What is the time complexity of Total Cost to Hire K Workers?

The optimal solution has a time complexity of O(k * log(candidates)).

What is the space complexity of Total Cost to Hire K Workers?

The space complexity is O(candidates).

What data structures are used to solve Total Cost to Hire K Workers?

This problem uses Array, Two Pointers, Heap (Priority Queue), Simulation. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Total Cost to Hire K Workers?

Insight 1: Using heaps (priority queues) is crucial for efficiently finding the minimum cost worker in each hiring session among the candidates. Insight 2: Maintaining two heaps (one for the beginning and one for the end of the array) allows you to consider candidates from both ends simultaneously without scanning the entire array each time.

Which companies ask LeetCode 2462. Total Cost to Hire K Workers in interviews?

This problem is frequently asked at BNY Mellon, GSA Capital, MathWorks, Zomato.

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Total Cost to Hire K Workers - LeetCode 2462 Solution

Total Cost to Hire K Workers - Complete Solution Guide

Total Cost to Hire K Workers is LeetCode problem 2462, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

You are given a 0-indexed integer array costs where costs[i] is the cost of hiring the i th worker. You are also given two integers k and candidates . We want to hire exactly k workers according to the following rules: You will run k sessions and hire exactly one worker in each session. In each hiring session, choose the worker with the lowest cost from either the first candidates workers or the last candidates workers. Break the tie by the smallest index. For example, if costs = [3,2,7,7,1,2] a

Detailed Explanation

The problem requires you to hire `k` workers from a pool of workers represented by the `costs` array. You can only hire one worker per session (there are `k` sessions). In each session, you must choose the worker with the minimum cost from either the first `candidates` workers or the last `candidates` workers. If there are ties in cost, choose the worker with the smaller index. Workers can only be hired once. The goal is to calculate the total cost of hiring `k` workers.

Solution Approach

The provided solutions utilize two min-heaps (priority queues), `left_heap` and `right_heap`, to store the costs of the candidate workers from the beginning and end of the `costs` array, respectively. In each hiring session, the solution extracts the worker with the minimum cost (and smallest index in case of ties) from the two heaps. The algorithm updates the total cost and re-populates the heaps as necessary, making sure to avoid overlapping selections by adjusting left and right pointers.

Step-by-Step Algorithm

Step 1: Initialize two min-heaps, `left_heap` and `right_heap`, to store the costs and indices of the candidate workers from the beginning and end of the array, respectively.
Step 2: Initialize `l` and `r` pointers to the beginning and end of the `costs` array.
Step 3: Iterate `k` times (for each hiring session).
Step 4: While the `left_heap` has fewer than `candidates` elements and `l` is not greater than `r`, add the cost and index of the worker at index `l` to the `left_heap` and increment `l`.
Step 5: While the `right_heap` has fewer than `candidates` elements and `l` is not greater than `r`, add the cost and index of the worker at index `r` to the `right_heap` and decrement `r`.
Step 6: Compare the minimum cost worker (and index if costs are equal) from the `left_heap` and `right_heap`.
Step 7: Add the cost of the chosen worker to the `total_cost` and remove the worker from the appropriate heap.
Step 8: Return the `total_cost`.

Key Insights

Insight 1: Using heaps (priority queues) is crucial for efficiently finding the minimum cost worker in each hiring session among the candidates.
Insight 2: Maintaining two heaps (one for the beginning and one for the end of the array) allows you to consider candidates from both ends simultaneously without scanning the entire array each time.
Insight 3: Handling the overlap between the first `candidates` and last `candidates` workers is important to avoid double-counting. This requires careful updating of the left and right pointers used to populate the heaps.

Complexity Analysis

Time Complexity: O(k * log(candidates))

Space Complexity: O(candidates)

Topics

This problem involves: Array, Two Pointers, Heap (Priority Queue), Simulation.

Companies

Asked at: BNY Mellon, GSA Capital, MathWorks, Zomato.

Total Cost to Hire K Workers | LeetCode 2462

Medium

Array Two Pointers Heap (Priority Queue)Simulation

Time: O(k * log(candidates))Space: O(candidates)

Solution Code

Available in:

import java.util.PriorityQueue;
import java.util.List;
import java.util.ArrayList;

class Solution {
    public long totalCost(int[] costs, int k, int candidates) {
        int n = costs.length;
        PriorityQueue<int[]> leftHeap = new PriorityQueue<>((a, b) -> {
            if (a[0] != b[0]) {
                return a[0] - b[0];
            } else {
                return a[1] - b[1];
            }
        });
        PriorityQueue<int[]> rightHeap = new PriorityQueue<>((a, b) -> {
            if (a[0] != b[0]) {
                return a[0] - b[0];
            } else {
                return a[1] - b[1];
            }
        });

        int l = 0, r = n - 1;
        long totalCost = 0;

        for (int i = 0; i < k; i++) {
            while (leftHeap.size() < candidates && l <= r) {
                leftHeap.offer(new int[]{costs[l], l});
                l++;
            }

            while (rightHeap.size() < candidates && l <= r) {
                rightHeap.offer(new int[]{costs[r], r});
                r--;
            }

            int[] leftCandidate = leftHeap.isEmpty() ? new int[]{Integer.MAX_VALUE, Integer.MAX_VALUE} : leftHeap.peek();
            int[] rightCandidate = rightHeap.isEmpty() ? new int[]{Integer.MAX_VALUE, Integer.MAX_VALUE} : rightHeap.peek();

            if (leftCandidate[0] <= rightCandidate[0]) {
                int[] temp = leftHeap.poll();
                totalCost += temp[0];
            } else {
                int[] temp = rightHeap.poll();
                totalCost += temp[0];
            }
        }

        return totalCost;
    }
}

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Problem Info

Description

Problem 2462: Total Cost to Hire K Workers - Detailed explanation and solution approaches.

