Description#
Design a Leaderboard class, which has 3 functions:
addScore(playerId, score): Update the leaderboard by adding score to the given player's score. If there is no player with such id in the leaderboard, add him to the leaderboard with the given score.top(K): Return the score sum of the top K players.reset(playerId): Reset the score of the player with the given id to 0 (in other words erase it from the leaderboard). It is guaranteed that the player was added to the leaderboard before calling this function.
Initially, the leaderboard is empty.
Example 1:
Input:
["Leaderboard","addScore","addScore","addScore","addScore","addScore","top","reset","reset","addScore","top"]
[[],[1,73],[2,56],[3,39],[4,51],[5,4],[1],[1],[2],[2,51],[3]]
Output:
[null,null,null,null,null,null,73,null,null,null,141]
Explanation:
Leaderboard leaderboard = new Leaderboard ();
leaderboard.addScore(1,73); // leaderboard = [[1,73]];
leaderboard.addScore(2,56); // leaderboard = [[1,73],[2,56]];
leaderboard.addScore(3,39); // leaderboard = [[1,73],[2,56],[3,39]];
leaderboard.addScore(4,51); // leaderboard = [[1,73],[2,56],[3,39],[4,51]];
leaderboard.addScore(5,4); // leaderboard = [[1,73],[2,56],[3,39],[4,51],[5,4]];
leaderboard.top(1); // returns 73;
leaderboard.reset(1); // leaderboard = [[2,56],[3,39],[4,51],[5,4]];
leaderboard.reset(2); // leaderboard = [[3,39],[4,51],[5,4]];
leaderboard.addScore(2,51); // leaderboard = [[2,51],[3,39],[4,51],[5,4]];
leaderboard.top(3); // returns 141 = 51 + 51 + 39;
Constraints:
1 <= playerId, K <= 10000- It's guaranteed that
K is less than or equal to the current number of players. 1 <= score <= 100- There will be at most
1000 function calls.
Solutions#
Solution 1: Hash Table + Ordered List#
We use a hash table $d$ to record the scores of each player, and an ordered list $rank$ to record the scores of all players.
When the addScore function is called, we first check if the player is in the hash table $d$. If not, we add their score to the ordered list $rank$. Otherwise, we first remove their score from the ordered list $rank$, then add their updated score to the ordered list $rank$, and finally update the score in the hash table $d$. The time complexity is $O(\log n)$.
When the top function is called, we directly return the sum of the first $K$ elements in the ordered list $rank$. The time complexity is $O(K \times \log n)$.
When the reset function is called, we first remove the player from the hash table $d$, then remove their score from the ordered list $rank$. The time complexity is $O(\log n)$.
The space complexity is $O(n)$, where $n$ is the number of players.
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| from sortedcontainers import SortedList
class Leaderboard:
def __init__(self):
self.d = defaultdict(int)
self.rank = SortedList()
def addScore(self, playerId: int, score: int) -> None:
if playerId not in self.d:
self.d[playerId] = score
self.rank.add(score)
else:
self.rank.remove(self.d[playerId])
self.d[playerId] += score
self.rank.add(self.d[playerId])
def top(self, K: int) -> int:
return sum(self.rank[-K:])
def reset(self, playerId: int) -> None:
self.rank.remove(self.d.pop(playerId))
# Your Leaderboard object will be instantiated and called as such:
# obj = Leaderboard()
# obj.addScore(playerId,score)
# param_2 = obj.top(K)
# obj.reset(playerId)
|
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| class Leaderboard {
private Map<Integer, Integer> d = new HashMap<>();
private TreeMap<Integer, Integer> rank = new TreeMap<>((a, b) -> b - a);
public Leaderboard() {
}
public void addScore(int playerId, int score) {
d.merge(playerId, score, Integer::sum);
int newScore = d.get(playerId);
if (newScore != score) {
rank.merge(newScore - score, -1, Integer::sum);
}
rank.merge(newScore, 1, Integer::sum);
}
public int top(int K) {
int ans = 0;
for (var e : rank.entrySet()) {
int score = e.getKey(), cnt = e.getValue();
cnt = Math.min(cnt, K);
ans += score * cnt;
K -= cnt;
if (K == 0) {
break;
}
}
return ans;
}
public void reset(int playerId) {
int score = d.remove(playerId);
if (rank.merge(score, -1, Integer::sum) == 0) {
rank.remove(score);
}
}
}
/**
* Your Leaderboard object will be instantiated and called as such:
* Leaderboard obj = new Leaderboard();
* obj.addScore(playerId,score);
* int param_2 = obj.top(K);
* obj.reset(playerId);
*/
|
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| class Leaderboard {
public:
Leaderboard() {
}
void addScore(int playerId, int score) {
d[playerId] += score;
int newScore = d[playerId];
if (newScore != score) {
rank.erase(rank.find(newScore - score));
}
rank.insert(newScore);
}
int top(int K) {
int ans = 0;
for (auto& x : rank) {
ans += x;
if (--K == 0) {
break;
}
}
return ans;
}
void reset(int playerId) {
int score = d[playerId];
d.erase(playerId);
rank.erase(rank.find(score));
}
private:
unordered_map<int, int> d;
multiset<int, greater<int>> rank;
};
/**
* Your Leaderboard object will be instantiated and called as such:
* Leaderboard* obj = new Leaderboard();
* obj->addScore(playerId,score);
* int param_2 = obj->top(K);
* obj->reset(playerId);
*/
|
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| use std::collections::BTreeMap;
#[allow(dead_code)]
struct Leaderboard {
/// This also keeps track of the top K players since it's implicitly sorted
record_map: BTreeMap<i32, i32>,
}
impl Leaderboard {
#[allow(dead_code)]
fn new() -> Self {
Self {
record_map: BTreeMap::new(),
}
}
#[allow(dead_code)]
fn add_score(&mut self, player_id: i32, score: i32) {
if self.record_map.contains_key(&player_id) {
// The player exists, just add the score
self.record_map.insert(player_id, self.record_map.get(&player_id).unwrap() + score);
} else {
// Add the new player to the map
self.record_map.insert(player_id, score);
}
}
#[allow(dead_code)]
fn top(&self, k: i32) -> i32 {
let mut cur_vec: Vec<(i32, i32)> = self.record_map
.iter()
.map(|(k, v)| (*k, *v))
.collect();
cur_vec.sort_by(|lhs, rhs| { rhs.1.cmp(&lhs.1) });
// Iterate reversely for K
let mut sum = 0;
let mut i = 0;
for (_, value) in &cur_vec {
if i == k {
break;
}
sum += value;
i += 1;
}
sum
}
#[allow(dead_code)]
fn reset(&mut self, player_id: i32) {
// The player is ensured to exist in the board
// Just set the score to 0
self.record_map.insert(player_id, 0);
}
}
|
