Leetcode #2102: Sequentially Ordinal Rank Tracker
In this guide, we solve Leetcode #2102 Sequentially Ordinal Rank Tracker in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
A scenic location is represented by its name and attractiveness score, where name is a unique string among all locations and score is an integer. Locations can be ranked from the best to the worst.
Quick Facts
- Difficulty: Hard
- Premium: No
- Tags: Design, Data Stream, Ordered Set, Heap (Priority Queue)
Intuition
We need to repeatedly access the smallest or largest element as the input changes.
A heap provides fast insertions and removals while keeping order.
Approach
Push candidates into the heap as you scan, and pop when you need the best element.
Keep the heap size bounded if the problem requires a top-k structure.
Steps:
- Push candidates into a heap.
- Pop the best candidate when needed.
- Maintain heap size or invariants.
Example
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".
Python Solution
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()
Complexity
The time complexity is O(n log n). The space complexity is .
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.