Dota2 Senate

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ArraysStringsGreedy AlgorithmsTwo PointersStacks

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".

For example:

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.

Dota2 Senate

Medium

Amazon

1 view

Topics:

ArraysStringsGreedy AlgorithmsTwo PointersStacks

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".

For example:

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.

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What is the maximum number of senators in the senate?
Can the input string `senate` be empty or null?
If neither faction can ban the other, meaning the game goes on indefinitely, what should be the return value?
Are the characters 'R' and 'D' the only possible characters in the input string?
Does the order of senators matter in the input string, or is it only the count of 'R' and 'D' that matters?

Brute Force Solution

Approach

The brute force strategy involves simulating the senate voting process round by round. We consider each senator's ability to ban the opposing party in each round, and continue until only one party remains capable of banning.

Here's how the algorithm would work step-by-step:

Look at the list of senators, one by one, in the order they appear.
If a senator can ban someone from the opposing party, do it, and mark that senator as having used their ability for this round.
If a senator is already banned, skip them and move to the next senator.
Repeat this process for all senators in the list.
If, after a round, there are still senators from both parties who can ban, start another round from the beginning of the list.
Keep doing this, round after round, until only senators from one party are left who can ban others.
The party of the remaining senators is the winner.

Code Implementation

def dota2_senate(senate_string):
    senators = list(senate_string)
    senator_count = len(senators)
    banned_senators = [False] * senator_count

    while True:
        radiant_present = False
        dire_present = False

        for i in range(senator_count):
            if not banned_senators[i]:
                if senators[i] == 'R':
                    radiant_present = True
                else:
                    dire_present = True

        if not radiant_present:
            return "Dire"
        if not dire_present:
            return "Radiant"

        for i in range(senator_count):
            if not banned_senators[i]:
                if senators[i] == 'R':
                    # Find the next available Dire senator to ban
                    for j in range(i + 1, senator_count) :
                        if not banned_senators[j] and senators[j] == 'D':
                            banned_senators[j] = True
                            break
                    else:
                        # If no Dire senators found after i, loop back
                        for j in range(i):
                            if not banned_senators[j] and senators[j] == 'D':
                                banned_senators[j] = True
                                break
                else:
                    # Find the next available Radiant senator to ban
                    for j in range(i + 1, senator_count):
                        if not banned_senators[j] and senators[j] == 'R':
                            banned_senators[j] = True
                            break
                    else:
                        # If no Radiant senators found after i, loop back
                        for j in range(i):
                            if not banned_senators[j] and senators[j] == 'R':
                                banned_senators[j] = True
                                break

Big(O) Analysis

Time Complexity

O(n^2) – The outer loop iterates through the senate array until only one party remains. In the worst-case scenario, each senator needs to ban another in multiple rounds. The inner loop iterates through all the senators in each round trying to find senators from the other party to ban. Because the outer loop can run up to n times (where n is the number of senators) and the inner loop iterates through n senators, the time complexity is O(n*n). Therefore, the overall time complexity is O(n^2).

Space Complexity

O(N) – The described brute force strategy implicitly maintains a queue of senators, as senators are processed round by round, and potentially revisited in later rounds if not banned. Although not explicitly stated, senators who haven't been banned are effectively kept in a queue for future rounds. In the worst-case scenario, all senators might need to be revisited multiple times until one party wins, implying a data structure of size proportional to the number of senators, N, where N is the length of the input string. Therefore, the auxiliary space complexity is O(N).

Optimal Solution

Approach

The key to this problem is simulating the senate process in a way that accurately reflects the power dynamics. Instead of naively trying to eliminate senators, we track their influence and efficiently determine the eventual winning party by simulating the rounds of banning.

Here's how the algorithm would work step-by-step:

Keep track of which senators are still active and can vote.
Go through the senators one by one, in order.
If a senator is still active, check if they are banned by a senator from the opposing party. Keep track of how many bans each party has available.
If the senator is not banned, they ban the next active senator from the opposing party, if one exists.
If a senator is banned, they do nothing and are skipped for the rest of the process, in effect removing them.
Repeat steps 2-5 until only senators from one party remain. The winning party is the one whose senators are the last ones standing.

Code Implementation

def dota2_senate(senate):
    senators_active = [True] * len(senate)
    radiant_bans = 0
    dire_bans = 0

    while True:
        for i in range(len(senate)): 
            if senators_active[i]:
                if senate[i] == 'R':
                    if radiant_bans > 0:
                        radiant_bans -= 1
                        senators_active[i] = False
                    else:
                        dire_bans += 1

                else:
                    if dire_bans > 0:
                        dire_bans -= 1
                        senators_active[i] = False
                    else:
                        radiant_bans += 1
        
        # Check if only one party remains
        radiant_present = False
        dire_present = False
        for i in range(len(senate)): 
            if senators_active[i]:
                if senate[i] == 'R':
                    radiant_present = True
                else:
                    dire_present = True

        if not dire_present:
            return "Radiant"
        if not radiant_present:
            return "Dire"

Big(O) Analysis

Time Complexity

O(n^2) – The algorithm iterates through the senators in rounds until only one party remains. In the worst case, each senator might need to ban another senator from the opposing party in each round. The outer loop iterates through the senators potentially multiple times, and in each iteration, it searches for a senator from the opposing party to ban. This inner search can take O(n) time in the worst case for each senator in the outer loop. Thus, the worst case scenario involves approximately n iterations of the senate, and each iteration does potentially O(n) work checking each senator, leading to O(n*n) complexity. Therefore the overall time complexity is O(n^2).

Space Complexity

O(N) – The algorithm keeps track of active senators. This could be done implicitly (by modifying the input string directly) or explicitly, for example, using an array or set of size N, where N is the number of senators. We also need to keep track of the number of bans each party has available, which uses constant space. However, the dominant space usage is tracking the active senators, which requires space proportional to the input size N. Therefore, the space complexity is O(N).

Edge Cases

Case	How to Handle
Empty senate string	Return an appropriate message (e.g., 'No senator to ban') or throw an exception, indicating invalid input.
Senate string with only one senator (e.g., 'R' or 'D')	Return the single senator's party as the winner immediately, as banning is impossible.
Senate string with all senators from the same party (e.g., 'RRRR' or 'DDDD')	Return the dominant party as the winner directly, since no banning can change the outcome.
Senate string with an equal number of 'R' and 'D' senators (e.g., 'RD' or 'RDRD')	The solution must correctly handle the banning process and determine the winner based on the banning order, typically the senate string order.
Large senate string exceeding reasonable memory limits	The solution should ideally be optimized for memory usage, and consider stream processing if input size is truly massive.
String contains characters other than 'R' and 'D'	Either throw an error indicating an invalid input or filter out the invalid characters before processing the senate string.
Order of senators drastically favoring one party initially (e.g., 'RRRRRDD')	The solution must fairly simulate the banning process irrespective of initial senator order to determine the winner correctly.
Alternating sequence of R and D (e.g. 'RDRDRDRD')	The simulation should correctly identify the winner based on the order senators are banned, where the last senator remaining will win.