# Online Election Problem

## Description

LeetCode Problem 911.

You are given two integer arrays persons and times. In an election, the i^th vote was cast for persons[i] at time times[i].

For each query at a time t, find 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.

Implement the TopVotedCandidate class:

• TopVotedCandidate(int[] persons, int[] times) Initializes the object with the persons and times arrays.
• int q(int t) Returns the number of the person that was leading the election at time t according to the mentioned rules.

Example 1:

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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
TopVotedCandidate topVotedCandidate = new TopVotedCandidate([0, 1, 1, 0, 0, 1, 0], [0, 5, 10, 15, 20, 25, 30]);
topVotedCandidate.q(3); // return 0, At time 3, the votes are [0], and 0 is leading.
topVotedCandidate.q(12); // return 1, At time 12, the votes are [0,1,1], and 1 is leading.
topVotedCandidate.q(25); // return 1, At time 25, the votes are [0,1,1,0,0,1], and 1 is leading (as ties go to the most recent vote.)
topVotedCandidate.q(15); // return 0
topVotedCandidate.q(24); // return 0
topVotedCandidate.q(8); // return 1
``````

Constraints:

• 1 <= persons.length <= 5000
• times.length == persons.length
• 0 <= persons[i] < persons.length
• 0 <= times[i] <= 10^9
• times is sorted in a strictly increasing order.
• times[0] <= t <= 10^9
• At most 10^4 calls will be made to q.

## Sample C++ Code

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class TopVotedCandidate {
public:
TopVotedCandidate(vector<int>& persons, vector<int>& times) : times(times) {
unordered_map<int, int> freq;
int pp = 0;
for (auto& p : persons) {
++freq[p];
if (freq[p] >= freq[pp]) pp = p;
}
}

int q(int t) {
int k = upper_bound(times.begin(), times.end(), t) - times.begin();
}
};
/**
* Your TopVotedCandidate object will be instantiated and called as such:
* TopVotedCandidate* obj = new TopVotedCandidate(persons, times);
* int param_1 = obj->q(t);
*/
``````