LeetCode 649: Dota2 Senate Solution in Python – A Step-by-Step Guide

What Is LeetCode 649: Dota2 Senate?

In LeetCode 649: Dota2 Senate, you’re given a string senate of length n where each character is either 'R' (Radiant) or 'D' (Dire), representing senators. The voting process works in rounds:

• Each senator bans one senator from the opposing party, starting from the left.
• Banned senators are skipped in future rounds.
• Rounds continue until one party has all remaining senators.

Your task is to predict the winning party, returning "Radiant" or "Dire". For example, with senate = "RD", Radiant bans Dire in round 1, leaving only Radiant, so "Radiant" wins. This problem tests your ability to simulate a process with strategic eliminations.

Problem Statement

• Input:
• senate: String of 'R' and 'D' characters.
• Output: A string—"Radiant" or "Dire" (winning party).
• Rules:
• Each senator bans one opponent per round, left to right.
• Banned senators lose voting rights.
• Rounds repeat until one party remains.

Constraints

• 1 ≤ senate.length ≤ 10⁴.
• senate[i] is either 'R' or 'D'.

Examples

• Input: senate = "RD"
• Output: "Radiant" (R bans D, R remains)
• Input: senate = "RDD"
• Output: "Dire" (R bans D, first D bans R, second D remains)
• Input: senate = "RRR"
• Output: "Radiant" (All R, no D to ban)

These examples set the senate—let’s solve it!

Understanding the Problem: Predicting the Winner

To solve LeetCode 649: Dota2 Senate in Python, we need to simulate rounds where senators ban opponents, continuing until one party dominates. A brute-force approach—simulating each ban with array modifications—could be O(n²) per round, inefficient for n ≤ 10⁴ with multiple rounds. Since bans follow a left-to-right order and affect future rounds, we can optimize with a queue or counting strategy. We’ll use:

• Best Solution (Greedy with Queue Simulation): O(n) time, O(n) space—fast and dynamic.
• Alternative Solution (Count-Based Simulation): O(n²) time, O(1) space—simpler but slower.

Let’s start with the queue solution, breaking it down for beginners.

Best Solution: Greedy with Queue Simulation

Why This Is the Best Solution

The greedy with queue simulation is the top choice for LeetCode 649 because it’s efficient—O(n) time with O(n) space—and uses two queues to simulate rounds greedily, ensuring each senator bans the nearest opponent from the opposing party in a single pass per round. It fits constraints (n ≤ 10⁴) perfectly and is intuitive once you see the queues in action. It’s like lining up senators and letting them take turns banning efficiently!

How It Works

Think of this as a ban lineup:

• Step 1: Initialize Queues:
• Two queues: one for Radiant indices, one for Dire indices.
• Populate with initial senator positions (0 to n-1).
• Step 2: Simulate Rounds:
• While both queues have elements:
• Compare front indices of R and D queues.
• Lower index bans higher; winner re-enters queue with index + n (next round).
• Step 3: Determine Winner:
• Whichever queue remains non-empty wins.
• Step 4: Return Result:
• Return "Radiant" or "Dire".

It’s like a ban queue—line up and eliminate!

Step-by-Step Example

Example: senate = "RDD"

• Step 1: Init:
• R_queue = [0], D_queue = [1, 2].
• n = 3.
• Step 2: Simulate:
• Round 1:
• R[0] vs D[1]: 0 < 1, R bans D, R_queue = [3], D_queue = [2].
• D[2] vs R[3]: 2 < 3, D bans R, R_queue = [], D_queue = [5].
• R_queue empty, D_queue = [5], stop.
• Step 3: D_queue has elements, Dire wins.
• Output: "Dire".

