911. Online Election LeetCode Solution | Easy Approach

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Online Election You are given two integer arrays persons and times. In an election, the ith 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
  • 0 <= times[i] <= 109
  • times is sorted in a strictly increasing order.
  • times[0] <= t <= 109
  • At most 104 calls will be made to q.

Online Election Solutions

Time: O(n)
Space: O(n)

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