We use cookies (including Google cookies) to personalize ads and analyze traffic. By continuing to use our site, you accept our Privacy Policy.

SimplyLeet
TagsCompaniesDisclaimersPrivacyTermsContactAbout Us

Number of Employees Who Met the Target

Number: 2876

Difficulty: Easy

Paid? No

Companies: Amazon, Adobe

Problem Description

Given an array representing the number of hours each employee has worked and a target number of hours required, count how many employees have met or exceeded the target hours.

Key Insights

  • You are given a list of non-negative integers representing work hours.
  • A simple iteration through the array is sufficient since the constraints are small.
  • Increment a counter each time an employee's hours are greater than or equal to the target.
  • The solution is straightforward with a single loop.

Space and Time Complexity

Time Complexity: O(n), where n is the number of employees. Space Complexity: O(1), only a counter is used.

Solution

The approach is to iterate over the list of work hours and count the number of employees whose hours are at least the target. Use a simple counter variable initialized to 0, and for every value in the array, increment the counter if the value is greater than or equal to the target. This yields a time-efficient and space-efficient solution. The key data structure is the array provided, and the algorithmic approach is a single linear scan.

Code Solutions

# Define a function to count employees meeting the target
def countEmployees(hours, target):
    count = 0  # Initialize counter to track qualifying employees
    for h in hours:  # Iterate over each employee's work hours
        if h >= target:  # Check if employee's hours are at least the target
            count += 1  # Increment the counter if condition is met
    return count  # Return the final count of employees who met the target

# Example usage:
# hours = [0, 1, 2, 3, 4]
# target = 2
# print(countEmployees(hours, target))  # Output: 3
← Back to All Questions