Dota2 Senate

class Solution {
public:
    string predictPartyVictory(string senate) {

        // Initialize counts for Radiant and Dire senators
        int radiantCount = 0, direCount = 0;

        // Initialize counters for the bans applied by each side
        int bansByDire = 0, bansByRadiant = 0;

        // Queue to handle the senators sequence
        queue<char> senatorQueue;
        for (char senator : senate) {
            senatorQueue.push(senator);
            if (senator == 'R') radiantCount++;
            else direCount++;
        }

        // Process the senators
        while (radiantCount > 0 && direCount > 0) {
            char currentSenator = senatorQueue.front();
            senatorQueue.pop();

            // Check senator party and apply rules for bans
            if (currentSenator == 'D') {
                if (bansByDire > 0) {
                    bansByDire--;
                    direCount--;
                } else {
                    bansByRadiant++;
                    senatorQueue.push('D');
                }
            } else {
                if (bansByRadiant > 0) {
                    bansByRadiant--;
                    radiantCount--;
                } else {
                    bansByDire++;
                    senatorQueue.push('R');
                }
            }
        }

        // Outcome based on which side has senators left
        return radiantCount > 0 ? "Radiant" : "Dire";
    }
};

The problem at hand involves simulating a political scenario in the Dota2 Senate, where each senator belongs to either the "Radiant" or the "Dire" faction. The challenge is to determine which faction ultimately wins the control of the senate based on the banning strategy applied by the senators.

To achieve this, the code makes use of the following methods and data structures:

• Use of a queue to manage the sequence of actions by each senator.
• Tracking the count of senators for each faction as bans are issued.
• Considering each senator's action in sequence and applying a ban strategy based on available bans from the opposing faction.

For senators from the "Dire" side:

• If there are any bans available from the "Dire" quota, the current "Dire" senator gets banned and the dire count decreases.
• If not, a ban is added to the "Radiant" quota for later use and the current senator rejoins the queue.

For senators from the "Radiant" side:

• If there are any bans available from the "Radiant" quota, the current "Radiant" senator gets banned and the radiant count decreases.
• If not, a ban is added to the "Dire" quota for later use and the senator reenters the queue.

The loop continues until one faction no longer has any senators left, indicating that all their senators have been banned. The faction with remaining senators is declared the winner.

At the end of execution, the function returns "Radiant" if any radiant senators are left, otherwise it returns "Dire", indicating which faction has taken control over the Senate. This is determined by checking the counts of senators still available in each faction after the queue processing is completed.

This solution effectively simulates the political dynamics and showcases how the strategy and sequence of banning can impact the outcome in a senate majority control situation.

class Solution {
    public String determineWinner(String senate) {
        int radiant = 0, dire = 0;
        int bansOnDire = 0, bansOnRadiant = 0;
        Queue<Character> senatorsQueue = new LinkedList<>();
        for (char senator : senate.toCharArray()) {
            senatorsQueue.offer(senator);
            if (senator == 'R') radiant++;
            else dire++;
        }

        while (radiant > 0 && dire > 0) {
            char currentSenator = senatorsQueue.poll();
            if (currentSenator == 'D') {
                if (bansOnDire > 0) {
                    bansOnDire--;
                    dire--;
                } else {
                    bansOnRadiant++;
                    senatorsQueue.offer('D');
                }
            } else {
                if (bansOnRadiant > 0) {
                    bansOnRadiant--;
                    radiant--;
                } else {
                    bansOnDire++;
                    senatorsQueue.offer('R');
                }
            }
        }
        return radiant > 0 ? "Radiant" : "Dire";
    }
}

The Java program provided simulates a scenario to determine the winner between two factions, Radiant and Dire, in a fictional environment based on the number of senators each faction can neutralize. The 'determineWinner' method accomplishes the final outcome through efficient use of data structures and control flow.

• Start with initializing counters for Radiant and Dire senators.
• Track the bans each faction can impose on the other using two variables, bansOnDire and bansOnRadiant.
• Convert the input string senate into a queue to efficiently manipulate and cycle through remaining senators.
• Iterate through the queue:
  • If the current senator is from Dire and has bans against them, decrease the Dire count and the ban count.
  • If no bans exist for Dire, increment the ban against Radiant and keep the senator in the rotation by adding them back into the queue.
  • Apply similar logic for Radiant senators.
• Continue processing until one faction has no senators left.
• Determine the winner by checking which faction still has senators remaining.

This solution ensures that even as senators ban opponents efficiently, the computation stays responsive by cycling through options with a queue structure. The logic smartly manages bans and re-entries into the debate which can dynamically change the power balance until a conclusion is reached. The provided code handles the synchronization of bans and senate re-entry smoothly, resulting in a fair outcome of either "Radiant" or "Dire".

char * determineWinner(char * senateStream) {
    
    // Track each party members
    int radiantCount = 0, direCount = 0;

    // Track pending bans for each party
    int pendingBansDire = 0, pendingBansRadiant = 0;

    // Create a circular queue of senators
    int len = strlen(senateStream);
    int queue[len];
    int head = 0, tail = 0;
    for (int i = 0; senateStream[i]; i++) {
        if (senateStream[i] == 'R') radiantCount++;
        else direCount++;
        queue[tail] = senateStream[i];
        tail = (tail + 1) % len;
    }

    // Process senators while both parties have members
    while (radiantCount && direCount) {
        
        // Process the next senator
        char currentSenator = queue[head];
        head = (head + 1) % len;

        // Apply or accrue bans based on senator party
        if (currentSenator == 'D') {
            if (pendingBansDire) {
                pendingBansDire--;
                direCount--;
            } else {
                pendingBansRadiant++;
                queue[tail] = 'D';
                tail = (tail + 1) % len;
            }
        } else {
            if (pendingBansRadiant) {
                pendingBansRadiant--;
                radiantCount--;
            } else {
                pendingBansDire++;
                queue[tail] = 'R';
                tail = (tail + 1) % len;
            }
        }
    }

    // Declare the winner
    return radiantCount ? "Radiant" : "Dire";
}

This C program determines the winner of a simulated game between two teams, Radiant and Dire, using a string representation where each character represents a senator from one of the teams. The string is processed in a circular manner, which means when the end of the string is reached, it continues processing from the beginning. Each senator has the capability to ban a senator from the opposite team from the next round, affecting the overall team counts. The winner is determined by checking which team has senators remaining at the end of the circular processing.

