Solution

Method 1 – Simple Linear Scan

Intuition

We just need to count how many students are doing homework at the given queryTime. For each student, if queryTime is between their start and end time (inclusive), we count them. This is a direct application of the problem statement.

Approach

Initialize a counter ans to 0.
For each student, check if startTime[i] <= queryTime <= endTime[i].
If true, increment ans.
Return ans.

Code

[{"language":"C++","code":"class Solution {\npublic:\n    int busyStudent(vector<int>& startTime, vector<int>& endTime, int queryTime) {\n        int ans = 0, n = startTime.size();\n        for (int i = 0; i < n; ++i) {\n            if (startTime[i] <= queryTime && queryTime <= endTime[i]) ++ans;\n        }\n        return ans;\n    }\n};","lang":"cpp","html":"<pre class=\"shiki shiki-themes github-light github-dark\" style=\"background-color:#fff;--shiki-dark-bg:#24292e;color:#24292e;--shiki-dark:#e1e4e8\" tabindex=\"0\"><code><span class=\"line\"><span style=\"color:#D73A49;--shiki-dark:#F97583\">class</span><span style=\"color:#6F42C1;--shiki-dark:#B392F0\"> Solution</sp

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