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Leetcode #1244: Design A Leaderboard

In this guide, we solve Leetcode #1244 Design A Leaderboard 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.

Leetcode

Problem Statement

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.

Quick Facts

  • Difficulty: Medium
  • Premium: Yes
  • Tags: Design, Hash Table, Sorting

Intuition

Fast membership checks and value lookups are the heart of this problem, which makes a hash map the natural choice.

By storing what we have already seen (or counts/indexes), we can answer the question in one pass without backtracking.

Approach

Scan the input once, using the map to detect when the condition is satisfied and to update state as you go.

This keeps the solution linear while remaining easy to explain in an interview setting.

Steps:

  • Initialize a hash map for seen items or counts.
  • Iterate through the input, querying/updating the map.
  • Return the first valid result or the final computed value.

Example

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;

Python Solution

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)

Complexity

The time complexity is O(log⁡n)O(\log n)O(logn). The space complexity is O(n)O(n)O(n), where nnn is the number of players.

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.


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