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. 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].

Solution: HashTable + Binary Search

Compute the leads for each t in times using a hash table.

binary search the upper bound of t, and return the lead of previous entry.

Time complexity: Constructor O(n), Query: O(logn)

Space complexity: O(n)

// Author: Huahua
class TopVotedCandidate {
public:
  TopVotedCandidate(vector<int> persons, vector<int> times) {
    vector<int> votes(persons.size() + 1);
    int last_lead = persons.front();
    for (int i = 0; i < persons.size(); ++i) {
      if (++votes[persons[i]] >= votes[last_lead])
        last_lead = persons[i];      
      leads_[times[i]] = last_lead;      
    }
  }

  int q(int t) {
    return prev(leads_.upper_bound(t))->second;
  }
private:
  map<int, int> leads_; // time -> lead
};