Start by clarifying the problem: strictly increasing means each next element is greater than the previous, and contiguous means adjacent elements. Then propose a single-pass linear scan that tracks the current run length and updates the best run when it ends, handling ties by keeping the earliest occurrence unless specified otherwise.
Pro tip: Explicitly state your tie-breaking rule (e.g., earliest start index) and justify it; this shows attention to detail and prevents ambiguity in real-world systems where deterministic output matters.
Confirm definitions: strictly increasing (a[i] < a[i+1]), contiguous subarray, and tie-breaking preference (e.g., earliest start). Discuss edge cases: empty array, single element, all equal, strictly decreasing.
Use a single pass with variables for current run start, current run length, best start, best length. Iterate from index 1, extending the run if a[i] > a[i-1], else compare and reset.
When current run length equals best length, decide whether to update based on the tie-breaking rule (e.g., keep earliest start). Implement by only updating best when current length > best length, or when equal and current start < best start.
Trace the algorithm on a sample array (e.g., [1,2,3,2,4,5,6]) to show correctness, and test edge cases like empty array, single element, and all decreasing.
State O(n) time and O(1) space. Mention that tie-breaking adds negligible overhead. If asked, discuss alternative approaches (e.g., two pointers) but emphasize the single-pass optimality.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Knew the patience sorting approach going in but explaining the correctness proof on the spot was rough.
Use the patience sorting algorithm with binary search to compute the LIS length in O(n log n), maintaining an array of the smallest tail values for increasing subsequences of each length. To reconstruct the subsequence, store parent pointers and indices during the process, then backtrack from the end of the longest subsequence.
Pro tip: Emphasize that the tails array does not represent the actual LIS but is a tool for length computation; reconstruction requires separate parent tracking. Also, mention that binary search must find the first element >= current (lower_bound) to maintain strict increase.
Confirm that the subsequence must be strictly increasing and that we need both length and one valid subsequence. Discuss handling of empty arrays, duplicates, and negative numbers.
Describe maintaining an array 'tails' where tails[i] is the smallest tail of an increasing subsequence of length i+1. For each element, use binary search to find its position and update tails accordingly.
Introduce parent pointers and an index array to track the predecessor of each element in the LIS. When updating tails, record the index of the previous element in the subsequence.
After processing all elements, backtrack from the index stored at the end of the longest subsequence using parent pointers to build the subsequence in reverse order.
Explain why the algorithm is correct: tails maintains the minimal possible tail for each length, ensuring optimality. State time O(n log n) due to binary search per element, and space O(n) for tails, parents, and indices.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
This is the classic memoized DFS on a grid.
Use DFS with memoization to compute the longest increasing path from each cell, caching results to avoid redundant work. For path reconstruction, store the next cell in the path for each cell. Analyze time and space complexity, emphasizing that each cell is processed once.
Pro tip: Mention that the problem is equivalent to finding the longest path in a DAG where edges go from smaller to larger values, so topological sort can also be used. Also, note that the grid size 500x500 means recursion depth could be up to 250,000, so an iterative approach or increasing recursion limit is necessary.
Restate the problem: find the length of the longest strictly increasing path in a grid, moving to orthogonal neighbors. Confirm if path reconstruction is required and discuss constraints (e.g., grid size up to 500x500).
Propose DFS with memoization: for each cell, recursively explore neighbors with larger values, caching the longest path length from that cell. For path reconstruction, store the next cell in the path.
Time complexity: O(m*n) because each cell is visited once. Space complexity: O(m*n) for memoization and recursion stack (or iterative stack).
Explain that memoization prevents revisiting states: once a cell's longest path is computed, it's stored and reused. Also, since paths are strictly increasing, cycles are impossible.
Discuss handling large grids (iterative DFS to avoid stack overflow), and potential optimizations like topological sort or using a 2D array for memoization.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.