LeetCode //C - 933. Number of Recent Calls
The problem involves tracking recent requests within a 3000-millisecond window. The solution uses a RecentCounter class with a dynamically expanding array to store request timestamps. The key optimiza
933. Number of Recent Calls
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.s the number of the person that was leading the election at time t according to the mentioned rules.
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 < = 1 0 9 1 <= t <= 10^9 1<=t<=109
- Each test case will call ping with strictly increasing values of t.
- At most 1 0 4 10^4 104 calls will be made to ping.
From: LeetCode
Link: 933. Number of Recent Calls
Solution:
Ideas:
-
Purpose: Count how many requests happened in the last 3000 milliseconds.
-
Data structure: Uses a dynamic array to store all request timestamps.
-
Optimization: Maintains a head index to mark the start of the valid 3000ms window (no need to scan all past requests).
-
On each ping(t):
- Add t to the array.
- Remove (logically) all timestamps older than t - 3000 by moving head forward.
- Return size - head, which is the number of timestamps in [t - 3000, t].
Code:
typedef struct {
int *requests; // Array to store the timestamps of requests
int size; // Number of timestamps stored (tail index)
int capacity; // Total capacity of the array
int head; // Index of the first timestamp within [t-3000, t]
} RecentCounter;
RecentCounter* recentCounterCreate() {
RecentCounter* obj = (RecentCounter*) malloc(sizeof(RecentCounter));
obj->capacity = 10; // Initial capacity
obj->size = 0;
obj->head = 0;
obj->requests = (int*) malloc(obj->capacity * sizeof(int));
return obj;
}
// Helper function to resize the array if needed
static void resizeIfNeeded(RecentCounter* obj) {
if (obj->size >= obj->capacity) {
obj->capacity *= 2;
obj->requests = (int*) realloc(obj->requests, obj->capacity * sizeof(int));
}
}
int recentCounterPing(RecentCounter* obj, int t) {
// Ensure capacity and append new timestamp
resizeIfNeeded(obj);
obj->requests[obj->size++] = t;
// Maintain sliding window: advance head while outside [t-3000, t]
int minTime = t - 3000;
while (obj->head < obj->size && obj->requests[obj->head] < minTime) {
obj->head++;
}
// Number of recent requests is the window length
return obj->size - obj->head;
}
void recentCounterFree(RecentCounter* obj) {
if (!obj) return;
free(obj->requests);
free(obj);
}
/**
* Your RecentCounter struct will be instantiated and called as such:
* RecentCounter* obj = recentCounterCreate();
* int param_1 = recentCounterPing(obj, t);
* recentCounterFree(obj);
*/
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