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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]],,,,,,]
Output: [null,0,1,1,0,0,1]
Explanation:
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.

## Solution 1.

• class TopVotedCandidate {
private int[] times;
private int[] wins;

public TopVotedCandidate(int[] persons, int[] times) {
int n = persons.length;
int mx = 0, cur = 0;
this.times = times;
wins = new int[n];
int[] counter = new int[n];
for (int i = 0; i < n; ++i) {
int p = persons[i];
if (++counter[p] >= mx) {
mx = counter[p];
cur = p;
}
wins[i] = cur;
}
}

public int q(int t) {
int left = 0, right = wins.length - 1;
while (left < right) {
int mid = (left + right + 1) >>> 1;
if (times[mid] <= t) {
left = mid;
} else {
right = mid - 1;
}
}
return wins[left];
}
}

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

• class TopVotedCandidate {
public:
vector<int> times;
vector<int> wins;

TopVotedCandidate(vector<int>& persons, vector<int>& times) {
int n = persons.size();
wins.resize(n);
int mx = 0, cur = 0;
this->times = times;
vector<int> counter(n);
for (int i = 0; i < n; ++i) {
int p = persons[i];
if (++counter[p] >= mx) {
mx = counter[p];
cur = p;
}
wins[i] = cur;
}
}

int q(int t) {
int left = 0, right = wins.size() - 1;
while (left < right) {
int mid = left + right + 1 >> 1;
if (times[mid] <= t)
left = mid;
else
right = mid - 1;
}
return wins[left];
}
};

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

• class TopVotedCandidate:
def __init__(self, persons: List[int], times: List[int]):
mx = cur = 0
counter = Counter()
self.times = times
self.wins = []
for i, p in enumerate(persons):
counter[p] += 1
if counter[p] >= mx:
mx, cur = counter[p], p
self.wins.append(cur)

def q(self, t: int) -> int:
left, right = 0, len(self.wins) - 1
while left < right:
mid = (left + right + 1) >> 1
if self.times[mid] <= t:
left = mid
else:
right = mid - 1
return self.wins[left]

# Your TopVotedCandidate object will be instantiated and called as such:
# obj = TopVotedCandidate(persons, times)
# param_1 = obj.q(t)


• type TopVotedCandidate struct {
times []int
wins  []int
}

func Constructor(persons []int, times []int) TopVotedCandidate {
mx, cur, n := 0, 0, len(persons)
counter := make([]int, n)
wins := make([]int, n)
for i, p := range persons {
counter[p]++
if counter[p] >= mx {
mx = counter[p]
cur = p
}
wins[i] = cur
}
}

func (this *TopVotedCandidate) Q(t int) int {
left, right := 0, len(this.wins)-1
for left < right {
mid := (left + right + 1) >> 1
if this.times[mid] <= t {
left = mid
} else {
right = mid - 1
}
}
return this.wins[left]
}

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