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.
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.
1 <= persons.length = times.length <= 5000
0 <= persons[i] <= persons.length
times
is a strictly increasing array with all elements in[0, 10^9]
.TopVotedCandidate.q
is called at most10000
times per test case.TopVotedCandidate.q(int t)
is always called witht >= times[0]
.
use std::collections::HashMap;
struct TopVotedCandidate {
winners: Vec<(i32, i32)>,
}
/**
* `&self` means the method takes an immutable reference.
* If you need a mutable reference, change it to `&mut self` instead.
*/
impl TopVotedCandidate {
fn new(persons: Vec<i32>, times: Vec<i32>) -> Self {
let mut counter = HashMap::new();
let mut max = 0;
let mut winner = 0;
let mut winners = vec![];
for (p, t) in persons.into_iter().zip(times.into_iter()) {
let c = counter.entry(p).or_insert(0);
*c += 1;
if *c >= max {
max = *c;
winner = p;
}
winners.push((t, winner));
}
Self { winners }
}
fn q(&self, t: i32) -> i32 {
match self.winners.binary_search_by_key(&t, |&(time, _)| time) {
Ok(i) => self.winners[i].1,
Err(i) => self.winners[i - 1].1,
}
}
}
/**
* Your TopVotedCandidate object will be instantiated and called as such:
* let obj = TopVotedCandidate::new(persons, times);
* let ret_1: i32 = obj.q(t);
*/