LeetCode 1244. Design A Leaderboard + StockTicker Problem
1 min readJul 29, 2022
Design a Leaderboard class, which has 3 functions:
addScore(playerId, score)
: Update the leaderboard by addingscore
to the given player's score. If there is no player with such id in the leaderboard, add him to the leaderboard with the givenscore
.top(K)
: Return the score sum of the topK
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.
Now How can we use the same logic for below problem?
Implement the following two functions such that they are optimal.
void addStocksVolume(string stockSymbol, int volume)
vector<string> topKstocks(int k)
addStocksVolume
receives a symbol (INTC, APPL, etc) plus a volume which accumulate over time.topKstocks
would return the k stocks with the highest volume.