How do you solve Sequentially Ordinal Rank Tracker on LeetCode?

Sequentially Ordinal Rank Tracker (LeetCode #2102) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log n) time complexity and O(n) space complexity. Key insight: Using a sorted list or priority queue can significantly reduce the time complexity for retrieval operations.

What is the brute force solution for Sequentially Ordinal Rank Tracker?

The brute force approach for Sequentially Ordinal Rank Tracker is Brute Force, with O(n²) time complexity and O(n) space complexity. The brute-force approach involves maintaining a list of locations and sorting it every time we need to get the ith best location. This is straightforward but inefficient as it requires sorting the list repeatedly. The optimal Optimal Solution approach improves this to O(n log n).

What algorithm or data structure is used to solve Sequentially Ordinal Rank Tracker?

Sequentially Ordinal Rank Tracker uses the following concepts: Design, Heap (Priority Queue), Data Stream, Ordered Set. The recommended patterns to study are: Heap (Priority Queue), Sorting, Binary Search (for maintaining order).

What are the interview tips for Sequentially Ordinal Rank Tracker?

Clearly explain your thought process and why you choose a specific data structure. Consider edge cases, such as adding locations with the same score. Practice implementing both brute-force and optimal solutions to understand trade-offs.

What are common mistakes when solving Sequentially Ordinal Rank Tracker?

Forgetting to handle the tie-breaking condition based on lexicographical order. Not considering the efficiency of sorting for each query, leading to performance issues.

#2102

Sequentially Ordinal Rank Tracker

Hard

DesignHeap (Priority Queue)Data StreamOrdered SetHeap (Priority Queue)SortingBinary Search (for maintaining order)

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n log n)
Space	O(n)	O(n)

💡

Intuition

Time O(n log n)Space O(n)

An optimal solution uses a priority queue (or a sorted list) to maintain the locations in order. This allows us to efficiently retrieve the ith best location without needing to sort the entire list every time.

⚙️

Algorithm

3 steps

1Step 1: Use a list to store the locations and a counter for queries.
2Step 2: For each 'add' operation, insert the new location into the list in a way that keeps it sorted.
3Step 3: For each 'get' operation, simply return the ith best location from the already sorted list.

solution.py13 lines

1import bisect
2
3class SORTracker:
4    def __init__(self):
5        self.locations = []
6        self.query_count = 0
7
8    def add(self, name: str, score: int) -> None:
9        bisect.insort(self.locations, (-score, name))
10
11    def get(self) -> str:
12        self.query_count += 1
13        return self.locations[self.query_count - 1][1]

ℹ

Complexity note: The time complexity is O(n log n) due to the sorting of the list after each addition, but we avoid the repeated full sort for each query, making it efficient overall.

1Using a sorted list or priority queue can significantly reduce the time complexity for retrieval operations.
2Maintaining order during insertion allows for efficient retrieval without needing to sort the entire list repeatedly.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.