator must consider both frequency and value to handle ties correctly.
Ask about input size, range of values, whether k is always valid, and if the output order matters. Confirm the tie-breaking rule: higher value first when frequencies are equal.
Use a hash map to count the occurrence of each element. This takes O(n) time and O(n) space.
Iterate through the frequency map and push each (element, frequency) pair into a min-heap ordered by frequency, then by value (so the smallest frequency and smallest value are at the top). If the heap size exceeds k, pop the top element.
After processing all elements, the heap contains the k most frequent elements. Extract them and reverse the order to get the correct sequence (highest frequency first, then highest value).
State the time complexity: O(n log k) due to heap operations, and space complexity: O(n) for the frequency map and O(k) for the heap. Walk through a small example to verify correctness.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Recognize that the problem reduces to making all root-to-leaf path sums equal by only increasing node costs. Use a bottom-up greedy approach: for each node, compute the maximum path sum from its children, then increment the node's cost to match the difference, accumulating the total increments. This ensures every path through the node has the same sum with minimal increments.
Pro tip: Emphasize that the greedy choice is optimal because increments at a node affect all paths through it equally, so matching the maximum child sum minimizes total increments. Also, mention that the tree is complete, so an array representation allows efficient bottom-up processing without explicit pointers.
Clarify that we can only increment costs, and we need all root-to-leaf path sums to be equal. Note that the tree is complete and stored as an array, so parent-child relationships are index-based.
For a node, the path sum from that node to any leaf in its subtree must be equal. Let maxChildSum be the maximum of the adjusted path sums from its children. The node's cost must be increased so that its path sum equals maxChildSum plus its own cost? Actually, the node's cost is fixed except we can increment it. We need to make all paths through the node equal, so we set the node's cost such that the sum from the node to leaves is uniform.
Process nodes in reverse level order (from leaves to root). For each node, compute the maximum path sum among its children (for leaves, this is 0). Then, the required increment for this node is the sum over children of (maxChildSum - childPathSum). Add this to total increments. The node's resulting path sum is its original cost plus maxChildSum.
Use an array to store the adjusted path sums. Iterate from the last parent down to the root. For each node, find its children (if any), compute maxChildSum, accumulate increments, and set the node's path sum. Return the total increments.
Time complexity is O(n) since each node is visited once. Space complexity is O(n) for the array of path sums. Discuss that this greedy approach is optimal because increments at a node affect all paths equally, so matching the maximum is necessary and sufficient.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.