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[leetcode] problem 911 – online election

admin 11月 13, 2020 0

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

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].

Code

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private int[] leads;
private int[] times;

public (int[] persons, int[] times) {
    int n = persons.length;
    int lead = 0;
    int[] votes = new int[n + 1];
    leads = new int[n];
    this.times = times;

    for (int i = 0; i < n; i++) {
        if (++votes[persons[i]] >= votes[lead])
            lead = persons[i];

        leads[i] = lead;
    }
}

public int q(int t) {
    int idx = Arrays.binarySearch(times, t);
    return idx >= 0 ? leads[idx] : leads[-idx-2];
}
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