LeetCode #3508 — MEDIUM

Implement Router

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

The Problem

Problem Statement

Design a data structure that can efficiently manage data packets in a network router. Each data packet consists of the following attributes:

source: A unique identifier for the machine that generated the packet.
destination: A unique identifier for the target machine.
timestamp: The time at which the packet arrived at the router.

Implement the Router class:

Router(int memoryLimit): Initializes the Router object with a fixed memory limit.

memoryLimit is the maximum number of packets the router can store at any given time.
If adding a new packet would exceed this limit, the oldest packet must be removed to free up space.

bool addPacket(int source, int destination, int timestamp): Adds a packet with the given attributes to the router.

A packet is considered a duplicate if another packet with the same source, destination, and timestamp already exists in the router.
Return true if the packet is successfully added (i.e., it is not a duplicate); otherwise return false.

int[] forwardPacket(): Forwards the next packet in FIFO (First In First Out) order.

Remove the packet from storage.
Return the packet as an array [source, destination, timestamp].
If there are no packets to forward, return an empty array.

int getCount(int destination, int startTime, int endTime):

Returns the number of packets currently stored in the router (i.e., not yet forwarded) that have the specified destination and have timestamps in the inclusive range [startTime, endTime].

Note that queries for addPacket will be made in non-decreasing order of timestamp.

Example 1:

Input:
["Router", "addPacket", "addPacket", "addPacket", "addPacket", "addPacket", "forwardPacket", "addPacket", "getCount"]
[[3], [1, 4, 90], [2, 5, 90], [1, 4, 90], [3, 5, 95], [4, 5, 105], [], [5, 2, 110], [5, 100, 110]]

Output:
[null, true, true, false, true, true, [2, 5, 90], true, 1]

Explanation

Router router = new Router(3); // Initialize Router with memoryLimit of 3.
router.addPacket(1, 4, 90); // Packet is added. Return True.
router.addPacket(2, 5, 90); // Packet is added. Return True.
router.addPacket(1, 4, 90); // This is a duplicate packet. Return False.
router.addPacket(3, 5, 95); // Packet is added. Return True
router.addPacket(4, 5, 105); // Packet is added, [1, 4, 90] is removed as number of packets exceeds memoryLimit. Return True.
router.forwardPacket(); // Return [2, 5, 90] and remove it from router.
router.addPacket(5, 2, 110); // Packet is added. Return True.
router.getCount(5, 100, 110); // The only packet with destination 5 and timestamp in the inclusive range [100, 110] is [4, 5, 105]. Return 1.

Example 2:

Input:
["Router", "addPacket", "forwardPacket", "forwardPacket"]
[[2], [7, 4, 90], [], []]

Output:
[null, true, [7, 4, 90], []]

Explanation

Router router = new Router(2); // Initialize Router with memoryLimit of 2.
router.addPacket(7, 4, 90); // Return True.
router.forwardPacket(); // Return [7, 4, 90].
router.forwardPacket(); // There are no packets left, return [].

Constraints:

2 <= memoryLimit <= 10⁵
1 <= source, destination <= 2 * 10⁵
1 <= timestamp <= 10⁹
1 <= startTime <= endTime <= 10⁹
At most 10⁵ calls will be made to addPacket, forwardPacket, and getCount methods altogether.
queries for addPacket will be made in non-decreasing order of timestamp.

Step 01

Brute Force Baseline

Problem summary: Design a data structure that can efficiently manage data packets in a network router. Each data packet consists of the following attributes: source: A unique identifier for the machine that generated the packet. destination: A unique identifier for the target machine. timestamp: The time at which the packet arrived at the router. Implement the Router class: Router(int memoryLimit): Initializes the Router object with a fixed memory limit. memoryLimit is the maximum number of packets the router can store at any given time. If adding a new packet would exceed this limit, the oldest packet must be removed to free up space. bool addPacket(int source, int destination, int timestamp): Adds a packet with the given attributes to the router. A packet is considered a duplicate if another packet with the same source, destination, and timestamp already exists in the router. Return true if the packet

Baseline thinking

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

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

Example 1

["Router","addPacket","addPacket","addPacket","addPacket","addPacket","forwardPacket","addPacket","getCount"]
[[3],[1,4,90],[2,5,90],[1,4,90],[3,5,95],[4,5,105],[],[5,2,110],[5,100,110]]

Example 2

["Router","addPacket","forwardPacket","forwardPacket"]
[[2],[7,4,90],[],[]]

Step 02

Core Insight

What unlocks the optimal approach

A deque can simulate the adding and forwarding of packets efficiently.
Use binary search for counting packets within a timestamp range.

