What is the time complexity of Dota2 Senate?

The optimal solution has a time complexity of O(n).

What data structures are used to solve Dota2 Senate?

This problem uses String, Greedy, Queue. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Dota2 Senate?

Insight 1: The problem can be modeled as a queue-based simulation, where senators take turns and are added back to the queue if they haven't been banned. Insight 2: The relative positions of 'R' and 'D' senators in the original senate string matter. Early senators have an advantage since they get to ban first.

Which companies ask LeetCode 649. Dota2 Senate in interviews?

This problem is frequently asked at Valve.

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Dota2 Senate - LeetCode 649 Solution

Dota2 Senate - Complete Solution Guide

Dota2 Senate is LeetCode problem 649, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

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

Detailed Explanation

The Dota2 Senate problem simulates a voting process between two parties, Radiant ('R') and Dire ('D'). The goal is to predict which party will ultimately win the vote given an initial arrangement of senators from both parties. Senators take turns, and each senator can either ban another senator from voting or announce victory if all remaining senators are from their party. The problem specifies that senators act strategically to maximize their party's chances, and we must predict the winning party.

Solution Approach

The provided solution uses queues to simulate the voting process. Each party ('R' and 'D') has its own queue, storing the indices of the senators. The algorithm iterates, taking one senator from each queue. The senator with the smaller index (appearing earlier in the original string) gets to ban the other. The surviving senator is then added back to the end of their queue, but with an offset to represent the next round. This continues until one of the queues is empty, indicating that the corresponding party has won.

Step-by-Step Algorithm

Step 1: Initialize two queues, `r_queue` and `d_queue`, to store the indices of 'R' and 'D' senators, respectively.
Step 2: Iterate through the input string `senate` and populate the queues with senator indices based on their party.
Step 3: Enter a `while` loop that continues as long as both queues are not empty.
Step 4: In each iteration of the loop, dequeue one senator's index from `r_queue` and one from `d_queue`.
Step 5: Compare the indices. The senator with the lower index gets to ban the other. The senator with the lower index's index is enqueued back to its queue with an offset of `n` (the length of the input string) to simulate the next round.
Step 6: The loop terminates when either `r_queue` or `d_queue` becomes empty.
Step 7: If `r_queue` is not empty, return "Radiant". Otherwise, return "Dire".

Key Insights

Insight 1: The problem can be modeled as a queue-based simulation, where senators take turns and are added back to the queue if they haven't been banned.
Insight 2: The relative positions of 'R' and 'D' senators in the original senate string matter. Early senators have an advantage since they get to ban first.
Insight 3: The core logic revolves around simulating rounds of voting and strategically eliminating opposing senators until only one party remains.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

This problem involves: String, Greedy, Queue.

Companies

Asked at: Valve.

Dota2 Senate | LeetCode 649

Medium

String Greedy Queue

Time: O(n)Space: O(n)

Solution Code

Available in:

import java.util.Deque;
import java.util.LinkedList;

class Solution {
    public String predictPartyVictory(String senate) {
        int n = senate.length();
        
        Deque<Integer> r_queue = new LinkedList<>();
        Deque<Integer> d_queue = new LinkedList<>();
        
        for (int i = 0; i < senate.length(); i++) {
            char s = senate.charAt(i);
            if (s == 'R') {
                r_queue.add(i);
            } else {
                d_queue.add(i);
            }
        }
        
        while (!r_queue.isEmpty() && !d_queue.isEmpty()) {
            int r_idx = r_queue.poll();
            int d_idx = d_queue.poll();
            
            if (r_idx < d_idx) {
                r_queue.add(r_idx + n);
            } else {
                d_queue.add(d_idx + n);
            }
        }
        
        if (!r_queue.isEmpty()) {
            return "Radiant";
        } else {
            return "Dire";
        }
    }
}

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Problem Info

Description

Problem 649: Dota2 Senate - Detailed explanation and solution approaches.

