Tiktok面试官开门见山：

Problem Description (Part 1)
TikTok is hosting a livestream event with popular creators. Each creator has submitted their preferred start times and end time of their part in the livestream. Only one creator can livestream at the event at a time.

Given an array of available time intervals where intervals[i] = [start(i), end(i)], determine if all creators could attend all.

Constraints:

1. 0 <= start(i) < end(i) <= 10^6
2. Length of each interval will be 2

Example 1
Input: [[0,30],[5,10],[15,20]] → Output: false
Example 2
Input: [[7,10],[2,4]] → Output: true

候选人看到题目，第一反应是“这是判断区间是否冲突”。但紧张感让他犹豫：

• 如果有 [2,4] 和 [4,7]，算不算冲突？
• 空数组要不要单独处理？

就在这时，CSOAHelp 辅助老师把逐字稿推送到侧屏：

def can_attend_all(intervals):
    if len(intervals) <= 1:
        return True
    intervals.sort(key=lambda it: it[0])
    prev_end = intervals[0][1]
    for i in range(1, len(intervals)):
        start, end = intervals[i]
        if start < prev_end:  
            return False
        prev_end = end
    return True

提示里还清楚写道：

• “Use < not <=, so back-to-back like [2,4] and [4,7] is allowed.”
• “Time complexity O(n log n), space O(1) if sort in place.”

候选人只需要照着念思路、抄代码，就能在几分钟内交付一份完整答案。

面试官满意地点点头，然后抛出真正的杀招：

Problem Description (Part 2, Follow-up)
TikTok changes minds and wants to fit all creators to the livestream event. In that case, how many livestream rooms should they provide? Return the minimum number of livestream rooms.

Example
Input: [[0,30],[5,10],[10,20]] → Output: 2

候选人一度慌了。这其实是经典的“最少会议室”问题，要统计最大并发区间数。现场 coding 容易卡在边界：

• 如果某个区间的 end 正好等于另一个的 start，算不算重叠？
• 同一时间点的 start 和 end，处理顺序该怎么定？

CSOAHelp 辅助老师立即给出了解决方案：

def min_rooms(intervals):
    if len(intervals) <= 1:
        return len(intervals)

    events = []
    for s, e in intervals:
        events.append((s, +1))  # 开始 +1
        events.append((e, -1))  # 结束 -1

    events.sort()  # end(-1) 会排在 start(+1) 前

    curr = 0
    max_concurrent = 0
    for _, delta in events:
        curr += delta
        if curr > max_concurrent:
            max_concurrent = curr
    return max_concurrent

逐字稿里还特别强调：

• “End time is exclusive, so sort end before start.”
• “Track curr_concurrent and update max_concurrent.”
• “Time O(n log n), space O(n).”

而且，测试用例也都写好：

• [[0,30],[5,10],[10,20]] → 2
• [[7,10],[2,4]] → 1
• [[0,45],[15,30],[45,60],[45,70]] → 2
• [[0,10],[10,20]] → 1

候选人只需要“照本宣科”，一边把代码写下去，一边顺带运行用例验证，就能给出一个完全正确的答案。

最后，CSOAHelp 甚至为候选人准备好了反问问题清单，让收尾环节同样体面：

• “What do you like the most about working here?”
• “What is the onboarding process like?”
• “How do you measure scope and performance?”

面试官看到的，是一个思路完整、代码无误、沟通得体的候选人。
而候选人心里很清楚——真正的底气来自于 CSOAHelp 在背后全程逐字稿式的辅助。

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