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Dota2 Senate - Solution & Explanation

MediumStringGreedyQueue17 min readAsked at: Amazon, Meta, American Airlines +3
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Problem Statement

In the world of Dota2, there are two parties: the Radiant and the Dire.

The Dota2 senate consists of senators coming from two parties. Now the Senate wants to decide on a change in the Dota2 game. The voting for this change is a round-based procedure. In each round, each senator can exercise one of the two rights:

  • Ban one senator's right: A senator can make another senator lose all his rights in this and all the following rounds.
  • Announce the victory: If this senator found the senators who still have rights to vote are all from the same party, he can announce the victory and decide on the change in the game.

Given a string senate representing each senator's party belonging. The character 'R' and 'D' represent the Radiant party and the Dire party. Then if there are n senators, the size of the given string will be n.

The round-based procedure starts from the first senator to the last senator in the given order. This procedure will last until the end of voting. All the senators who have lost their rights will be skipped during the procedure.

Suppose every senator is smart enough and will play the best strategy for his own party. Predict which party will finally announce the victory and change the Dota2 game. The output should be "Radiant" or "Dire".

 

Example 1:

Input: senate = "RD"
Output: "Radiant"
Explanation: 
The first senator comes from Radiant and he can just ban the next senator's right in round 1. 
And the second senator can't exercise any rights anymore since his right has been banned. 
And in round 2, the first senator can just announce the victory since he is the only guy in the senate who can vote.

Example 2:

Input: senate = "RDD"
Output: "Dire"
Explanation: 
The first senator comes from Radiant and he can just ban the next senator's right in round 1. 
And the second senator can't exercise any rights anymore since his right has been banned. 
And the third senator comes from Dire and he can ban the first senator's right in round 1. 
And in round 2, the third senator can just announce the victory since he is the only guy in the senate who can vote.

 

Constraints:

  • n == senate.length
  • 1 <= n <= 104
  • senate[i] is either 'R' or 'D'.

Approach Overview

Problem Overview: You are given a string where each character represents a senator from the Radiant (R) or Dire (D) party. Senators take turns banning opponents. Once banned, a senator cannot vote in future rounds. The process continues in rounds until only one party remains with active senators. Your task is to predict which party wins.

Approach 1: Direct Round Simulation (O(n^2) time, O(n) space)

The most straightforward idea is to simulate the voting rounds exactly as described. Iterate through the string in a circular manner. When a senator appears, check whether they still have the right to vote. If so, ban the next available opponent by marking them inactive. Continue looping until all remaining active senators belong to the same party.

This approach mirrors the problem statement but repeatedly scans the array to find the next valid opponent. In the worst case, each senator might trigger a long search across the list. That repeated scanning pushes the time complexity to O(n^2). The space cost stays O(n) because you maintain the senate state and banned markers.

Approach 2: Queue Simulation with Indices (O(n) time, O(n) space)

The optimized strategy treats the process as a turn-based competition using two queues. Store the indices of Radiant senators in one queue and Dire senators in another. Each round compares the front elements of both queues. The senator with the smaller index acts first and bans the opponent.

After banning, the acting senator returns to the queue with an updated index of currentIndex + n. This represents the next round because the senate order wraps around. The banned senator simply disappears from the process. Continue until one queue becomes empty.

This method avoids scanning for opponents. Each senator enters and leaves the queue a limited number of times, giving an overall time complexity of O(n) and space complexity of O(n). The behavior is essentially a greedy turn scheduling problem using a queue. The input itself is a string, and the decision rule follows a greedy principle: whoever appears earlier in the current round gets priority to ban.

Recommended for interviews: Interviewers typically expect the queue simulation approach. The brute-force round simulation shows that you understand the mechanics of the problem, but the index-based queue solution demonstrates stronger algorithmic thinking. It converts a circular voting process into a clean greedy scheduling problem and reduces the complexity from quadratic to linear.

