# 911 Online Election

## 911. [Online Election](https://leetcode.com/problems/online-election/description/)

## 1. 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]`.

## 2. Implementation

**(1) HashMap + Binary Search**

思路: 对于给定times数组，我们需要做的就是对于times数组里每个对应的时间点，找到对应的winner是谁，这个可以直接用hashmap完成。而对于调用q()这个函数时，我们需要知道的是，对于任意的时刻t，我们要根据我们已知的times数组上的时间，**用二分法找到不大于t的最大的时间。**

```java
class TopVotedCandidate {
    Map<Integer, Integer> map = new HashMap(); // key： time, value : winner
    int[] times;

    public TopVotedCandidate(int[] persons, int[] times) {
        map = new HashMap();
        this.times = times;

        int[] count = new int[persons.length];
        int winner = -1;

        for (int i = 0; i < times.length; i++) {
            ++count[persons[i]];

            if (map.isEmpty() || count[winner] <= count[persons[i]]) {
                winner = persons[i];
            }
            map.put(times[i], winner);
        }
    }

    public int q(int t) {
        int index = binarySearch(times, t);
        return map.get(times[index]);
    }

    public int binarySearch(int[] array, int key) {
        int start = 0, end = array.length - 1;

        while (start + 1 < end) {
            int mid = start + (end - start) / 2;

            if (array[mid] > key) {
                end = mid - 1;
            }
            else {
                start = mid;
            }
        }
        return array[end] <= key ? end : start;
    }
}

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

## 3. Time & Space Complexity

**HashMap + Binary Search:** 时间复杂度: Constructor O(n), q(): O(logn), n是输入数组persons的长度，即人数。空间复杂度: O(n)