LeetCode 2004 - The Number of Seniors and Juniors to Join the Company

This problem requires determining how many candidates a company can hire as seniors and juniors under a fixed budget of $70000, following a strict priority: hire as many seniors as possible first, then use the remaining budget to hire juniors.

leetcodeharddatabase

LeetCode Problem 2004

Difficulty: 🔴 Hard
Topics: Database

Solution

Problem Understanding

This problem requires determining how many candidates a company can hire as seniors and juniors under a fixed budget of $70000, following a strict priority: hire as many seniors as possible first, then use the remaining budget to hire juniors. The input is a table Candidates containing employee_id (unique identifier), experience (either 'Senior' or 'Junior'), and salary (monthly salary). The output should be a table showing how many seniors and juniors can be hired, according to the priority and budget constraint.

Constraints indicate that each candidate has a positive integer salary, and that experience is guaranteed to be either 'Senior' or 'Junior'. Important edge cases include scenarios where all seniors are too expensive, or when the budget allows hiring only part of the candidates. The problem guarantees no negative salaries and that all employee IDs are unique.

Approaches

A brute-force approach would attempt to enumerate all possible combinations of seniors and juniors to check which combination maximizes seniors first and juniors second, summing their salaries to see if they fit within the budget. While correct, this approach is inefficient because it grows exponentially with the number of candidates and is unnecessary for the problem constraints.

The key insight for an optimal solution is to process seniors and juniors separately by sorting each group by salary. By sorting, we can sequentially pick the lowest-salary seniors first, maximizing the number of seniors that fit in the budget. After hiring seniors, we can repeat the process for juniors using the remaining budget. This greedy approach ensures that we always hire the maximum number of seniors and then juniors without exploring unnecessary combinations.

Approach Time Complexity Space Complexity Notes
Brute Force O(2^n) O(n) Check all combinations of seniors and juniors to find maximum hires under budget
Optimal O(n log n) O(n) Sort seniors and juniors separately by salary and use greedy accumulation to maximize hires

Algorithm Walkthrough

  1. Initialize two separate lists for seniors and juniors by scanning the Candidates table and splitting based on the experience column.
  2. Sort both lists by salary in ascending order. Sorting ensures that the cheapest candidates are considered first, which maximizes the number of hires.
  3. Initialize two counters for accepted seniors and juniors, and a variable for the remaining budget starting at $70000.
  4. Iterate over the sorted senior list. For each senior, check if their salary is less than or equal to the remaining budget. If so, subtract their salary from the budget and increment the senior counter. Stop when the next senior cannot be afforded.
  5. Repeat the process for the juniors using the remaining budget. This ensures that after maximizing seniors, the remaining budget is used optimally to hire juniors.
  6. Return a table with the counts of accepted seniors and juniors. The order of the rows does not matter.

Why it works: Sorting candidates by salary and picking the lowest-cost candidates first guarantees that the maximum possible number of seniors is hired. The same logic applies for juniors after the senior selection. This is a classic greedy strategy that works because the problem constraints allow independent selection within each group and there are no additional conditions linking the hires of seniors and juniors.

Python Solution

class Solution:
    def numberOfSeniorAndJunior(self, candidates: list[dict]) -> list[dict]:
        BUDGET = 70000
        seniors = []
        juniors = []
        
        for c in candidates:
            if c['experience'] == 'Senior':
                seniors.append(c['salary'])
            else:
                juniors.append(c['salary'])
        
        seniors.sort()
        juniors.sort()
        
        accepted_seniors = 0
        accepted_juniors = 0
        remaining_budget = BUDGET
        
        for salary in seniors:
            if salary <= remaining_budget:
                remaining_budget -= salary
                accepted_seniors += 1
            else:
                break
        
        for salary in juniors:
            if salary <= remaining_budget:
                remaining_budget -= salary
                accepted_juniors += 1
            else:
                break
        
        return [
            {'experience': 'Senior', 'accepted_candidates': accepted_seniors},
            {'experience': 'Junior', 'accepted_candidates': accepted_juniors}
        ]

The implementation first separates candidates into two lists, sorts them by salary, then greedily picks the lowest-cost candidates from each group until the budget is exhausted. Counters track the number of hires, and the remaining budget ensures we never exceed the limit.

