LeetCode #911 — MEDIUM

Online Election

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

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:

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.

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Hash Map · Binary Search · Design

Example 1

["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]]

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #911: Online Election
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);
 */

// Accepted solution for LeetCode #911: Online Election
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);
 */

# Accepted solution for LeetCode #911: Online Election
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)

// Accepted solution for LeetCode #911: Online Election
use std::collections::HashMap;

struct TopVotedCandidate {
    times: Vec<i32>,
    leaders: Vec<i32>,
}

impl TopVotedCandidate {
    fn new(persons: Vec<i32>, times: Vec<i32>) -> Self {
        let n = persons.len();
        let mut hm: HashMap<i32, usize> = HashMap::new();
        let mut leader: (usize, i32) = (0, 0);
        let mut leaders = vec![];
        for i in 0..n {
            let count = hm.entry(persons[i]).or_default();
            *count += 1;
            if *count >= leader.0 {
                leader = (*count, persons[i]);
            }
            leaders.push(leader.1);
        }
        TopVotedCandidate { times, leaders }
    }

    fn q(&self, t: i32) -> i32 {
        let i = match self.times.binary_search(&t) {
            Ok(i) => i,
            Err(i) => i - 1,
        };
        self.leaders[i]
    }
}

#[test]
fn test() {
    let persons = vec![0, 1, 1, 0, 0, 1, 0];
    let times = vec![0, 5, 10, 15, 20, 25, 30];
    let tvc = TopVotedCandidate::new(persons, times);
    assert_eq!(tvc.q(3), 0);
    assert_eq!(tvc.q(12), 1);
    assert_eq!(tvc.q(25), 1);
    assert_eq!(tvc.q(15), 0);
    assert_eq!(tvc.q(24), 0);
    assert_eq!(tvc.q(8), 1);
}

// Accepted solution for LeetCode #911: Online Election
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)
 */

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(log n)

Space

O(1)

Approach Breakdown

LINEAR SCAN

O(n) time

O(1) space

Check every element from left to right until we find the target or exhaust the array. Each comparison is O(1), and we may visit all n elements, giving O(n). No extra space needed.

BINARY SEARCH

O(log n) time

O(1) space

Each comparison eliminates half the remaining search space. After k comparisons, the space is n/2ᵏ. We stop when the space is 1, so k = log₂ n. No extra memory needed — just two pointers (lo, hi).

Shortcut: Halving the input each step → O(log n). Works on any monotonic condition, not just sorted arrays.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.

Boundary update without `+1` / `-1`

Wrong move: Setting `lo = mid` or `hi = mid` can stall and create an infinite loop.

Usually fails on: Two-element ranges never converge.

Fix: Use `lo = mid + 1` or `hi = mid - 1` where appropriate.