What is the time complexity of Alert Using Same Key-Card Three or More Times in a One Hour Period?

The optimal solution for Alert Using Same Key-Card Three or More Times in a One Hour Period runs in O(n log n) time and uses O(n) space. The time complexity is O(n log n) due to sorting the times for each worker. The space complexity is O(n) because we store the times in a dictionary and the alerts in a set.

What is the brute force solution for Alert Using Same Key-Card Three or More Times in a One Hour Period?

The brute force approach for Alert Using Same Key-Card Three or More Times in a One Hour Period is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach checks every possible combination of key-card usage times for each worker. This means we will look at all pairs of times for each worker to see if they are within an hour of each other. The optimal Optimal Solution approach improves this to O(n log n).

What algorithm or data structure is used to solve Alert Using Same Key-Card Three or More Times in a One Hour Period?

Alert Using Same Key-Card Three or More Times in a One Hour Period uses the following concepts: Array, Hash Table, String, Sorting. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Alert Using Same Key-Card Three or More Times in a One Hour Period?

Always clarify the input format and constraints before diving into the solution. Think about edge cases, such as when a worker uses their key-card exactly three times at the boundary of one hour. Practice explaining your thought process clearly as you code, as communication is key in interviews.

What are common mistakes when solving Alert Using Same Key-Card Three or More Times in a One Hour Period?

Not sorting the times before checking for the one-hour condition. Confusing the time format or incorrectly calculating the time difference.

#1604

Alert Using Same Key-Card Three or More Times in a One Hour Period

Medium

ArrayHash TableStringSortingHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n log n)
Space	O(1)	O(n)

💡

Intuition

Time O(n log n)Space O(n)

The optimal solution leverages sorting and a sliding window technique to efficiently check for key-card usage within a one-hour period. By sorting the times and using two pointers, we can quickly determine if three usages fall within the required time frame.

⚙️

Algorithm

3 steps

1Step 1: Create a dictionary to group times by worker names.
2Step 2: For each worker, sort their usage times.
3Step 3: Use a sliding window approach to check if there are three usages within any one-hour period.

solution.py16 lines

1def alertNames(keyName, keyTime):
2    from collections import defaultdict
3    times = defaultdict(list)
4    for name, time in zip(keyName, keyTime):
5        times[name].append(time)
6    alerts = set()
7    for name, time_list in times.items():
8        time_list.sort()
9        n = len(time_list)
10        for i in range(n):
11            j = i
12            while j < n and (int(time_list[j][:2]) - int(time_list[i][:2])) * 60 + (int(time_list[j][3:]) - int(time_list[i][3:])) <= 60:
13                j += 1
14            if j - i >= 3:
15                alerts.add(name)
16    return sorted(alerts)

ℹ

Complexity note: The time complexity is O(n log n) due to sorting the times for each worker. The space complexity is O(n) because we store the times in a dictionary and the alerts in a set.

1Sorting the times allows for efficient checking of time ranges using a sliding window.
2Using a set to store alerts ensures unique names without duplicates.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.