Design Log Storage System

Published: Oct 17, 2022

Medium Hash Table String Design

Introduction

When a problem is related to logging, how to handle timestamp is a key to solve the problem. Sometime it is already an integer, float, but sometime is is given as a string. Whether the timestamp string should be converted to specific datetime object or not is a design decision. In this case, using it just as a string makes the code simpler.

Problem Description

You are given several logs, where each log contains a unique ID and timestamp. Timestamp is a string that has the following format: Year:Month:Day:Hour:Minute:Second, for example, 2017:01:01:23:59:59. All domains are zero-padded decimal numbers.

Implement the LogSystem class:

  • LogSystem() Initializes the LogSystem object.
  • void put(int id, string timestamp) Stores the given log (id, timestamp)` in your storage system.
  • int[] retrieve(string start, string end, string granularity) Returns the IDs of the logs whose timestamps are within the range from start to end inclusive. The start and end all have the same format as timestamp, and granularity means how precise the range should be (i.e. to the exact Day, Minute, etc.). For example, start = "2017:01:01:23:59:59", end = "2017:01:02:23:59:59", and granularity = "Day" means that we need to find the logs within the inclusive range from Jan. 1st 2017 to Jan. 2nd 2017, and the Hour, Minute, and Second for each log entry can be ignored.

Constraints:

  • 1 <= id <= 500
  • 2000 <= Year <= 2017
  • 1 <= Month <= 12
  • 1 <= Day <= 31
  • 0 <= Hour <= 23
  • 0 <= Minute, Second <= 59
  • granularity is one of the values ["Year", "Month", "Day", "Hour", "Minute", "Second"].
  • At most 500 calls will be made to put and retrieve.

https://leetcode.com/problems/design-log-storage-system/

Examples

Example 1

Input
["LogSystem", "put", "put", "put", "retrieve", "retrieve"]
[[], [1, "2017:01:01:23:59:59"], [2, "2017:01:01:22:59:59"], [3, "2016:01:01:00:00:00"],
["2016:01:01:01:01:01", "2017:01:01:23:00:00", "Year"], ["2016:01:01:01:01:01", "2017:01:01:23:00:00", "Hour"]]
Output
[null, null, null, null, [3, 2, 1], [2, 1]]

Explanation
LogSystem logSystem = new LogSystem();
logSystem.put(1, "2017:01:01:23:59:59");
logSystem.put(2, "2017:01:01:22:59:59");
logSystem.put(3, "2016:01:01:00:00:00");

// return [3,2,1], because you need to return all logs between 2016 and 2017.
logSystem.retrieve("2016:01:01:01:01:01", "2017:01:01:23:00:00", "Year");

// return [2,1], because you need to return all logs between Jan. 1, 2016 01:XX:XX and Jan. 1, 2017 23:XX:XX.
// Log 3 is not returned because Jan. 1, 2016 00:00:00 comes before the start of the range.
logSystem.retrieve("2016:01:01:01:01:01", "2017:01:01:23:00:00", "Hour");

Analysis

The given timestamp strings are like: ‘yyyy:mm:dd:hh:MM:ss’ always. Additionally, all are zero padded. This means comparison by string is easy. The granularity parameter specified the length of string comparison. If it is ‘Year’, the first 4 letters should be compared. If it is ‘Hour’, the first 13 letters should be compared. Given that, the solution here has a granularity to index table. To retrieve the ids in the range, timestamp strings are compared up to the granularity index. The answer is straightforward.

Solution

class LogSystem:

    def __init__(self):
        self.logs = []
        self.indices = {
            'Year': 5, 'Month': 8, 'Day': 11,
            'Hour': 14, 'Minute': 17, 'Second': 20
        }


    def put(self, id: int, timestamp: str) -> None:
        self.logs.append((id, timestamp))
        

    def retrieve(self, start: str, end: str, granularity: str) -> List[int]:
        idx = self.indices[granularity]
        s, e = start[:idx], end[:idx]
        return [id_ for id_, timestamp in self.logs if s <= timestamp[:idx] <= e]

Complexities

  • Time: put – O(1), retrieve – O(n): n is a total number of logs
  • Space: O(n)