Maximum Requests in a Time Window (OA)

Second problem on the same recent Salesforce OA (HackerRank, 2 problems / 60 min). Given a list of integer timestamps and a window size W, find the maximum number of timestamps that fit inside any contiguous window of length W (window = `[start, start + W - 1]`). Requirements Input: array timestamp of length n (1 ≤ n ≤ 2·10⁵) with 1 ≤ timestamp[i] ≤ 10⁹, and integer windowSize (1 ≤ W ≤ 10⁹). Output: maximum count of timestamp[i] values inside any inclusive window [s, s + W - 1] for some integer s. The optimal window always has its left edge at one of the input timestamps (otherwise sliding right loses no points and may gain some). Examples timestamp = [1, 3, 7, 5], W = 4 → 2 # e.g. [1,4], [3,6], or [5,8] timestamp = [2, 2, 3], W = 1 → 2 # window [2,2] catches both Notes Canonical sliding-window solution: Sort timestamp ascending. O(n log n). Two-pointer sweep: keep lo at the left edge; advance hi while timestamp[hi] - timestamp[lo] <= W - 1 (inclusive window of length W = W-1 span). When the condition breaks, advance lo. Track the running maximum (hi - lo + 1). Answer is the maximum count observed. O(n) after the sort. Equivalent formulation: for each i, binary search for the largest j such that timestamp[j] <= timestamp[i] + W - 1. Max of (j - i + 1). Same complexity, slightly more code. Duplicate timestamps are fine — the sort puts them adjacent and the two-pointer counts them naturally. Edge cases: W = 1 (each window holds only the timestamps equal to its single value — count duplicates of the most-common value); all timestamps equal (answer = n); n = 1 (answer = 1). Be careful with the inclusive window definition. "Length W" in this prompt means span W-1; an off-by-one in the inner comparison silently produces wrong answers on the boundary cases. Preparation Write the sliding-window version twice — once expanding hi, once shrinking lo — and confirm both pass the two examples plus a 1-element and a duplicate-heavy case. Verify your inequality carefully: timestamp[hi] - timestamp[lo] <= W - 1 vs < W are equivalent here; pick one and stick to it. Time the implementation — interviewers expect <10 minutes given the 60-minute / 2-problem budget. The bulk of OA time goes to the palindrome problem above; this one should be fast. Be ready for the variation "return the actual window" — track (lo, hi) at the moment the maximum was set, not just the count.