Question

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. 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].

Difficulty:Medium

Category:Binary-Search

Analyze

Understand Question

一次选举中,在time[i]的时候会投票给person[i],然后求在t时刻得票最多的候选人. 如果存在票数相等的情况,就选择最新获得投票的候选人.

Solution

// Time complexity: Constructor O(n), Query: O(logn)
// Space complexity: O(n)
class TopVotedCandidate {
 public:
  // the i-th person at the times[i] time vote the person[i]
  TopVotedCandidate(vector<int> persons, vector<int> times) {
    vector<int> votes(persons.size() + 1, 0);
    int last_lead = persons.front();
    for (int i = 0; i < persons.size(); ++i) {
      if (++votes[persons[i]] >= votes[last_lead]) last_lead = persons[i];
      // The i-th vote at time times[i]
      lead_[times[i]] = last_lead;
    }
  }

  int q(int t) { return prev(lead_.upper_bound(t))->second; }

 private:
  // time -> leader (IN this time, who lead this vote)
  map<int, int> lead_;
};