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 #3508: Implement Router
class Router {
    private int lim;
    private Set<Long> vis = new HashSet<>();
    private Deque<int[]> q = new ArrayDeque<>();
    private Map<Integer, Integer> idx = new HashMap<>();
    private Map<Integer, List<Integer>> d = new HashMap<>();

    public Router(int memoryLimit) {
        this.lim = memoryLimit;
    }

    public boolean addPacket(int source, int destination, int timestamp) {
        long x = f(source, destination, timestamp);
        if (vis.contains(x)) {
            return false;
        }
        vis.add(x);
        if (q.size() >= lim) {
            forwardPacket();
        }
        q.offer(new int[] {source, destination, timestamp});
        d.computeIfAbsent(destination, k -> new ArrayList<>()).add(timestamp);
        return true;
    }

    public int[] forwardPacket() {
        if (q.isEmpty()) {
            return new int[] {};
        }
        int[] packet = q.poll();
        int s = packet[0], d_ = packet[1], t = packet[2];
        vis.remove(f(s, d_, t));
        idx.merge(d_, 1, Integer::sum);
        return new int[] {s, d_, t};
    }

    private long f(int a, int b, int c) {
        return ((long) a << 46) | ((long) b << 29) | (long) c;
    }

    public int getCount(int destination, int startTime, int endTime) {
        List<Integer> ls = d.getOrDefault(destination, List.of());
        int k = idx.getOrDefault(destination, 0);
        int i = lowerBound(ls, startTime, k);
        int j = lowerBound(ls, endTime + 1, k);
        return j - i;
    }

    private int lowerBound(List<Integer> list, int target, int fromIndex) {
        int l = fromIndex, r = list.size();
        while (l < r) {
            int m = (l + r) >>> 1;
            if (list.get(m) < target) {
                l = m + 1;
            } else {
                r = m;
            }
        }
        return l;
    }
}

/**
 * Your Router object will be instantiated and called as such:
 * Router obj = new Router(memoryLimit);
 * boolean param_1 = obj.addPacket(source,destination,timestamp);
 * int[] param_2 = obj.forwardPacket();
 * int param_3 = obj.getCount(destination,startTime,endTime);
 */

// Accepted solution for LeetCode #3508: Implement Router
type Router struct {
	lim int
	vis map[int64]struct{}
	q   [][3]int
	idx map[int]int
	d   map[int][]int
}

func Constructor(memoryLimit int) Router {
	return Router{
		lim: memoryLimit,
		vis: make(map[int64]struct{}),
		q:   make([][3]int, 0),
		idx: make(map[int]int),
		d:   make(map[int][]int),
	}
}

func (this *Router) f(a, b, c int) int64 {
	return int64(a)<<46 | int64(b)<<29 | int64(c)
}

func (this *Router) AddPacket(source int, destination int, timestamp int) bool {
	x := this.f(source, destination, timestamp)
	if _, ok := this.vis[x]; ok {
		return false
	}
	this.vis[x] = struct{}{}
	if len(this.q) >= this.lim {
		this.ForwardPacket()
	}
	this.q = append(this.q, [3]int{source, destination, timestamp})
	this.d[destination] = append(this.d[destination], timestamp)
	return true
}

func (this *Router) ForwardPacket() []int {
	if len(this.q) == 0 {
		return []int{}
	}
	packet := this.q[0]
	this.q = this.q[1:]
	s, d, t := packet[0], packet[1], packet[2]
	delete(this.vis, this.f(s, d, t))
	this.idx[d]++
	return []int{s, d, t}
}

func (this *Router) GetCount(destination int, startTime int, endTime int) int {
	ls := this.d[destination]
	k := this.idx[destination]
	i := sort.Search(len(ls)-k, func(i int) bool { return ls[i+k] >= startTime }) + k
	j := sort.Search(len(ls)-k, func(i int) bool { return ls[i+k] >= endTime+1 }) + k
	return j - i
}

/**
 * Your Router object will be instantiated and called as such:
 * obj := Constructor(memoryLimit)
 * param_1 := obj.AddPacket(source,destination,timestamp)
 * param_2 := obj.ForwardPacket()
 * param_3 := obj.GetCount(destination,startTime,endTime)
 */

# Accepted solution for LeetCode #3508: Implement Router
class Router:

    def __init__(self, memoryLimit: int):
        self.lim = memoryLimit
        self.vis = set()
        self.q = deque()
        self.idx = defaultdict(int)
        self.d = defaultdict(list)

    def addPacket(self, source: int, destination: int, timestamp: int) -> bool:
        x = self.f(source, destination, timestamp)
        if x in self.vis:
            return False
        self.vis.add(x)
        if len(self.q) >= self.lim:
            self.forwardPacket()
        self.q.append((source, destination, timestamp))
        self.d[destination].append(timestamp)
        return True

    def forwardPacket(self) -> List[int]:
        if not self.q:
            return []
        s, d, t = self.q.popleft()
        self.vis.remove(self.f(s, d, t))
        self.idx[d] += 1
        return [s, d, t]

    def f(self, a: int, b: int, c: int) -> int:
        return a << 46 | b << 29 | c

    def getCount(self, destination: int, startTime: int, endTime: int) -> int:
        ls = self.d[destination]
        k = self.idx[destination]
        i = bisect_left(ls, startTime, k)
        j = bisect_left(ls, endTime + 1, k)
        return j - i


