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 :
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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