Topics

String Greedy Queue

Companies

Valve

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:649

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

7 steps

1
Step 1: Initialize two queues, `r_queue` and `d_queue`, to store the indices of 'R' and 'D' senators, respectively.
2
Step 2: Iterate through the input string `senate` and populate the queues with senator indices based on their party.
3
Step 3: Enter a `while` loop that continues as long as both queues are not empty.
4
Step 4: In each iteration of the loop, dequeue one senator's index from `r_queue` and one from `d_queue`.
5
Step 5: Compare the indices. The senator with the lower index gets to ban the other. The senator with the lower index's index is enqueued back to its queue with an offset of `n` (the length of the input string) to simulate the next round.
6
Step 6: The loop terminates when either `r_queue` or `d_queue` becomes empty.
7
Step 7: If `r_queue` is not empty, return "Radiant". Otherwise, return "Dire".

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The time complexity is O(n) because we iterate through the string once to initialize the queues, and then the while loop processes each senator at most twice (once when initially added and once when being re-added to the queue with an offset). The space complexity is O(n) because the queues store, in the worst case, all the senators of one type (R or D).

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Simpler Simulation with Boolean Array

Instead of using queues, a boolean array could be used to track which senators are still active. The simulation proceeds in rounds, iterating through the array and banning senators until only one party remains. This involves keeping track of 'ban' counts for each party.

Time: O(n^2) in the worst case. The outer loop could iterate n times and the inner loop will iterate over n senators in each turn.Space: O(n) for the boolean array

Trade-offs: This approach avoids queues, but it can be less efficient due to the nested loops needed to simulate each round and find eligible senators to ban. The queue-based approach is generally better for its O(n) time complexity.

✨

Key Insights

1Insight 1: The problem can be modeled as a queue-based simulation, where senators take turns and are added back to the queue if they haven't been banned.
2Insight 2: The relative positions of 'R' and 'D' senators in the original senate string matter. Early senators have an advantage since they get to ban first.
3Insight 3: The core logic revolves around simulating rounds of voting and strategically eliminating opposing senators until only one party remains.

📚

Key Terms Explained

For Students

Queue

FIFO structure - First In, First Out. Like a line at a store!

Greedy Algorithm

Making the best choice at each step without looking ahead.

String

A sequence of characters. Like a word or sentence in programming!

Edge Cases

Edge case 1: Input string contains only one type of senator (e.g., "RRR" or "DDD"). The algorithm correctly identifies the winning party as there will only be one queue populated and the loop will terminate immediately.
Edge case 2: Input string where 'R' and 'D' alternate frequently (e.g., "RDRDRD"). The algorithm correctly simulates the banning process, and the winner depends on which party gets the initial advantage.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrectly handling the round-based procedure. Forgetting to add senators back to the queue (or equivalent data structure) after they ban another senator.

✗Common mistake 2: Failing to consider the advantage of senators appearing earlier in the input string. Banning senators randomly or based on some other criterion instead of prioritizing banning senators with later indices leads to incorrect results.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Resource allocation and conflict resolution. Simulating the allocation of resources between competing entities or resolving conflicts where entities take turns to eliminate each other (e.g., in a network or game).

🏦Real-world scenario 2: Process scheduling in operating systems. Modeling the execution of processes, where processes can block each other (analogous to banning senators). Determining which process will ultimately complete its execution based on priority and dependencies.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

String problems often involve sliding windows or two pointers. Consider if the character set is limited (lowercase letters = only 26 possibilities).

🎯 Interview Tip

If stuck, try working through a small example by hand. The pattern often reveals itself.

⚠️ Watch Out

Mutating strings directly in languages where strings are immutable. Use StringBuilder or arrays.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: September 18, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 13

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 6

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJun 29

❓

Frequently Asked Questions

What is the sliding window technique for string problems?

Sliding window maintains a "window" of characters with two pointers. Expand the right pointer to include characters, shrink from left when window violates constraints. Useful for substring problems, anagrams, and longest/shortest substring with properties.

What are common greedy problem patterns?

Activity selection (sort by end time), interval scheduling (sort by start/end), Huffman coding (always combine smallest), fractional knapsack (sort by value/weight ratio), and minimum spanning tree. Look for problems where local decisions don't affect future options.

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Dota2 Senate"?

For this medium-level String problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether greedy techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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