# ---
# title: 911. Online Election
# id: problem911
# author: Tian Jun
# date: 2020-10-31
# difficulty: Medium
# categories: Binary Search
# link: <https://leetcode.com/problems/online-election/description/>
# hidden: true
# ---
# 
# 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]`.
# 
# 
## @lc code=start
using LeetCode

## add your code here:
## @lc code=end
