Recursive Voting - Facebook Top Interview Questions


Problem Statement :


You are given a list of integers votes representing election votes for candidates a and b. 

Each element in votes represents a vote to one of the two candidates:

If votes[i] < 0, then it's a vote for a.

If votes[i] ≥ n, then it's a vote for b.

Otherwise, the vote at index i made the same vote as votes[votes[i]].

Return the number of votes that candidate a received. You can assume all votes are eventually made to 
either a or b.

Constraints

0 ≤ n ≤ 100,000 where n is the length of votes

Example 1

Input

votes = [2, -1, 5, 1, 3]

Output

3

Explanation

We can see that votes[1], votes[3], and votes[4] were made to candidate a.


Solution :



title-img 




                        Solution in C++ :

int find(int x, int n, vector<int>& votes) {
    if (votes[x] < 0 || votes[x] >= n) return votes[x];
    return votes[x] = find(votes[x], n, votes);
}

int solve(vector<int>& votes) {
    int a = 0;
    int n = votes.size();

    for (int i = 0; i < n; ++i) {
        if (votes[i] >= 0 && votes[i] < n) votes[i] = find(i, n, votes);
        if (votes[i] < 0) ++a;
    }

    return a;
}
                    


                        Solution in Java :

import java.util.*;

class Solution {
    public int solve(int[] votes) {
        int N = votes.length;
        DisjointSetUnion dsu = new DisjointSetUnion(votes);
        for (int i = 0; i < N; i++) {
            if (0 <= votes[i] && votes[i] < N) {
                dsu.connect(i, votes[i]);
            }
        }
        int ans = 0;
        for (int r = 0; r < N; r++) {
            if (r == dsu.root(r) && dsu.votes[r] == -1)
                ans += dsu.weight[r];
        }
        return ans;
    }

    static class DisjointSetUnion {
        public int[] parent;
        public int[] weight;
        public int[] votes; //-1 is a vote for a, 1 is a vote for b
        public int count;

        public DisjointSetUnion(int[] v) {
            int N = v.length;
            count = N;
            parent = new int[N];
            votes = new int[N];
            for (int i = 0; i < N; i++) {
                parent[i] = i;
                if (v[i] < 0)
                    votes[i] = -1;
                else if (v[i] >= N)
                    votes[i] = 1;
            }
            weight = new int[N];
            Arrays.fill(weight, 1);
        }

        //"find"
        public int root(int p) {
            if (p == parent[p])
                return p;
            return parent[p] = root(parent[p]);
        }

        //"union"
        public void connect(int p, int q) {
            p = root(p);
            q = root(q);
            if (p == q)
                return;
            if (weight[p] < weight[q]) {
                parent[p] = q;
                weight[q] += weight[p];
                if (votes[p] != 0)
                    votes[q] = votes[p];
            } else {
                parent[q] = p;
                weight[p] += weight[q];
                if (votes[q] != 0)
                    votes[p] = votes[q];
            }
            count--;
        }

        public boolean connected(int p, int q) {
            return root(p) == root(q);
        }
    }
}
                    


                        Solution in Python : 
                            
class Solution:
    def solve(self, votes):
        def find(votes, x):
            if x < 0 or x >= len(votes):
                return x
            votes[x] = find(votes, votes[x])
            return votes[x]

        return sum([find(votes, i) < 0 for i in range(len(votes))])


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