Approach 1: Queue Simulation Approach

We utilize two separate queues to maintain indices of Radiant and Dire senators. In each round, senators from both parties execute their rights and effect bans on their opponent. The process continues until one queue becomes empty, indicating that all senators from one party have been eliminated.

In the solution, we employ a queue to separately track Radiant and Dire senators' indices. For each senate round, we compare indices of both parties. The senator with a smaller index, indicating an earlier position, bans the other and gets scheduled for the next round with an increased index (index + senate length). The queue that first becomes empty indicates the losing party.

Code

CC++JavaPythonC#JavaScript

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C++

Java

Python

C#

JavaScript

Complexity

Time Complexity: O(n), where n is the length of the senate, since each senator is processed a finite number of times.
Space Complexity: O(n), for storing indices in queues.

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Approach 2: Queue + Simulation

We create two queues qr and qd to record the indices of the Radiant and Dire senators, respectively. Then we start the simulation, where in each round we dequeue one senator from each queue and perform different operations based on their factions:

  • If the Radiant senator's index is less than the Dire senator's index, the Radiant senator can permanently ban the voting rights of the Dire senator. We add n to the Radiant senator's index and enqueue it back to the end of the queue, indicating that this senator will participate in the next round of voting.
  • If the Dire senator's index is less than the Radiant senator's index, the Dire senator can permanently ban the voting rights of the Radiant senator. We add n to the Dire senator's index and enqueue it back to the end of the queue, indicating that this senator will participate in the next round of voting.

Finally, when there are only senators from one faction left in the queues, the senators from that faction win.

The time complexity is O(n), and the space complexity is O(n). Here, n is the number of senators.

Code

PythonJavaC++GoTypeScriptRust

Python

Java

C++

Go

TypeScript

Rust

Try this approach in the editor →

Complexity Comparison

ApproachComplexity
Queue Simulation Approach

Time Complexity: O(n), where n is the length of the senate, since each senator is processed a finite number of times.
Space Complexity: O(n), for storing indices in queues.

Queue + Simulation—

Detailed Complexity Analysis

ApproachTimeSpaceWhen to Use
Direct Round SimulationO(n^2)O(n)Useful for understanding the rules and basic simulation before optimizing
Queue Simulation with IndicesO(n)O(n)Best approach for interviews and large inputs; avoids repeated scanning

Video Solution

Dota2 Senate - Leetcode 649 - Python • NeetCodeIO • 36,086 views views

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Frequently Asked Questions

Is Dota2 Senate easy or hard?
Dota2 Senate is classified as a medium problem. The challenge lies in modeling the circular voting process efficiently. Once you recognize that two queues can simulate the turns, the implementation becomes straightforward.
Dota2 Senate Python/Java solution
Most implementations follow the same logic across languages: maintain two queues of indices, simulate bans, and reinsert the winning senator with index + n. The approach translates cleanly to Python, Java, C++, C#, and JavaScript.
How to solve Dota2 Senate in O(n)?
Store indices of 'R' and 'D' senators in two queues. At each step, compare the front elements. The senator with the smaller index bans the opponent and re-enters the queue with index + n to represent the next round. Continue until one queue becomes empty.
What is the best approach for Dota2 Senate?
The queue simulation approach is the most efficient and widely expected solution. Maintain two queues storing indices of Radiant and Dire senators. Compare the front indices each round, allow the earlier senator to ban the opponent, and push the winner back with index + n. This runs in O(n) time and O(n) space.
Is Dota2 Senate asked at Google/Amazon/Meta?
Dota2 Senate appears in interview preparation sets for companies like Amazon, Google, and Meta because it tests queue simulation, greedy reasoning, and handling circular processes. It is a common medium-level interview practice problem.
What data structure is used in Dota2 Senate?
The optimal solution uses queues to track the order of Radiant and Dire senators. Each queue stores indices so the algorithm can determine which senator acts first in each round.
What is the time complexity of Dota2 Senate?
The optimal queue-based solution runs in O(n) time because each senator is processed a limited number of times in the queues. Space complexity is O(n) to store the indices of senators from both parties.

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