# Your Router object will be instantiated and called as such:
# obj = Router(memoryLimit)
# param_1 = obj.addPacket(source,destination,timestamp)
# param_2 = obj.forwardPacket()
# param_3 = obj.getCount(destination,startTime,endTime)

// Accepted solution for LeetCode #3508: Implement Router
use std::collections::{HashSet, HashMap, VecDeque};

struct Router {
    lim: usize,
    vis: HashSet<i64>,
    q: VecDeque<(i32, i32, i32)>,
    idx: HashMap<i32, usize>,
    d: HashMap<i32, Vec<i32>>,
}

impl Router {
    fn new(memory_limit: i32) -> Self {
        Router {
            lim: memory_limit as usize,
            vis: HashSet::new(),
            q: VecDeque::new(),
            idx: HashMap::new(),
            d: HashMap::new(),
        }
    }

    fn f(a: i32, b: i32, c: i32) -> i64 {
        ((a as i64) << 46) | ((b as i64) << 29) | (c as i64)
    }

    fn add_packet(&mut self, source: i32, destination: i32, timestamp: i32) -> bool {
        let x = Self::f(source, destination, timestamp);
        if self.vis.contains(&x) {
            return false;
        }
        self.vis.insert(x);
        if self.q.len() >= self.lim {
            self.forward_packet();
        }
        self.q.push_back((source, destination, timestamp));
        self.d.entry(destination).or_default().push(timestamp);
        true
    }

    fn forward_packet(&mut self) -> Vec<i32> {
        if let Some((s, d, t)) = self.q.pop_front() {
            self.vis.remove(&Self::f(s, d, t));
            *self.idx.entry(d).or_insert(0) += 1;
            vec![s, d, t]
        } else {
            vec![]
        }
    }

    fn get_count(&self, destination: i32, start_time: i32, end_time: i32) -> i32 {
        let ls = self.d.get(&destination);
        if ls.is_none() {
            return 0;
        }
        let ls = ls.unwrap();
        let k = *self.idx.get(&destination).unwrap_or(&0);
        let i = k + ls[k..].partition_point(|&x| x < start_time);
        let j = k + ls[k..].partition_point(|&x| x < end_time + 1);
        (j - i) as i32
    }
}

// Accepted solution for LeetCode #3508: Implement Router
class Router {
    private lim: number;
    private vis: Set<number>;
    private q: [number, number, number][];
    private idx: Map<number, number>;
    private d: Map<number, number[]>;

    constructor(memoryLimit: number) {
        this.lim = memoryLimit;
        this.vis = new Set();
        this.q = [];
        this.idx = new Map();
        this.d = new Map();
    }

    private f(a: number, b: number, c: number): number {
        return ((BigInt(a) << 46n) | (BigInt(b) << 29n) | BigInt(c)) as unknown as number;
    }

    addPacket(source: number, destination: number, timestamp: number): boolean {
        const x = this.f(source, destination, timestamp);
        if (this.vis.has(x)) {
            return false;
        }
        this.vis.add(x);
        if (this.q.length >= this.lim) {
            this.forwardPacket();
        }
        this.q.push([source, destination, timestamp]);
        if (!this.d.has(destination)) {
            this.d.set(destination, []);
        }
        this.d.get(destination)!.push(timestamp);
        return true;
    }

    forwardPacket(): number[] {
        if (this.q.length === 0) {
            return [];
        }
        const [s, d, t] = this.q.shift()!;
        this.vis.delete(this.f(s, d, t));
        this.idx.set(d, (this.idx.get(d) ?? 0) + 1);
        return [s, d, t];
    }

    getCount(destination: number, startTime: number, endTime: number): number {
        const ls = this.d.get(destination) ?? [];
        const k = this.idx.get(destination) ?? 0;
        const i = this.lowerBound(ls, startTime, k);
        const j = this.lowerBound(ls, endTime + 1, k);
        return j - i;
    }

    private lowerBound(arr: number[], target: number, from: number): number {
        let l = from,
            r = arr.length;
        while (l < r) {
            const m = Math.floor((l + r) / 2);
            if (arr[m] < target) {
                l = m + 1;
            } else {
                r = m;
            }
        }
        return l;
    }
}

/**
 * Your Router object will be instantiated and called as such:
 * var obj = new Router(memoryLimit)
 * var param_1 = obj.addPacket(source,destination,timestamp)
 * var param_2 = obj.forwardPacket()
 * var param_3 = obj.getCount(destination,startTime,endTime)
 */

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.