# 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.

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

### Solution :

```                        ```Solution in C++ :

int find(int x, int n, vector<int>& votes) {
}

int a = 0;

for (int i = 0; i < n; ++i) {
}

return a;
}```
```

```                        ```Solution in Java :

import java.util.*;

class Solution {
for (int i = 0; i < N; 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];
for (int i = 0; i < N; i++) {
parent[i] = i;
if (v[i] < 0)
else if (v[i] >= N)
}
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];
} else {
parent[q] = p;
weight[p] += weight[q];
}
count--;
}

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

```                        ```Solution in Python :

class Solution:
if x < 0 or x >= len(votes):
return x

```

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