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 3, the votes are [0], and 0 is leading. At time 12, the votes are [0,1,1], and 1 is leading. 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 <= persons.length = times.length <= 50000 <= persons[i] <= persons.lengthtimes is a strictly increasing array with all elements in [0, 10^9].TopVotedCandidate.q is called at most 10000 times per test case.TopVotedCandidate.q(int t) is always called with t >= times[0].use std::collections::HashMap;
struct TopVotedCandidate {
times: Vec<i32>,
leaders: 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);
let mut leaders = vec![];
for i in 0..n {
let count = hm.entry(persons[i]).or_default();
*count += 1;
if *count >= leader.0 {
leader = (*count, persons[i]);
}
leaders.push(leader.1);
}
TopVotedCandidate { times, leaders }
}
fn q(&self, t: i32) -> i32 {
let i = match self.times.binary_search(&t) {
Ok(i) => i,
Err(i) => i - 1,
};
self.leaders[i]
}
}
#[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);
}