Question 28 of 75 - LeetCode Weekly Challenge

LeetCode Challenge #649. Dota2 Senate

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.

 

Constraints:

  • n == senate.length
  • 1 <= n <= 104
  • senate[i] is either 'R' or 'D'.
Video Solution
C++ Solution
				
					class Solution {
public:
    string predictPartyVictory(string senate) {
        queue<int> rad,dir;
        int n=senate.length();

        for(int i=0; i<senate.size();i++)
        {
            if(senate[i]=='R') rad.push(i);
            else dir.push(i);
        }

        while(!rad.empty() and !dir.empty())
        {
            if(rad.front()>dir.front())dir.push(n++);
            else rad.push(n++);
            rad.pop();
            dir.pop();
        }

        if(rad.empty())return "Dire";
        return "Radiant";
    }
};
				
			
Code Explanation

This problem involves predicting the winner of a vote in the Senate between two parties, Radiant (R) and Dire (D). Each senator can cast a ban vote against the senator immediately following them in the list. 

Here’s the step-by-step explanation of the solution:

  1. We will initialise two queues, ‘rad’ and ‘dir’ so that the last one will fight at last and the first one will fight at first ; (FIFO) 
  2. Int ‘n’ is taken for storing the size and tracking the length of the Senate. Now, we’ll use a loop that continues as long as both the Radiant and Dire queues are not empty. This means the fight will continue until and unless there are elements present. 
  3. If the front of Radiant is greater than Dire then Dire will get a chance and a new Dire senator is pushed to the back of the queue with an increased index (n++). 
  • if the Dire Senator is greater, it means that Radiant Senator will get a chance and Radiant is pushed to the back of the queue with an increased index (n++). 
  1. Now we will pop both of them and then remove them from both queues. 
  2. For the Result Determination- after the voting process, if the Radiant queue is empty, the winner will be Dire so, the function returns “Dire” otherwise, it returns “radian

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