2102. Sequentially Ordinal Rank Tracker

Description

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. The higher the score, the better the location. If the scores of two locations are equal, then the location with the lexicographically smaller name is better.

You are building a system that tracks the ranking of locations with the system initially starting with no locations. It supports:

• Adding scenic locations, one at a time.
• Querying the ith best location of all locations already added, where i is the number of times the system has been queried (including the current query).
  • For example, when the system is queried for the 4th time, it returns the 4th best location of all locations already added.

Note that the test data are generated so that at any time, the number of queries does not exceed the number of locations added to the system.

Implement the SORTracker class:

• SORTracker() Initializes the tracker system.
• void add(string name, int score) Adds a scenic location with name and score to the system.
• string get() Queries and returns the ith best location, where i is the number of times this method has been invoked (including this invocation).

 

Example 1:

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.

Solutions

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 $i = -1$.

When calling the add method, we take the negative of the attraction's rating, so that we can use the ordered set to sort by rating in descending order. If the ratings are the same, sort by the dictionary order of the attraction names in ascending order.

When calling the get method, we increment $i$ by one, and then return the name of the $i$-th attraction in the ordered set.

The time complexity of each operation is $O(\log n)$, where $n$ is the number of added attractions. The space complexity is $O(n)$.

Python3C++

from sortedcontainers import SortedList


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()

#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/hash_policy.hpp>
using namespace __gnu_pbds;

template <class T>
using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;

class SORTracker {
public:
    SORTracker() {
    }

    void add(string name, int score) {
        st.insert({-score, name});
    }

    string get() {
        return st.find_by_order(++i)->second;
    }

private:
    ordered_set<pair<int, string>> st;
    int i = -1;
};

/**
 * Your SORTracker object will be instantiated and called as such:
 * SORTracker* obj = new SORTracker();
 * obj->add(name,score);
 * string 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.

Each time the add method is called, we add the attraction's rating and name to good, and then add the worst attraction in good to bad.

Each time the get method is called, we add the best attraction in bad to good, and then return the worst attraction in good.

The time complexity of each operation is $O(\log n)$, where $n$ is the number of added attractions. The space complexity is $O(n)$.

Python3JavaC++

class Node:
    def __init__(self, s: str):
        self.s = s

    def __lt__(self, other):
        return self.s > other.s


class SORTracker:

    def __init__(self):
        self.good = []
        self.bad = []

    def add(self, name: str, score: int) -> None:
        score, node = heappushpop(self.good, (score, Node(name)))
        heappush(self.bad, (-score, node.s))

    def get(self) -> str:
        score, name = heappop(self.bad)
        heappush(self.good, (-score, Node(name)))
        return self.good[0][1].s


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

class SORTracker {
    private PriorityQueue<Map.Entry<Integer, String>> good = new PriorityQueue<>(
        (a, b)
            -> a.getKey().equals(b.getKey()) ? b.getValue().compareTo(a.getValue())
                                             : a.getKey() - b.getKey());
    private PriorityQueue<Map.Entry<Integer, String>> bad = new PriorityQueue<>(
        (a, b)
            -> a.getKey().equals(b.getKey()) ? a.getValue().compareTo(b.getValue())
                                             : b.getKey() - a.getKey());

    public SORTracker() {
    }

    public void add(String name, int score) {
        good.offer(Map.entry(score, name));
        bad.offer(good.poll());
    }

    public String get() {
        good.offer(bad.poll());
        return good.peek().getValue();
    }
}

/**
 * Your SORTracker object will be instantiated and called as such:
 * SORTracker obj = new SORTracker();
 * obj.add(name,score);
 * String param_2 = obj.get();
 */

using pis = pair<int, string>;

class SORTracker {
public:
    SORTracker() {
    }

    void add(string name, int score) {
        good.push({-score, name});
        bad.push(good.top());
        good.pop();
    }

    string get() {
        good.push(bad.top());
        bad.pop();
        return good.top().second;
    }

private:
    priority_queue<pis, vector<pis>, less<pis>> good;
    priority_queue<pis, vector<pis>, greater<pis>> bad;
};

/**
 * Your SORTracker object will be instantiated and called as such:
 * SORTracker* obj = new SORTracker();
 * obj->add(name,score);
 * string param_2 = obj->get();
 */

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