Code with Detailed Line-by-Line Explanation

Here’s the Python code, explained clearly:

Copy codefrom collections import deque

class Solution:
    def predictPartyVictory(self, senate: str) -> str:
        # Step 1: Initialize queues
        n = len(senate)
        r_queue = deque([i for i in range(n) if senate[i] == 'R'])
        d_queue = deque([i for i in range(n) if senate[i] == 'D'])

        # Step 2: Simulate rounds
        while r_queue and d_queue:
            r_idx = r_queue.popleft()
            d_idx = d_queue.popleft()

            if r_idx < d_idx:
                # R bans D, R re-enters next round
                r_queue.append(r_idx + n)
            else:
                # D bans R, D re-enters next round
                d_queue.append(d_idx + n)

        # Step 3: Determine winner
        return "Radiant" if r_queue else "Dire"

• Line 1: Import deque for queues.
• Lines 6-9: Init:
• Create queues with indices of R and D senators.
• Lines 12-19: Simulate:
• Pop front indices, lower bans higher, winner re-enters with +n.
• Line 22: Return winner based on non-empty queue.
• Time Complexity: O(n)—each senator processed once per round, amortized.
• Space Complexity: O(n)—queue sizes.

This is like a senate simulator—queue and ban!

Alternative Solution: Count-Based Simulation

Why an Alternative Approach?

The count-based simulation tracks party counts and bans—O(n²) time, O(1) space. It’s simpler, simulating rounds by iterating and banning directly in the string, but slower due to repeated passes. It’s like counting votes and crossing out opponents round by round!

How It Works

Picture this as a vote counter:

• Step 1: Count initial R and D senators.
• Step 2: Simulate rounds:
• Scan string left to right.
• Each senator bans one opponent, update counts, mark bans.
• Repeat until one party’s count hits 0.
• Step 3: Determine winner from remaining count.
• Step 4: Return result.

It’s like a ban tallier—count and eliminate!

Step-by-Step Example

Example: senate = "RDD"

• Step 1: R = 1, D = 2, senate = "RDD".
• Step 2: Simulate:
• Round 1:
• R[0]: Bans D[1], senate = "R-D", R = 1, D = 1.
• D[2]: Bans R[0], senate = "--D", R = 0, D = 1.
• R = 0, stop.
• Step 3: D > 0, Dire wins.
• Output: "Dire".

Code for Count-Based Approach

Copy codeclass Solution:
    def predictPartyVictory(self, senate: str) -> str:
        # Step 1: Initialize counts and array
        n = len(senate)
        senate = list(senate)
        r_count = senate.count('R')
        d_count = n - r_count

        # Step 2: Simulate rounds
        while r_count > 0 and d_count > 0:
            r_bans = d_bans = 0
            for i in range(n):
                if senate[i] == 'R' and r_bans < d_count:
                    senate[i] = '-'  # Ban this R
                    r_count -= 1
                    d_bans += 1
                elif senate[i] == 'D' and d_bans < r_count:
                    senate[i] = '-'  # Ban this D
                    d_count -= 1
                    r_bans += 1
            if r_count == 0 or d_count == 0:
                break

        # Step 3: Determine winner
        return "Radiant" if r_count > 0 else "Dire"

• Lines 4-7: Init: Convert to list, count R and D.
• Lines 10-22: Simulate:
• Track bans per round, update counts, mark banned.
• Line 25: Return winner.
• Time Complexity: O(n²)—n iterations per round.
• Space Complexity: O(1)—modifies input in-place.

It’s a vote eliminator—count and ban!

Comparing the Two Solutions

• Queue-Based (Best):
• Pros: O(n) time, O(n) space, fast and elegant.
• Cons: Queue management less intuitive.
• Count-Based (Alternative):
• Pros: O(n²) time, O(1) space, straightforward.
• Cons: Slower due to multiple passes.

Queue-based wins for efficiency.

Additional Examples and Edge Cases

• Input: senate = "RRR"
• Output: "Radiant".
• Input: senate = "DDR"
• Output: "Dire".

Both handle these well.

Complexity Breakdown

• Queue-Based: Time O(n), Space O(n).
• Count-Based: Time O(n²), Space O(1).

Queue-based is optimal.

Key Takeaways

• Queue-Based: Greedy queues—smart!
• Count-Based: Simple counts—clear!
• Simulation: Voting is fun.
• Python Tip: Queues rock—see Python Basics.

Final Thoughts: Crown the Victor

LeetCode 649: Dota2 Senate in Python is a fun simulation challenge. Queue-based simulation offers speed and dynamics, while count-based provides a clear baseline. Want more? Try LeetCode 621: Task Scheduler or LeetCode 406: Queue Reconstruction by Height. Ready to vote? Head to Solve LeetCode 649 on LeetCode and predict that winner today!