911. Online Election

Description

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].

Problem URL

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Solution

构建一个TopVotedCandidate类，给一个person数组和一个times数组，表示第ith投票投给的人和时间，然后query问某个时间那个人领先。

Approach1：Using TreeMap to store the map from time to current highest votes person. Then in construction, we use a hash map to count each person’s votes. For each person in persons, update his votes then if the votes is bigger than highest count, updata highest count and highest person. After that store them in treemap.

In query, return the floorKey of t, it is the key before value t.

Code

class TopVotedCandidate {
    //a map from time to current highest votes person
    private TreeMap<Integer, Integer> map = new TreeMap<>();
    public TopVotedCandidate(int[] persons, int[] times) {
        //count each person's votes
        HashMap<Integer, Integer> count = new HashMap<>();
        int highestCount = 0, highestPerson = 0;
        for (int i = 0; i < persons.length; i++){
            int person = persons[i];
            count.put(person, count.getOrDefault(person, 0) + 1);
            
            if (count.get(person) >= highestCount){
                highestCount = count.get(person);
                highestPerson = person;
            }
            map.put(times[i], highestPerson);
        }
    }
    
    public int q(int t) {
        return map.get(map.floorKey(t));
    }
}

/**
 * Your TopVotedCandidate object will be instantiated and called as such:
 * TopVotedCandidate obj = new TopVotedCandidate(persons, times);
 * int param_1 = obj.q(t);
 */

construct: O(NlogN)
query: O(logN)

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Review

Approach 2 Binary search
Count the top candidate and store it in a max[], then perform binary search using Arrays.binarySearch();

class TopVotedCandidate {
    private int[] max;
    private int[] times;
    public TopVotedCandidate(int[] persons, int[] times) {
        this.times = times;
        max = new int[persons.length];
        HashMap<Integer, Integer> count = new HashMap<>();
        int highestCount = 0, highestPerson = 0;
        
        for (int i = 0; i < persons.length; i++){
            int person = persons[i];
            count.put(person, count.getOrDefault(person, 0) + 1);
            
            if (count.get(person) >= highestCount){
                highestCount = count.get(person);
                highestPerson = person;
            }
            max[i] = highestPerson;
        }
    }
    
    public int q(int t) {
        int timePoint = Arrays.binarySearch(times, t);
        return max[timePoint < 0 ? -timePoint - 2: timePoint];
    }
}

/**
 * Your TopVotedCandidate object will be instantiated and called as such:
 * TopVotedCandidate obj = new TopVotedCandidate(persons, times);
 * int param_1 = obj.q(t);
 */

construct: O(N)
query: O(logN)