Formatted question description: https://leetcode.ca/all/911.html

911. Online Election (Medium)

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].

Solution 1.

// OJ: https://leetcode.com/problems/online-election/

// Time:
//     TopVotedCandidate: O(N)
//     q: O(logN)
// Space: O(N)
class TopVotedCandidate {
private:
    vector<int> winners;
    vector<int> times;
public:
    TopVotedCandidate(vector<int> persons, vector<int> times): times(times) {
        unordered_map<int, int> m;
        int winner = 0;
        for (int p : persons) {
            m[p]++;
            if (m[p] >= m[winner]) winner = p;
            winners.push_back(winner);
        }
    }
    
    int q(int t) {
        int n = upper_bound(times.begin(), times.end(), t) - times.begin();
        return winners[n - 1];
    }
};

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