Go Solution

func numberOfSeniorAndJunior(candidates []map[string]interface{}) []map[string]int {
    const BUDGET = 70000
    seniors := []int{}
    juniors := []int{}
    
    for _, c := range candidates {
        experience := c["experience"].(string)
        salary := c["salary"].(int)
        if experience == "Senior" {
            seniors = append(seniors, salary)
        } else {
            juniors = append(juniors, salary)
        }
    }
    
    sort.Ints(seniors)
    sort.Ints(juniors)
    
    acceptedSeniors := 0
    acceptedJuniors := 0
    remainingBudget := BUDGET
    
    for _, salary := range seniors {
        if salary <= remainingBudget {
            remainingBudget -= salary
            acceptedSeniors++
        } else {
            break
        }
    }
    
    for _, salary := range juniors {
        if salary <= remainingBudget {
            remainingBudget -= salary
            acceptedJuniors++
        } else {
            break
        }
    }
    
    return []map[string]int{
        {"experience": "Senior", "accepted_candidates": acceptedSeniors},
        {"experience": "Junior", "accepted_candidates": acceptedJuniors},
    }
}

The Go implementation mirrors Python closely. We handle type assertions from map[string]interface{} and use slices. Sorting is done using the standard library, and integers handle the budget calculations.

Worked Examples

Example 1: Budget $70000, candidates [Senior 20000, Senior 20000, Senior 50000, Junior 10000, Junior 10000, Junior 40000]

  • Sort seniors: [20000, 20000, 50000]
  • Hire seniors: first 20000 -> budget 50000, second 20000 -> budget 30000, third 50000 -> cannot hire
  • Sort juniors: [10000, 10000, 40000]
  • Hire juniors: first 10000 -> budget 20000, second 10000 -> budget 10000, third 40000 -> cannot hire
  • Output: Senior 2, Junior 2

Example 2: Budget $70000, candidates [Senior 80000, Senior 80000, Senior 80000, Junior 10000, Junior 10000, Junior 40000]

  • Sort seniors: [80000, 80000, 80000], cannot hire any
  • Sort juniors: [10000, 10000, 40000], hire all three -> budget 70000 - 60000 = 10000
  • Output: Senior 0, Junior 3

Complexity Analysis

Measure Complexity Explanation
Time O(n log n) Sorting seniors and juniors dominates time complexity, where n is the total number of candidates
Space O(n) Two separate lists are created for seniors and juniors

Sorting ensures optimal candidate selection, and the greedy pass over each sorted list is linear.

Test Cases

# Provided examples
assert Solution().numberOfSeniorAndJunior([
    {'employee_id': 1, 'experience': 'Junior', 'salary': 10000},
    {'employee_id': 9, 'experience': 'Junior', 'salary': 10000},
    {'employee_id': 2, 'experience': 'Senior', 'salary': 20000},
    {'employee_id': 11, 'experience': 'Senior', 'salary': 20000},
    {'employee_id': 13, 'experience': 'Senior', 'salary': 50000},
    {'employee_id': 4, 'experience': 'Junior', 'salary': 40000},
]) == [{'experience': 'Senior', 'accepted_candidates': 2}, {'experience': 'Junior', 'accepted_candidates': 2}]

assert Solution().numberOfSeniorAndJunior([
    {'employee_id': 1, 'experience': 'Junior', 'salary': 10000},
    {'employee_id': 9, 'experience': 'Junior', 'salary': 10000},
    {'employee_id': 2, 'experience': 'Senior', 'salary': 80000},
    {'employee_id': 11, 'experience': 'Senior', 'salary': 80000},
    {'employee_id': 13, 'experience': 'Senior', 'salary': 80000},
    {'employee_id': 4, 'experience': 'Junior', 'salary': 40000},
]) == [{'experience': 'Senior', 'accepted_candidates': 0}, {'experience': 'Junior', 'accepted_candidates': 3}]

# Edge cases
assert Solution().numberOfSeniorAndJunior([]) == [{'experience': 'Senior', 'accepted_candidates': 0}, {'experience': 'Junior', 'accepted_candidates': 0}] # no candidates
assert Solution().numberOfSeniorAndJunior([
    {'employee_id': 1, 'experience': 'Senior', 'salary': 70000},
]) == [{'experience': 'Senior', 'accepted_candidates': 1}, {'experience': 'Junior', 'accepted_candidates': 0}] # single senior fits exactly
assert Solution().numberOfSeniorAndJunior([