LeetCode 1244 Design A Leaderboard - Medium

1244. Design A Leaderboard — Medium

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Problem

1244. Design A Leaderboard -- Medium

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.

Solution

# Using Hashmap and Sort:
class Leaderboard:
    def __init__(self):
        self.leaderboad = dict()
    def addScore(self, playerId: int, score: int) -> None:
        if playerId in self.leaderboad:
            self.leaderboad[playerId] += score
        else:
            self.leaderboad[playerId] = score
    def top(self, K: int) -> int:
        ranks = sorted(self.leaderboad.values(), reverse = True)
        sum = 0
        for i in range(K):
            sum += ranks[i]
        return sum
    def reset(self, playerId: int) -> None:
        del self.leaderboad[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)
 
### Using Hashmap and Heap:
import heapq
class Leaderboard:
 
    def __init__(self):
        self.leaderboad = dict()
 
    def addScore(self, playerId: int, score: int) -> None:
        if playerId in self.leaderboad:
            self.leaderboad[playerId] += score
        else:
            self.leaderboad[playerId] = score
 
    def top(self, K: int) -> int:
        ranks = [-i for i in self.leaderboad.values()]
        heapq.heapify(ranks)
        sum = 0
        for i in range(K):
            sum += heapq.heappop(ranks)
        return sum * -1
        
 
    def reset(self, playerId: int) -> None:
        del self.leaderboad[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)