LeetCode: Number of Recent Calls

Problem:

https://leetcode.com/problems/number-of-recent-calls/

Code:

class RecentCounter:

    def __init__(self):
        self.timestamps = deque()

    def ping(self, t: int) -> int:
        self.timestamps.append(t)

        while self.timestamps[0]<t-3000: # we directly remove te timestamps that won't fall in the new range, older ones, we pop them out.
            self.timestamps.popleft()

        return len(self.timestamps)

Key Points:

Problem in Simple Words

You keep calling .ping(t) with increasing timestamps t. You must return how many calls happened in the last 3000 ms: that means in the range: [t - 3000, t]

Any older calls are no longer counted.

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Core Idea (Sliding Window Using Queue)

Use a queue (deque) to store timestamps of recent calls. Since t is always increasing:

• Newest calls go to the right

• Remove old timestamps from the left

This automatically keeps only the timestamps that are still valid.

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Why Removing from the Left Works

deque pops from the left efficiently → O(1) And since timestamps are sorted by arrival:

• The oldest timestamps are always in front

• Just remove them while they fall outside the valid range: any timestamp < t - 3000 must be discarded

After cleanup, the queue size = number of valid recent calls

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Execution Flow of ping(t)

1️⃣ Append the new timestamp
2️⃣ Remove timestamps older than 3000 ms
3️⃣ Return how many timestamps remain inside the window

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Why This Approach Is Good

• Every timestamp enters and leaves once → efficient

• No need to re-check entire history

• Maintains a clean sliding window of recent calls

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Complexity

Time: O(1) amortized per ping
Space: O(n) → number of timestamps in the last 3000 ms