How do you solve Number of Employees Who Met the Target on LeetCode?

Number of Employees Who Met the Target (LeetCode #2798) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: Iterating through the array is a direct way to solve the problem.

What is the time complexity of Number of Employees Who Met the Target?

The optimal solution for Number of Employees Who Met the Target runs in O(n) time and uses O(1) space. The time complexity remains O(n) as we still iterate through the list once, and the space complexity is O(1) since we only use a counter.

What is the brute force solution for Number of Employees Who Met the Target?

The brute force approach for Number of Employees Who Met the Target is Brute Force, with O(n) time complexity and O(1) space complexity. The brute-force approach involves checking each employee's hours one by one to see if they meet the target. This is straightforward and easy to understand, but it can be inefficient for larger datasets. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Number of Employees Who Met the Target?

Number of Employees Who Met the Target uses the following concepts: Array. The recommended patterns to study are: Array, Counting.

What are the interview tips for Number of Employees Who Met the Target?

Always clarify the problem statement before jumping into coding. Think about edge cases, such as all employees meeting or not meeting the target. Discuss your thought process with the interviewer to demonstrate your understanding.

What are common mistakes when solving Number of Employees Who Met the Target?

Not initializing the counter correctly. Forgetting to check the condition properly.

#2798

Number of Employees Who Met the Target

Easy

ArrayArrayCounting

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n)Space O(1)

The optimal solution is essentially the same as the brute-force approach but emphasizes efficiency. Given the constraints, we can still iterate through the list once, which is efficient and straightforward.

⚙️

Algorithm

4 steps

1Step 1: Initialize a count variable to zero.
2Step 2: Loop through each element in the hours array.
3Step 3: If the current employee's hours are greater than or equal to the target, increment the count.
4Step 4: Return the count after the loop.

solution.py11 lines

1# Full working Python code
2
3def countEmployees(hours, target):
4    count = 0
5    for hour in hours:
6        if hour >= target:
7            count += 1
8    return count
9
10# Example usage
11print(countEmployees([5, 1, 4, 2, 2], 6))  # Output: 0

ℹ

Complexity note: The time complexity remains O(n) as we still iterate through the list once, and the space complexity is O(1) since we only use a counter.

1Iterating through the array is a direct way to solve the problem.
2The problem can be solved efficiently with a single pass through the data.

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