Quick Overview

Given an in-memory log sorted by integer timestamp, return every entry whose timestamp lies within an inclusive start and end time. The problem tests exact boundary handling with duplicate timestamps, empty results when the range misses the log, and locating a range efficiently in sorted data.

Return Log Entries Within an Inclusive Time Range From a Sorted Log

Company: Attentive

Role: Software Engineer

Category: Coding & Algorithms

Difficulty: medium

Interview Round: Onsite

You are given a log that has already been loaded into memory as a list of entries. Each entry is a pair `(timestamp, message)`, where `timestamp` is an integer and `message` is a string. The entries are sorted by timestamp in non-decreasing order, and several entries may share the same timestamp. Given `start_time` and `end_time`, return every entry whose timestamp `t` satisfies `start_time <= t <= end_time`. ### Function Signature ```python def logs_in_range(logs: list[tuple[int, str]], start_time: int, end_time: int) -> list[tuple[int, str]]: ``` ### Rules - Both bounds are inclusive. - Return the matching entries in the same order as they appear in `logs`. Because `logs` is sorted, the matching entries form one contiguous run of `logs`, and that run, in its original order, is the unique correct answer. Entries with equal timestamps keep their input order. - Return each matching entry unchanged, as the same `(timestamp, message)` pair. Do not remove duplicates, even if two entries have the same timestamp and the same message. - If no entry falls in the range, return an empty list. This includes the cases where `start_time` is greater than the largest timestamp, where `end_time` is smaller than the smallest timestamp, and where `logs` is empty. - Do not modify `logs`. ### Constraints - `0 <= len(logs) <= 10^5` - `0 <= timestamp <= 10^9` for every entry - `logs[i][0] <= logs[i + 1][0]` for every valid `i` - `0 <= start_time <= end_time <= 10^9` - Each message is a string of 1 to 100 printable ASCII characters. Messages need not be distinct. - Every timestamp and bound fits in a 32-bit signed integer. ### Examples **Example 1** ```text Input: logs = [(1, "boot"), (3, "connect"), (3, "retry"), (5, "ready"), (8, "shutdown")] start_time = 3, end_time = 5 Output: [(3, "connect"), (3, "retry"), (5, "ready")] ``` Both entries at timestamp `3` are included, and the entry at `5` is included because the end bound is inclusive. **Example 2** ```text Input: logs = [(2, "a"), (4, "b"), (6, "c")] start_time = 7, end_time = 10 Output: [] ``` `start_time` is greater than the largest timestamp, `6`, so nothing matches. **Example 3** ```text Input: logs = [(2, "a"), (2, "b"), (4, "c"), (6, "d")] start_time = 0, end_time = 2 Output: [(2, "a"), (2, "b")] ``` The end bound equals the smallest timestamp, so both entries at timestamp `2` match.

Overview: Given an in-memory log sorted by integer timestamp, return every entry whose timestamp lies within an inclusive start and end time. The problem tests exact boundary handling with duplicate timestamps, empty results when the range misses the log, and locating a range efficiently in sorted data.

