#2102

Hard

auto_awesomeDesign plus Heap (Priority Queue)

LeetCode Problem Workspace

Sequentially Ordinal Rank Tracker

Track rankings of locations with names and scores, adding new locations and retrieving top-ranked ones efficiently.

design heap priority queue data stream ordered set

Problem Statement

You are building a ranking system for locations based on their attractiveness scores. Each location has a unique name and an associated score. Locations are ranked from best to worst, with the higher score considered better. In the case of equal scores, locations with lexicographically smaller names are ranked higher.

The system supports adding locations and querying the best location. Initially, the system starts with no locations. The query mechanism returns the top-ranked location after each addition. The problem involves handling location additions and retrieving the best location in an efficient manner using heap structures.

Examples

Example 1

Input: See original problem statement.

Output: See original problem statement.

Input ["SORTracker", "add", "add", "get", "add", "get", "add", "get", "add", "get", "add", "get", "get"] [[], ["bradford", 2], ["branford", 3], [], ["alps", 2], [], ["orland", 2], [], ["orlando", 3], [], ["alpine", 2], [], []] Output [null, null, null, "branford", null, "alps", null, "bradford", null, "bradford", null, "bradford", "orland"]

Explanation SORTracker tracker = new SORTracker(); // Initialize the tracker system. tracker.add("bradford", 2); // Add location with name="bradford" and score=2 to the system. tracker.add("branford", 3); // Add location with name="branford" and score=3 to the system. tracker.get(); // The sorted locations, from best to worst, are: branford, bradford. // Note that branford precedes bradford due to its higher score (3 > 2). // This is the 1st time get() is called, so return the best location: "branford". tracker.add("alps", 2); // Add location with name="alps" and score=2 to the system. tracker.get(); // Sorted locations: branford, alps, bradford. // Note that alps precedes bradford even though they have the same score (2). // This is because "alps" is lexicographically smaller than "bradford". // Return the 2nd best location "alps", as it is the 2nd time get() is called. tracker.add("orland", 2); // Add location with name="orland" and score=2 to the system. tracker.get(); // Sorted locations: branford, alps, bradford, orland. // Return "bradford", as it is the 3rd time get() is called. tracker.add("orlando", 3); // Add location with name="orlando" and score=3 to the system. tracker.get(); // Sorted locations: branford, orlando, alps, bradford, orland. // Return "bradford". tracker.add("alpine", 2); // Add location with name="alpine" and score=2 to the system. tracker.get(); // Sorted locations: branford, orlando, alpine, alps, bradford, orland. // Return "bradford". tracker.get(); // Sorted locations: branford, orlando, alpine, alps, bradford, orland. // Return "orland".

Constraints

name consists of lowercase English letters, and is unique among all locations.
1 <= name.length <= 10
1 <= score <= 105
At any time, the number of calls to get does not exceed the number of calls to add.
At most 4 * 104 calls in total will be made to add and get.

Solution Approach

Use a Min-Heap for Efficient Querying

To maintain the rankings efficiently, use a min-heap (priority queue). Each time a location is added, the heap is adjusted to maintain the correct order based on score and name. The heap will store the locations and automatically ensure that the lowest-ranked location can be easily popped when necessary.

Store Locations in a Data Structure for Sorting

Use a custom data structure that can store location name and score pairs. Each time a new location is added, the data structure should handle sorting by score, and in case of ties, by lexicographic order of names. This ensures that the heap operations remain efficient and the top location is always accessible.

Efficient Querying

After each addition, use the heap to query the best location. By maintaining the heap's order, you can quickly retrieve the top location, even as new locations are continuously added. This guarantees an efficient querying mechanism for a high-volume data stream.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity for adding a location is O(log N) due to heap insertion, where N is the number of locations. Querying the top location is O(1), as it is always at the root of the heap. The space complexity is O(N), where N is the number of locations stored in the heap.

What Interviewers Usually Probe

Look for understanding of efficient data structure usage like heaps for ranking systems.
Expect the candidate to connect the problem with handling dynamic data streams effectively.
Candidates should mention managing both numeric scores and string-based lexicographic comparisons in their solution.

Common Pitfalls or Variants

Common pitfalls

Mismanaging ties in score rankings could lead to incorrect outputs. Remember to use lexicographical order when scores are equal.
Overcomplicating the design by introducing unnecessary data structures when a simple heap would suffice.
Failing to consider the size of the data set and optimizing for query performance could result in inefficiency.

Follow-up variants

Tracking the top-k locations instead of just the best location.
Handling locations with additional attributes besides name and score.
Modifying the problem to track and return the rankings for multiple queries at once.

FAQ

What data structure is optimal for tracking rankings in this problem?

A min-heap (priority queue) is optimal for tracking rankings because it allows efficient addition and removal of elements while maintaining order.

How do I handle ties in location scores?

In the case of ties, locations with lexicographically smaller names should be ranked higher. This can be achieved by storing both score and name in the heap and using lexicographical comparisons when scores are equal.

What is the time complexity for adding a location?

The time complexity for adding a location is O(log N), where N is the number of locations already in the system, due to the heap insertion operation.

Can the system handle a large number of locations?

Yes, the system can efficiently handle up to 4 * 10^4 calls to add and get operations, thanks to the efficient heap operations for both insertion and querying.

How does GhostInterview help with this problem?

GhostInterview provides a clear path for designing efficient ranking systems using heaps, offering step-by-step guidance on how to optimize for both score and lexicographical order.

terminal

Solution

Solution 1: Ordered Set

We can use an ordered set to store the attractions, and a variable $i$ to record the current number of queries, initially $i = -1$.

class SORTracker:

    def __init__(self):
        self.sl = SortedList()
        self.i = -1

    def add(self, name: str, score: int) -> None:
        self.sl.add((-score, name))

    def get(self) -> str:
        self.i += 1
        return self.sl[self.i][1]


# Your SORTracker object will be instantiated and called as such:
# obj = SORTracker()
# obj.add(name,score)
# param_2 = obj.get()

Solution 2: Double Priority Queue (Min-Max Heap)

We notice that the query operations in this problem are performed in strictly increasing order. Therefore, we can use a method similar to the median in the data stream. We define two priority queues `good` and `bad`. `good` is a min-heap, storing the current best attractions, and `bad` is a max-heap, storing the current $i$-th best attraction.

class SORTracker:

    def __init__(self):
        self.sl = SortedList()
        self.i = -1

    def add(self, name: str, score: int) -> None:
        self.sl.add((-score, name))

    def get(self) -> str:
        self.i += 1
        return self.sl[self.i][1]


# Your SORTracker object will be instantiated and called as such:
# obj = SORTracker()
# obj.add(name,score)
# param_2 = obj.get()

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