## 911. Online Election

In an election, the `i`-th vote was cast for `persons[i]` at time `times[i]`.

Now, we would like to implement the following query function: `TopVotedCandidate.q(int t)` will return the number of the person that was leading the election at time `t`.

Votes cast at time `t` will count towards our query.  In the case of a tie, the most recent vote (among tied candidates) wins.

Example 1:

```Input: ["TopVotedCandidate","q","q","q","q","q","q"], [[[0,1,1,0,0,1,0],[0,5,10,15,20,25,30]],[3],[12],[25],[15],[24],[8]]
Output: [null,0,1,1,0,0,1]
Explanation:
At time 25, the votes are [0,1,1,0,0,1], and 1 is leading (as ties go to the most recent vote.)
This continues for 3 more queries at time 15, 24, and 8.
```

Note:

1. `1 <= persons.length = times.length <= 5000`
2. `0 <= persons[i] <= persons.length`
3. `times` is a strictly increasing array with all elements in `[0, 10^9]`.
4. `TopVotedCandidate.q` is called at most `10000` times per test case.
5. `TopVotedCandidate.q(int t)` is always called with `t >= times[0]`.

## Rust Solution

``````use std::collections::HashMap;

struct TopVotedCandidate {
times: Vec<i32>,
}

impl TopVotedCandidate {
fn new(persons: Vec<i32>, times: Vec<i32>) -> Self {
let n = persons.len();
let mut hm: HashMap<i32, usize> = HashMap::new();
let mut leader: (usize, i32) = (0, 0);
for i in 0..n {
let count = hm.entry(persons[i]).or_default();
*count += 1;
}
}
}

fn q(&self, t: i32) -> i32 {
let i = match self.times.binary_search(&t) {
Ok(i) => i,
Err(i) => i - 1,
};
}
}

#[test]
fn test() {
let persons = vec![0, 1, 1, 0, 0, 1, 0];
let times = vec![0, 5, 10, 15, 20, 25, 30];
let tvc = TopVotedCandidate::new(persons, times);
assert_eq!(tvc.q(3), 0);
assert_eq!(tvc.q(12), 1);
assert_eq!(tvc.q(25), 1);
assert_eq!(tvc.q(15), 0);
assert_eq!(tvc.q(24), 0);
assert_eq!(tvc.q(8), 1);
}
``````

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