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 <= 5000 0 <= persons[i] <= persons.length times 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].
// 320ms, 95%
class TopVotedCandidate {
vector<int> votes;
map<int,int> maxVotes;
public:
TopVotedCandidate(vector<int>& persons, vector<int>& times) {
votes.resize(persons.size()+1, 0);
int maxV = 0, maxP = -1;
for (int i = 0; i < persons.size(); i++) {
auto pi = persons[i];
votes[pi]++;
if (votes[pi] >= maxV) {
maxV = votes[pi];
maxP = pi;
maxVotes[times[i]] = maxP;
}
}
}
int q(int t) {
if (maxVotes.empty())
return 0;
auto it = maxVotes.lower_bound(t);
if (it == maxVotes.end() ||
it->first > t) {
if (it != maxVotes.begin())
--it;
}
return it->second;
}
};
/**
* Your TopVotedCandidate object will be instantiated and called as such:
* TopVotedCandidate* obj = new TopVotedCandidate(persons, times);
* int param_1 = obj->q(t);
*/