This one took me longer than it should have.
Model the grid as a dependency graph where each block depends on all blocks to its right in the same row. Use a topological sort (e.g., Kahn's algorithm) to generate a valid removal order, processing blocks from right to left. Alternatively, simply iterate each row from right to left and remove blocks in that order, which naturally satisfies the condition.
Pro tip: Clarify that the problem is equivalent to topological sorting on a DAG, and mention that a simple right-to-left row-wise traversal is sufficient and optimal. This shows you recognize the underlying structure and avoid overcomplicating the solution.
Restate the rule: a block can be removed only if no block exists to its right in the same row. This means removal must proceed from rightmost to leftmost within each row.
Treat each block as a node with directed edges from a block to all blocks to its left in the same row (since left blocks depend on right blocks). This forms a DAG.
Use topological sorting (Kahn's or DFS) to produce a valid order. Alternatively, note that iterating each row from right to left and collecting coordinates yields a valid order directly.
Write code that iterates rows and columns appropriately, ensuring no block is removed before its right neighbors. Test with small grids and edge cases (empty rows, all 1s, etc.).
State that the simple row-wise approach runs in O(m*n) time and O(1) extra space (excluding output), which is optimal since every block must be visited.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Use a hash map to store each element's value and index as you iterate through the array. For each element, check if the complement (target - current element) exists in the map; if so, return the two indices. This achieves O(n) time and O(n) space.
Pro tip: Clarify upfront that you assume exactly one solution exists and that you cannot use the same element twice. Mention that if the array is sorted, a two-pointer approach could achieve O(1) space, but the hash map is optimal for unsorted input.
Confirm with the interviewer that there is exactly one valid answer, that indices must be distinct, and that the array is unsorted. Ask about edge cases like empty array or no solution.
Select a hash map (dictionary) to store values and their indices, enabling O(1) average-time lookups for complements.
Loop through the array; for each element, compute the complement and check if it exists in the map. If found, return the stored index and the current index.
If no pair is found after the loop, return an empty list or as specified. Ensure you don't use the same element twice by checking the map before inserting the current element.
State that time complexity is O(n) and space complexity is O(n). Mention alternative approaches like sorting with two pointers (O(n log n) time, O(1) space) if the array were sorted.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Start by clarifying the problem constraints (e.g., array size, element range, duplicates allowed) and then propose sorting the array first, as the minimum difference will be between adjacent elements in sorted order. Alternatively, mention a hash-based approach if the range is small, but emphasize that sorting is generally optimal with O(n log n) time.
Pro tip: After presenting the sorting solution, briefly discuss trade-offs: if the array is huge and elements are bounded, a counting sort or bucket approach could achieve O(n) time. This shows you consider scalability and constraints, which is valued at Upstart.
Ask about input size, element range, duplicates, and whether the array can be modified. This ensures you choose the right algorithm and handle edge cases.
Explain that sorting the array brings close elements together, so the minimum absolute difference must be between adjacent elements. Then a single pass computes the minimum difference.
State time complexity O(n log n) due to sorting and space O(1) if in-place. Handle edge cases: array with fewer than 2 elements, all elements equal, large differences.
Mention that if the range of elements is small, a counting sort or bucket approach can achieve O(n) time. Also note that a brute-force O(n^2) is inefficient and not recommended.
Implement the solution with clear variable names, then walk through a small example to verify correctness, including edge cases.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.