LeetCode #933 — EASY

Number of Recent Calls

Build confidence with an intuition-first walkthrough focused on design fundamentals.

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The Problem

Problem Statement

You have a RecentCounter class which counts the number of recent requests within a certain time frame.

Implement the RecentCounter class:

RecentCounter() Initializes the counter with zero recent requests.
int ping(int t) Adds a new request at time t, where t represents some time in milliseconds, and returns the number of requests that has happened in the past 3000 milliseconds (including the new request). Specifically, return the number of requests that have happened in the inclusive range [t - 3000, t].

It is guaranteed that every call to ping uses a strictly larger value of t than the previous call.

Example 1:

Input
["RecentCounter", "ping", "ping", "ping", "ping"]
[[], [1], [100], [3001], [3002]]
Output
[null, 1, 2, 3, 3]

Explanation
RecentCounter recentCounter = new RecentCounter();
recentCounter.ping(1);     // requests = [1], range is [-2999,1], return 1
recentCounter.ping(100);   // requests = [1, 100], range is [-2900,100], return 2
recentCounter.ping(3001);  // requests = [1, 100, 3001], range is [1,3001], return 3
recentCounter.ping(3002);  // requests = [1, 100, 3001, 3002], range is [2,3002], return 3

Constraints:

1 <= t <= 10⁹
Each test case will call ping with strictly increasing values of t.
At most 10⁴ calls will be made to ping.

Step 01

Brute Force Baseline

Problem summary: You have a RecentCounter class which counts the number of recent requests within a certain time frame. Implement the RecentCounter class: RecentCounter() Initializes the counter with zero recent requests. int ping(int t) Adds a new request at time t, where t represents some time in milliseconds, and returns the number of requests that has happened in the past 3000 milliseconds (including the new request). Specifically, return the number of requests that have happened in the inclusive range [t - 3000, t]. It is guaranteed that every call to ping uses a strictly larger value of t than the previous call.

Baseline thinking

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

Pattern signal: Design

Example 1

["RecentCounter","ping","ping","ping","ping"]
[[],[1],[100],[3001],[3002]]

Step 02

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 #933: Number of Recent Calls
class RecentCounter {
    private int[] s = new int[10010];
    private int idx;

    public RecentCounter() {
    }

    public int ping(int t) {
        s[idx++] = t;
        return idx - search(t - 3000);
    }

    private int search(int x) {
        int left = 0, right = idx;
        while (left < right) {
            int mid = (left + right) >> 1;
            if (s[mid] >= x) {
                right = mid;
            } else {
                left = mid + 1;
            }
        }
        return left;
    }
}

/**
 * Your RecentCounter object will be instantiated and called as such:
 * RecentCounter obj = new RecentCounter();
 * int param_1 = obj.ping(t);
 */

// Accepted solution for LeetCode #933: Number of Recent Calls
type RecentCounter struct {
	q []int
}

func Constructor() RecentCounter {
	return RecentCounter{[]int{}}
}

func (this *RecentCounter) Ping(t int) int {
	this.q = append(this.q, t)
	for this.q[0] < t-3000 {
		this.q = this.q[1:]
	}
	return len(this.q)
}

/**
 * Your RecentCounter object will be instantiated and called as such:
 * obj := Constructor();
 * param_1 := obj.Ping(t);
 */

# Accepted solution for LeetCode #933: Number of Recent Calls
class RecentCounter:
    def __init__(self):
        self.q = deque()

    def ping(self, t: int) -> int:
        self.q.append(t)
        while self.q[0] < t - 3000:
            self.q.popleft()
        return len(self.q)


# Your RecentCounter object will be instantiated and called as such:
# obj = RecentCounter()
# param_1 = obj.ping(t)

// Accepted solution for LeetCode #933: Number of Recent Calls
use std::collections::VecDeque;
struct RecentCounter {
    queue: VecDeque<i32>,
}

/**
 * `&self` means the method takes an immutable reference.
 * If you need a mutable reference, change it to `&mut self` instead.
 */
impl RecentCounter {
    fn new() -> Self {
        Self {
            queue: VecDeque::new(),
        }
    }

    fn ping(&mut self, t: i32) -> i32 {
        self.queue.push_back(t);
        while let Some(&v) = self.queue.front() {
            if v >= t - 3000 {
                break;
            }
            self.queue.pop_front();
        }
        self.queue.len() as i32
    }
}

// Accepted solution for LeetCode #933: Number of Recent Calls
class RecentCounter {
    private queue: number[];

    constructor() {
        this.queue = [];
    }

    ping(t: number): number {
        this.queue.push(t);
        while (this.queue[0] < t - 3000) {
            this.queue.shift();
        }
        return this.queue.length;
    }
}

/**
 * Your RecentCounter object will be instantiated and called as such:
 * var obj = new RecentCounter()
 * var param_1 = obj.ping(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(1) per op

Space

O(n)

Approach Breakdown

NAIVE

O(n) per op time

O(n) space

Use a simple list or array for storage. Each operation (get, put, remove) requires a linear scan to find the target element — O(n) per operation. Space is O(n) to store the data. The linear search makes this impractical for frequent operations.

OPTIMIZED DESIGN

O(1) per op time

O(n) space

Design problems target O(1) amortized per operation by combining data structures (hash map + doubly-linked list for LRU, stack + min-tracking for MinStack). Space is always at least O(n) to store the data. The challenge is achieving constant-time operations through clever structure composition.

Shortcut: Combine two data structures to get O(1) for each operation type. Space is always O(n).

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.