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Merge overlapping corrections | Grammarly Coding Question

Quick Overview

This question evaluates a candidate's ability to manage and manipulate overlapping intervals with multi-criteria tie-breaking, preserve original metadata across splits, and design efficient data structures and algorithms for producing non-overlapping correction spans.

Merge overlapping corrections

Company: Grammarly

Role: Software Engineer

Category: Coding & Algorithms

Difficulty: medium

Interview Round: Onsite

Implement `correctionMerge(corrections)`. You are given a list of text-correction spans. Each correction applies to an **inclusive** interval `[start, end]` and has the following fields: - `start: int` — inclusive start index - `end: int` — inclusive end index - `priority: int` — higher value means higher priority - `change: enum` — the modification to apply, such as `DELETE`, `DEDUP`, `UPDATE`, `UPPERCASE`, or `LOWERCASE` Return a list of corrections such that no two output corrections overlap. When two corrections overlap, resolve the overlap using these rules for every covered position: 1. The correction with the higher `priority` wins. 2. If priorities are equal, the correction with the **smaller original span length** `(end - start)` wins. 3. If both are still equal, preserve the earlier correction in the input order. If a correction loses only part of its interval, split it into the remaining non-overlapping pieces. Important detail: if an interval is split, every fragment must keep the metadata of the **original** correction for future comparisons, including its original priority, input order, and original span length. In other words, after splitting `(1,10,1,UPPERCASE)` into `(1,2,...)` and `(7,10,...)`, later tie-breaking must still treat both fragments as coming from the original span `(1,10)`, not from shorter spans `(1,2)` or `(7,10)`. Example: - input: `(1,10,1,UPPERCASE)`, `(3,6,1,LOWERCASE)` - output: `(1,2,1,UPPERCASE)`, `(3,6,1,LOWERCASE)`, `(7,10,1,UPPERCASE)` Assume the number of corrections can be large, so iterating over every character position is not allowed. Return the minimal sorted list of non-overlapping output segments.

Quick Answer: This question evaluates a candidate's ability to manage and manipulate overlapping intervals with multi-criteria tie-breaking, preserve original metadata across splits, and design efficient data structures and algorithms for producing non-overlapping correction spans.

Implement `solution(corrections)`. Each correction is a 4-tuple `(start, end, priority, change)` describing an inclusive interval `[start, end]` in a text. Return a sorted list of non-overlapping correction segments. For every covered position, choose the winning correction using these rules: 1. Higher `priority` wins. 2. If priorities tie, the correction with the smaller original span length `(end - start)` wins. 3. If they are still tied, the correction that appeared earlier in the input wins. If a correction loses only part of its range, it is split into the remaining pieces. Important: every split piece must keep the original correction's tie-break metadata for all later comparisons, including its original priority, original span length, and input order. Do not recompute span length from a fragment. You must not iterate over every character position. The input can be large. Return intervals as 4-tuples `(start, end, priority, change)`, still using inclusive ends. Omit uncovered positions. After all conflicts are resolved, merge adjacent output segments when their returned fields are identical.

Constraints

0 <= len(corrections) <= 2 * 10^5
0 <= start <= end <= 10^9
-10^9 <= priority <= 10^9
A solution that scans every position in every interval will time out

Examples

Input: ([(1, 10, 1, "UPPERCASE"), (3, 6, 1, "LOWERCASE")],)

Expected Output: [(1, 2, 1, "UPPERCASE"), (3, 6, 1, "LOWERCASE"), (7, 10, 1, "UPPERCASE")]

Explanation: Both corrections have priority 1, but `(3, 6)` has the smaller original span length, so it wins on the overlap. The larger interval survives on both sides.

Input: ([(0, 4, 1, "DELETE"), (2, 3, 5, "UPDATE")],)

Expected Output: [(0, 1, 1, "DELETE"), (2, 3, 5, "UPDATE"), (4, 4, 1, "DELETE")]

Explanation: The `UPDATE` correction has higher priority on positions 2 and 3, so the `DELETE` correction is split into the remaining left and right pieces.

Hints

Convert each inclusive interval `[start, end]` into a half-open interval `[start, end + 1)` and sweep only across event coordinates where something starts or stops.
Maintain the currently active corrections in a priority queue ordered by `(-priority, original_length, input_index)`. Use lazy removal for intervals that have already ended.

Quick Overview