LeetCode 3060: User Activities within Time Bounds Solution

Master LeetCode problem 3060 (User Activities within Time Bounds), a hard challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

3060. User Activities within Time Bounds

Hard

Database

Problem Explanation

Explanation:

To solve this problem, we can use a sliding window approach. We will maintain two pointers start and end to represent the start and end of the window. We will keep expanding the window to the right until the time difference between the activities at start and end is greater than the given time bounds. When this happens, we will move the start pointer to the right until the time difference falls within the bounds again. At each step, we will update the maximum number of activities within the time bounds.

Here is the detailed algorithm:

Initialize start and end pointers to 0, and initialize a variable maxActivities to store the maximum number of activities within the time bounds.
Iterate through the activities using the end pointer:
- Calculate the time difference between the activities at start and end.
- If the time difference is within the bounds, update maxActivities if the current number of activities in the window is greater.
- If the time difference exceeds the bounds, move the start pointer to the right until the time difference falls within the bounds.
Return the maximum number of activities within the time bounds.

Time complexity:

The time complexity of this approach is O(N), where N is the number of activities.

Space complexity:

The space complexity is O(1) as we are using constant extra space.

Solution Code

class Solution {
    public int maxUserActivities(int[] activities, int timeBounds) {
        int n = activities.length;
        int start = 0, end = 0;
        int maxActivities = 0;
        
        while (end < n) {
            if (activities[end] - activities[start] <= timeBounds) {
                maxActivities = Math.max(maxActivities, end - start + 1);
                end++;
            } else {
                start++;
            }
        }
        
        return maxActivities;
    }
}

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