Topics

Array Two Pointers Heap (Priority Queue)Simulation

Companies

BNY MellonGSA CapitalMathWorksZomato

Stats

Difficulty:Medium

Languages:4

Companies:4

Problem #:2462

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

8 steps

1
Step 1: Initialize two min-heaps, `left_heap` and `right_heap`, to store the costs and indices of the candidate workers from the beginning and end of the array, respectively.
2
Step 2: Initialize `l` and `r` pointers to the beginning and end of the `costs` array.
3
Step 3: Iterate `k` times (for each hiring session).
4
Step 4: While the `left_heap` has fewer than `candidates` elements and `l` is not greater than `r`, add the cost and index of the worker at index `l` to the `left_heap` and increment `l`.
5
Step 5: While the `right_heap` has fewer than `candidates` elements and `l` is not greater than `r`, add the cost and index of the worker at index `r` to the `right_heap` and decrement `r`.
6
Step 6: Compare the minimum cost worker (and index if costs are equal) from the `left_heap` and `right_heap`.
7
Step 7: Add the cost of the chosen worker to the `total_cost` and remove the worker from the appropriate heap.
8
Step 8: Return the `total_cost`.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(k * log(candidates))

💾

Space Complexity

O(candidates)

📝

Why?

The time complexity comes from the fact that we iterate k times and in each iteration we perform heap operations (push and pop), which take O(log n) time where n is the size of the heap. In our case the heaps sizes are bounded by candidates. The space complexity comes from storing at most 2*candidates elements in the heaps.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Brute Force (without heaps)

Instead of using heaps, we could iterate through the candidate sections in each hiring session to find the minimum cost. We would also need to keep track of hired workers to avoid double-counting.

Time: O(k * candidates)Space: O(n) to keep track of which indexes are hired

Trade-offs: This approach is simpler to implement but less efficient than the heap-based solution, particularly when `candidates` is large. The heap approach gives O(log candidates) per hiring round compared to O(candidates) for the brute force. The brute force approach also needs to keep track of hired workers with extra space.

✨

Key Insights

1Insight 1: Using heaps (priority queues) is crucial for efficiently finding the minimum cost worker in each hiring session among the candidates.
2Insight 2: Maintaining two heaps (one for the beginning and one for the end of the array) allows you to consider candidates from both ends simultaneously without scanning the entire array each time.
3Insight 3: Handling the overlap between the first `candidates` and last `candidates` workers is important to avoid double-counting. This requires careful updating of the left and right pointers used to populate the heaps.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Two Pointers

Using two indices to traverse data from different positions simultaneously.

Edge Cases

Edge case 1: `candidates` is greater than or equal to `costs.length`. In this case, both heaps will contain all the elements, and the algorithm still works correctly because the heap will always prioritize the smallest cost.
Edge case 2: `k` is greater than `costs.length`. The code implicitly handles this by ensuring that the loops terminate when either `k` iterations have completed or all available workers have been considered. The `while` loops within the main loop correctly manage the remaining workers.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not handling the overlapping region between the first and last `candidates` workers correctly. This can lead to the same worker being selected twice, resulting in an incorrect total cost. The code avoids this by carefully updating the `l` and `r` pointers.

✗Common mistake 2: Not breaking ties correctly by considering the index of the worker. This can lead to incorrect selection of workers when multiple workers have the same cost. The use of `(cost, index)` tuples in the heaps ensures that the worker with the smaller index is prioritized in case of a tie in cost.

✗Common mistake 3: Integer overflow. When accumulating the total cost for a large number of workers, the result might exceed the maximum value for an integer. The Java and C++ solutions use 'long' to avoid this.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Resource allocation. Imagine assigning servers to tasks, where `costs` represents the cost of assigning a server to a task. You need to choose the `k` most cost-effective servers from a pool of potential candidates, prioritizing servers with lower utilization (represented by the index).

🏦Real-world scenario 2: Staffing. Consider hiring employees for different projects, where `costs` represents the salary of each candidate. You need to hire `k` employees from a group of applicants, selecting the candidates with the lowest salary from the most promising candidates (first and last candidates), taking into consideration the employee's application order (index).

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

Consider multiple approaches before coding. Mentioning alternatives shows breadth of knowledge.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: October 1, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 13

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 6

❓

Frequently Asked Questions

What is the most efficient way to solve array problems?

Most array problems can be solved efficiently using techniques like two-pointer approach, sliding window, or hash maps. The key is identifying whether the array is sorted (enabling binary search) or if tracking seen elements (using hash maps) can help achieve O(n) time complexity.

When is the two-pointer technique most effective?

Two pointers work best on sorted arrays or when searching for pairs/triplets that satisfy a condition. Common patterns include: moving pointers inward from both ends (like in two-sum variants), fast/slow pointers for cycle detection, or left/right pointers for partitioning.

What techniques are commonly tested in medium-difficulty problems?

Medium problems often require combining multiple techniques: two-pointer with hash maps, BFS/DFS with state tracking, dynamic programming with optimization, or divide-and-conquer approaches. They test your ability to recognize patterns and choose the right combination of techniques.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Total Cost to Hire K Workers"?

For this medium-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether two pointers techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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