• First, the code initializes counters for both Radiant and Dire team members and variables to keep track of pending bans for each team.
• It populates a circular queue with senators from the input string and initializes the team counts based on the number of 'R' and 'D' characters found.
• In each iteration, the code processes senators from the head of the queue: if a senator can ban an opposing senator (based on pending bans), they do so and decrease the opposing team count; otherwise, the senator applies a ban to the opposing team for the next round.
• The loop continues until one of the teams has no remaining members, at which point the team with remaining senators is declared the winner by returning either "Radiant" or "Dire".

This logic ensures the processing adheres to the rules of the simulation, producing the correct output based on the structure and actions within the senate string.

var calculateVictory = function(senate) {
    let countR = 0, countD = 0;
    let bansD = 0, bansR = 0;
    let queue = [];
    for (let i = 0; i < senate.length; i++) {
        queue.push(senate[i]);
        if (senate[i] === 'R') countR++;
        else countD++;
    }

    while (countR > 0 && countD > 0) {
        let current = queue.shift();
        if (current === 'D') {
            if (bansD > 0) {
                bansD--;
                countD--;
            } else {
                bansR++;
                queue.push('D');
            }
        } else {
            if (bansR > 0) {
                bansR--;
                countR--;
            } else {
                bansD++;
                queue.push('R');
            }
        }
    }

    return countR > 0 ? "Radiant" : "Dire";
};

Here's a comprehensive solution summary for the "Dota2 Senate" problem implemented in JavaScript. This function simulates the decision-making process in the Dota2 game using a queue-based approach to determine which faction, Radiant or Dire, ultimately wins.

• Initialize counters for Radiants (countR) and Dires (countD) to count the members of each faction.
• Create variables bansD and bansR to keep track of the bans each faction can impose.
• Populate a queue with all the members from the input string senate and simultaneously update the counts for each faction.

Proceed with a loop that continues as long as both factions have members:

1. Dequeue an element, representing the faction member taking an action.
2. If the current member is from the Dire ('D'):
   • Check if any bans are left for Dire. If yes, decrease the Dire's ban count and the Dire's count (a member gets banned).
   • If no bans are left to use, increment the ban counter for Radiant and re-enqueue the Dire member to act in future rounds.
3. If the current member is from the Radiant ('R'):
   • Similarly, check and use any bans available against Radiant members or increment Dire's ban counter and re-enqueue the Radiant member if no bans are available.

At the end of the while loop, check which faction's count is greater:

• Return "Radiant" if Radiant members (countR) remain, otherwise return "Dire" indicating that the Dire faction has prevailed.

This logic effectively models the ban strategy in a loop until one side loses all its members, mirroring the strategic component of the Dota2 Senate gameplay.

class Solution:
    def determineWinner(self, senators: str) -> str:

        # Count of each party's senators
        radiants = senators.count('R')
        dires = senators.count('D')

        # Bans that can be imposed by one party onto another
        bans_on_dires = 0
        bans_on_radiants = 0

        # Processing the senators in sequence
        sequenced_senators = deque(senators)

        # Process until one party runs out of its senators
        while radiants and dires:

            # Consider the front senator in the sequence
            current_senator = sequenced_senators.popleft()

            # Apply the logic based on the senator's party
            if current_senator == 'D':
                if bans_on_dires:
                    bans_on_dires -= 1
                    dires -= 1
                else:
                    bans_on_radiants += 1
                    sequenced_senators.append('D')
            elif current_senator == 'R':
                if bans_on_radiants:
                    bans_on_radiants -= 1
                    radiants -= 1
                else:
                    bans_on_dires += 1
                    sequenced_senators.append('R')

        # Determine the winning party
        return 'Radiant' if radiants else 'Dire'

This Python program defines a class Solution with a method determineWinner to solve the Dota2 Senate problem. The method takes a string of senators (represented by 'R' for Radiants and 'D' for Dires) and returns the winning party.

Here's an outline of the solution steps:

• First, the program counts how many senators belong to each party (radiants for 'R' and dires for 'D').
• It initializes counters for the bans each party can impose on the other (bans_on_dires and bans_on_radiants).
• The senators string is converted to a dequeue to optimize the removal and addition of elements during processing.
• A while loop runs as long as both parties have members left.
  • The loop dequeues a senator and checks its party.
  • If the dequeued senator is from the Dire party ('D'):
    • If Dires have bans left, one is decremented, and a Dire senator is removed.
    • If no bans are left on Dires, they impose a ban on Radiants and the Dire senator is re-enqueued.
  • The process is similar when the dequeued senator is a Radiant ('R').
• The loop continues until one of the parties loses all members.
• The winner is determined based on which party has senators remaining.

Outputs whether "Radiant" or "Dire" wins depending on the remaining senators after all possible bans have been enacted.