How do you solve Dota2 Senate on LeetCode?

Dota2 Senate (LeetCode #649) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: The order of senators matters greatly in determining the outcome.

What is the brute force solution for Dota2 Senate?

The brute force approach for Dota2 Senate is Brute Force, with O(n²) time complexity and O(1) space complexity. In the brute-force approach, we simulate each round of voting. Each senator takes turns either banning an opponent or declaring victory. This method is straightforward but inefficient as it checks all senators repeatedly. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Dota2 Senate?

Dota2 Senate uses the following concepts: String, Greedy, Queue. The recommended patterns to study are: Queue, Greedy.

What are the interview tips for Dota2 Senate?

Explain your thought process clearly while coding. Discuss edge cases, such as all senators from one party. Practice simulating the process with small examples before coding.

What are common mistakes when solving Dota2 Senate?

Not considering the circular nature of the voting process. Assuming senators will always ban the nearest opponent without considering indices.

#649

Dota2 Senate

Medium

StringGreedyQueueQueueGreedy

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

The optimal approach uses a queue to efficiently manage the senators' actions. Each senator's index is stored, allowing us to process their actions in a round-robin fashion, ensuring we only process each senator once per round.

⚙️

Algorithm

4 steps

1Step 1: Initialize two queues, one for each party (Radiant and Dire).
2Step 2: Enqueue the indices of senators based on their party.
3Step 3: Process senators in a loop: dequeue one from each party, the one with the smaller index bans the other.
4Step 4: Enqueue the winner back to their respective queue with an updated index.

solution.py18 lines

1from collections import deque
2
3def predictPartyVictory(senate):
4    radiant = deque()
5    dire = deque()
6    for i, s in enumerate(senate):
7        if s == 'R':
8            radiant.append(i)
9        else:
10            dire.append(i)
11    while radiant and dire:
12        r_index = radiant.popleft()
13        d_index = dire.popleft()
14        if r_index < d_index:
15            radiant.append(r_index + len(senate))
16        else:
17            dire.append(d_index + len(senate))
18    return 'Radiant' if radiant else 'Dire'

ℹ

Complexity note: The time complexity is O(n) because each senator is processed once, and the space complexity is O(n) due to the storage of indices in the queues.

1The order of senators matters greatly in determining the outcome.
2Using queues allows us to efficiently manage the round-based voting process.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.