911 - Online Election.

Medium

You are given two integer arrays persons and times. In an election, the <code>i^th</code> 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:

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

• <code>0 <= timesi<= 10⁹</code>

• times is sorted in a strictly increasing order.

• <code>times0<= t <= 10⁹</code>

• At most <code>10⁴</code> calls will be made to q.