649. Dota2 Senate

[songyy5517]

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.

Intuition The problem is basically to find the part that finally exists after the voting process. From the description, we know that a unbanned senator can ban a subsequent senator of the other party. Therefore, we can define two queues to store all the indices of senators of each party respectively. Then simulate the voting process with these two queues. Specifically, when the two queues are not empty, we compare the top element of each queue, the smaller index would vote first. In this way, we remove the top element of the other queue, and add the smaller index plus the size of the whole senators in the same queue. Repeat the above steps until there is one empty queue. Therefore, the winner is the non-empty party.

[songyy5517]

Approach: Queue

1. Exception handling;
2. Define two queues to store the index of each senator of every party;
3. Loop through the array and store the index of each senator to its corresponding queue;
4. While the two queues are not empty: （1）Compare the top elements of two queues, the one with smaller index would vote first; （2）Remove the top element of greater index from its queue; （3）Add the smaller element plus the total number of all the senatores to its queue;
5. Return the party whose queue is non-empty.

Complexity Analysis

• Time complexity: $O(N)$ , looping through the whole string takes $O(N)$ time. In the worst case, we need to remove $N - 1$ elements from two queues, which also takes $O(N)$ time.
• Space complexity: $O(N)$ , additionally define two queues to store the index of all the senators, occupying $O(N)$ space.

[songyy5517]

class Solution {
    public String predictPartyVictory(String senate) {
        // Intuition: Queue & Greedy
        // 1. Exception handling
        if (senate == null || senate.length() == 0)
            return "";

        // 2. Define two queues to store the index of radient and dire
        Queue<Integer> radiant = new LinkedList();
        Queue<Integer> dire = new LinkedList();

        // 3. Add the index of each senator to their queues
        for (int i = 0; i < senate.length(); i++){
            if (senate.charAt(i) == 'R')
                radiant.add(i);
            else 
                dire.add(i);
        }

        // 3. Simulate the voting process
        while (!radiant.isEmpty() && !dire.isEmpty()){
            int r = radiant.remove(), d = dire.remove();
            if (r > d)
                dire.add(d + senate.length());
            else 
                radiant.add(r + senate.length());

        }

        return dire.isEmpty() ? "Radiant" : "Dire";
    }
}

[songyy5517]

2024/5/25