LeetCode 0911. Online Election Solution in Java, C++, Python & More | Explanation + Code

Description

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.

Solutions

Solution 1: Binary Search

We can record the winner at each moment during initialization, and then use binary search to find the largest moment less than or equal to t during the query, and return the winner at that moment.

During initialization, we use a counter cnt to record the votes of each candidate, and a variable cur to record the current leading candidate. Then we traverse each moment, update cnt and cur, and record the winner at each moment.

During the query, we use binary search to find the largest moment less than or equal to t, and return the winner at that moment.

In terms of time complexity, during initialization, we need O(n) time, and during the query, we need O(log n) time. The space complexity is O(n).

PythonJavaC++GoTypeScript

class TopVotedCandidate: def __init__(self, persons: List[int], times: List[int]): cnt = Counter() self.times = times self.wins = [] cur = 0 for p in persons: cnt[p] += 1 if cnt[cur] <= cnt[p]: cur = p self.wins.append(cur) def q(self, t: int) -> int: i = bisect_right(self.times, t) - 1 return self.wins[i] # Your TopVotedCandidate object will be instantiated and called as such: # obj = TopVotedCandidate(persons, times) # param_1 = obj.q(t)(code-box)

class TopVotedCandidate { private int[] times; private int[] wins; public TopVotedCandidate(int[] persons, int[] times) { int n = persons.length; wins = new int[n]; this.times = times; int[] cnt = new int[n]; int cur = 0; for (int i = 0; i < n; ++i) { int p = persons[i]; ++cnt[p]; if (cnt[cur] <= cnt[p]) { cur = p; } wins[i] = cur; } } public int q(int t) { int i = Arrays.binarySearch(times, t + 1); i = i < 0 ? -i - 2 : i - 1; return wins[i]; } } /** * Your TopVotedCandidate object will be instantiated and called as such: * TopVotedCandidate obj = new TopVotedCandidate(persons, times); * int param_1 = obj.q(t); */(code-box)

class TopVotedCandidate { public: TopVotedCandidate(vector<int>& persons, vector<int>& times) { int n = persons.size(); this->times = times; wins.resize(n); vector<int> cnt(n); int cur = 0; for (int i = 0; i < n; ++i) { int p = persons[i]; ++cnt[p]; if (cnt[cur] <= cnt[p]) { cur = p; } wins[i] = cur; } } int q(int t) { int i = upper_bound(times.begin(), times.end(), t) - times.begin() - 1; return wins[i]; } private: vector<int> times; vector<int> wins; }; /** * Your TopVotedCandidate object will be instantiated and called as such: * TopVotedCandidate* obj = new TopVotedCandidate(persons, times); * int param_1 = obj->q(t); */(code-box)

type TopVotedCandidate struct { times []int wins []int } func Constructor(persons []int, times []int) TopVotedCandidate { n := len(persons) wins := make([]int, n) cnt := make([]int, n) cur := 0 for i, p := range persons { cnt[p]++ if cnt[cur] <= cnt[p] { cur = p } wins[i] = cur } return TopVotedCandidate{times, wins} } func (this *TopVotedCandidate) Q(t int) int { i := sort.SearchInts(this.times, t+1) - 1 return this.wins[i] } /** * Your TopVotedCandidate object will be instantiated and called as such: * obj := Constructor(persons, times); * param_1 := obj.Q(t); */(code-box)

class TopVotedCandidate { private times: number[]; private wins: number[]; constructor(persons: number[], times: number[]) { const n = persons.length; this.times = times; this.wins = new Array<number>(n).fill(0); const cnt: Array<number> = new Array<number>(n).fill(0); let cur = 0; for (let i = 0; i < n; ++i) { const p = persons[i]; cnt[p]++; if (cnt[cur] <= cnt[p]) { cur = p; } this.wins[i] = cur; } } q(t: number): number { const search = (t: number): number => { let l = 0, r = this.times.length; while (l < r) { const mid = (l + r) >> 1; if (this.times[mid] > t) { r = mid; } else { l = mid + 1; } } return l; }; const i = search(t) - 1; return this.wins[i]; } } /** * Your TopVotedCandidate object will be instantiated and called as such: * var obj = new TopVotedCandidate(persons, times) * var param_1 = obj.q(t) */(code-box)