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You are given a log that has already been loaded into memory as a list `logs` of entries. Each entry is a pair `(timestamp, message)`, where `timestamp` is an integer and `message` is a string. The entries are sorted by timestamp in non-decreasing order, and several entries may share the same timestamp. Given two integers `start_time` and `end_time`, return every entry whose timestamp `t` satisfies `start_time <= t <= end_time`. Implement `logs_in_range(logs, start_time, end_time)`. ### Rules - Both bounds are inclusive. - Return the matching entries in the same order as they appear in `logs`. Because `logs` is sorted, the matching entries form one contiguous run of `logs`, and that run, in its original order, is the unique correct answer. Entries with equal timestamps keep their input order. - Return each matching entry unchanged, as the same `(timestamp, message)` pair. Do not remove duplicates, even if two entries have the same timestamp and the same message. - If no entry falls in the range, return an empty list. This includes the cases where `start_time` is greater than the largest timestamp, where `end_time` is smaller than the smallest timestamp, and where `logs` is empty. - Do not modify `logs`. ### Entry representation `logs` is passed as a single argument, a list of two-element entries, and the result uses the same entry shape: - Python: a list of `(timestamp, message)` tuples. - JavaScript: an array of `[timestamp, message]` arrays. - Java: a `java.util.List<java.util.List<Object>>`; in each entry, element 0 is the timestamp (read it through `Number`) and element 1 is the `String` message. - C++: a `std::vector<std::pair<int, std::string>>`. Every timestamp and bound is at most 10^9 and fits in a 32-bit signed integer; no value exceeds 2^31 - 1, so `int` is sufficient in Java and C++. ### Constraints - `0 <= len(logs) <= 10^5` - `0 <= timestamp <= 10^9` for every entry - `logs[i][0] <= logs[i + 1][0]` for every valid `i` - `0 <= start_time <= end_time <= 10^9` - Each message is a string of 1 to 100 printable ASCII characters. Messages need not be distinct. - Every timestamp and bound fits in a 32-bit signed integer. ### Examples **Example 1** ```text Input: logs = [(1, "boot"), (3, "connect"), (3, "retry"), (5, "ready"), (8, "shutdown")] start_time = 3, end_time = 5 Output: [(3, "connect"), (3, "retry"), (5, "ready")] ``` Both entries at timestamp `3` are included, and the entry at `5` is included because the end bound is inclusive. **Example 2** ```text Input: logs = [(2, "a"), (4, "b"), (6, "c")] start_time = 7, end_time = 10 Output: [] ``` `start_time` is greater than the largest timestamp, `6`, so nothing matches.

Constraints

  • 0 <= len(logs) <= 10^5
  • 0 <= timestamp <= 10^9 for every entry
  • logs[i][0] <= logs[i + 1][0] for every valid i
  • 0 <= start_time <= end_time <= 10^9
  • Each message is a string of 1 to 100 printable ASCII characters. Messages need not be distinct.
  • Every timestamp and bound fits in a 32-bit signed integer (no value exceeds 2^31 - 1).

Examples

Input: ([(1, 'boot'), (3, 'connect'), (3, 'retry'), (5, 'ready'), (8, 'shutdown')], 3, 5)

Expected Output: [(3, 'connect'), (3, 'retry'), (5, 'ready')]

Explanation: Source example 1: both entries at 3 and the entry at 5 match because both bounds are inclusive.

Input: ([(2, 'a'), (4, 'b'), (6, 'c')], 7, 10)

Expected Output: []

Explanation: Source example 2: start_time is greater than the largest timestamp.

Hints

  1. Because the entries are already sorted by timestamp, all of the matching entries sit next to each other in `logs`.
  2. Several entries can share a timestamp, including timestamps that equal `start_time` or `end_time` exactly; both bounds are inclusive, so none of those entries may be dropped.
  3. Return the matching entries exactly as they appear: same order, same pairs, duplicates included, and an empty list when nothing matches.

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Approach

Because logs is sorted by timestamp in non-decreasing order, the entries with start_time <= t <= end_time form one contiguous block, so it suffices to find the block's two boundaries. The first binary search finds left, the first index whose timestamp is >= start_time; its invariant is that every index before lo has a timestamp < start_time and every index from hi on has a timestamp >= start_time, so when lo == hi it is exactly the boundary. The second binary search finds right, the first index whose timestamp is > end_time, with the analogous invariant. It may start at left because every entry before left has a timestamp < start_time <= end_time, which also guarantees left <= right. Every entry in [left, right) is then >= start_time and <= end_time, and every entry outside it fails one bound, so the slice logs[left:right] is exactly the answer; slicing keeps the original order of equal timestamps and keeps every duplicate untouched, and logs itself is never modified. Edge cases: empty logs gives left = right = 0 and an empty list; a window entirely before the first timestamp, after the last one, or inside a gap gives left == right and an empty list; a bound equal to a timestamp includes every entry with that timestamp because both comparisons are inclusive. A single linear pass that keeps entries inside the window is also correct, in O(n) time.

Time complexity:
O(log n + k), where k is the number of returned entries
Space complexity:
O(k) for the returned list; O(1) extra