Number of Employees Who Met the Target - Practice Coding Problems

Number of Employees Who Met the Target - Problem

Array Easy

There are n employees in a company, numbered from 0 to n - 1. Each employee i has worked for hours[i] hours in the company.

The company requires each employee to work for at least target hours.

You are given a 0-indexed array of non-negative integers hours of length n and a non-negative integer target.

Return the integer denoting the number of employees who worked at least target hours.

Input & Output

Example 1 — Basic Case

$ Input: hours = [0,1,2,3,4], target = 2

› Output: 3

💡 Note: Employees with indices 2, 3, and 4 worked at least 2 hours: hours[2] = 2, hours[3] = 3, hours[4] = 4

Example 2 — All Meet Target

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

› Output: 5

💡 Note: All employees worked at least 1 hour, so all 5 employees meet the target

Example 3 — None Meet Target

$ Input: hours = [1,2,3], target = 5

› Output: 0

💡 Note: No employee worked 5 or more hours, so 0 employees meet the target

Constraints

1 ≤ n ≤ 50
0 ≤ hours[i], target ≤ 10⁵

Visualization

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Understanding the Visualization

Input

Array of employee hours and target requirement

Process

Count employees with hours >= target

Output

Number of qualifying employees

Key Takeaway

🎯 Key Insight: Simply iterate through the array once and count elements >= target

Asked in

G Google 25 a Amazon 20

The key insight is to simply count how many array elements are greater than or equal to the target value. Best approach is a single linear scan. Time: O(n), Space: O(1)

Common Approaches

Approach	Time	Space	Notes
✓ Functional Approach	O(n)	O(1)	Use built-in filter/count functions for cleaner code
Linear Scan	O(n)	O(1)	Iterate through each employee and count those meeting the target

Functional Approach — Algorithm Steps

Use filter/count function with condition
Return the count directly

Visualization

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Step-by-Step Walkthrough

Filter

Apply condition h >= target to all elements

Count

Count elements that pass the filter

Result

Return the count

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int solution(int* hours, int hoursSize, int target) {
    int count = 0;
    for (int i = 0; i < hoursSize; i++) {
        if (hours[i] >= target) {
            count++;
        }
    }
    return count;
}

int* parseArray(char* s, int* size) {
    int* result = malloc(1000 * sizeof(int));
    *size = 0;
    
    // Skip opening bracket
    s++;
    
    char* token = strtok(s, ",]");
    while (token != NULL) {
        result[(*size)++] = atoi(token);
        token = strtok(NULL, ",]");
    }
    
    return result;
}

int main() {
    char line[10000];
    
    // Read hours array
    fgets(line, sizeof(line), stdin);
    line[strcspn(line, "\n")] = 0;
    int hoursSize;
    int* hours = parseArray(line, &hoursSize);
    
    // Read target
    fgets(line, sizeof(line), stdin);
    int target = atoi(line);
    
    // Call solution and print result
    int result = solution(hours, hoursSize, target);
    printf("%d\n", result);
    
    free(hours);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n)

Still need to check every element once

✓ Linear Growth

Space Complexity

O(1)

Built-in functions use constant extra space

✓ Linear Space

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Functional Approach — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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