Leetcode 2798: Number of Employees Who Met the Target

Input: You are given a 0-indexed array 'hours' representing the hours worked by employees and a non-negative integer 'target'.

Example: Input: hours = [0, 1, 3, 5, 7], target = 3

Constraints:

• 1 <= n == hours.length <= 50

• 0 <= hours[i], target <= 10^5

Output: Return the number of employees who have worked at least 'target' hours.

Example: Output: 3

Constraints:

Goal: To determine how many employees worked at least the target number of hours.

Steps:

• Iterate through the array of hours.

• For each employee, check if their worked hours are greater than or equal to the target.

• Count the employees who meet the target hours.

Goal: Ensure that the solution adheres to the given constraints.

Steps:

• 1 <= n == hours.length <= 50

• 0 <= hours[i], target <= 10^5

Assumptions:

• The input array contains non-negative integers.

• The number of employees is small enough to allow a straightforward solution.

• Input: Input: hours = [0, 1, 3, 5, 7], target = 3

• Explanation: Here, the company wants employees to work for at least 3 hours. Employees with hours >= 3 are employee 2, 3, and 4, so the output is 3.

• Input: Input: hours = [4, 2, 1, 6], target = 5

• Explanation: Here, the company wants employees to work for at least 5 hours. Only employee 3 worked 6 hours, so the output is 1.

Approach: We can simply iterate through the list of hours and count how many employees worked hours greater than or equal to the target.

Observations:

• We need to check each employee's worked hours to determine if they meet the target.

• The solution requires a single pass through the list, making it efficient.

Steps:

• Initialize a counter variable to 0.

• Iterate through the 'hours' array.

• For each element, check if it is greater than or equal to the target. If so, increment the counter.

• Return the counter as the final result.

Empty Inputs:

• The input array is guaranteed to contain at least one element.

Large Inputs:

• The solution should handle arrays of up to 50 elements efficiently.

Special Values:

• If all employees have worked fewer hours than the target, the result will be 0.

Constraints:

• The input size and values are small enough that a simple linear scan will suffice.

int numberOfEmployeesWhoMetTarget(vector<int>& hours, int target) {
    int cnt = 0;
    for(int i = 0; i < hours.size(); i++)
        if(hours[i] >= target) cnt++;
    return cnt;
}

1 : Function Declaration

int numberOfEmployeesWhoMetTarget(vector<int>& hours, int target) {

This line declares the function `numberOfEmployeesWhoMetTarget`, which takes a reference to a vector of integers (`hours`) and an integer `target` as inputs. It returns the number of employees who met or exceeded the target hours.

2 : Variable Initialization

    int cnt = 0;

The variable `cnt` is initialized to 0. This will count how many employees have met or exceeded the target.

3 : Loop Initialization

    for(int i = 0; i < hours.size(); i++)

A `for` loop is initialized to iterate through the `hours` vector. The loop will process each employee's hours worked.

4 : Condition Check

        if(hours[i] >= target) cnt++;

Inside the loop, an `if` condition checks if the current employee's hours (i.e., `hours[i]`) meet or exceed the target. If the condition is true, the counter `cnt` is incremented.

5 : Return Result

    return cnt;

After the loop completes, the function returns the value of `cnt`, which represents the total number of employees who met or exceeded the target hours.

Best Case: O(n)

Average Case: O(n)

Worst Case: O(n)

Description: The time complexity is O(n) because we iterate through the array once.

Best Case: O(1)

Worst Case: O(1)

Description: The space complexity is O(1) because we only use a constant amount of extra space.

Leetcode 2798: Number of Employees Who Met the Target

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