Minimum Cost to Hire K Workers

🎯

Minimum Cost to Hire K Workers

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Imagine you are managing a project and need to hire exactly K workers from a pool, each demanding a minimum wage and having a quality rating. You want to pay them fairly based on their quality but minimize the total cost.

💡 This problem combines sorting, ratio comparisons, and heap data structures. Beginners often struggle because it requires understanding how to fairly pay workers based on ratios and then efficiently selecting the best subset using a heap.

📋

Problem Statement

You have n workers. Each worker has a quality and a minimum wage expectation. You want to hire exactly K workers to form a group. To pay them fairly, every worker in the group must be paid at least their minimum wage, and the pay must be proportional to their quality (i.e., if a worker has twice the quality, they get twice the pay). Find the minimum total wage to hire exactly K workers under these conditions.

1 ≤ n ≤ 10^51 ≤ K ≤ n1 ≤ quality[i], wage[i] ≤ 10^4

💡

Example

Input"quality = [10, 20, 5], wage = [70, 50, 30], K = 2"

Output105.00000

Hire workers 0 and 2. Pay ratio is max(wage[i]/quality[i]) = max(70/10, 30/5) = 7. Total cost = 7 * (10 + 5) = 105.

• All workers have the same quality and wage → output is K times the minimum wage
• K equals n → must hire all workers, sum all wages proportionally
• One worker has extremely high wage but low quality → avoid if possible
• Workers with identical wage/quality ratio → any subset with lowest sum quality is best

⚠️

Common Mistakes

Not sorting workers by wage-to-quality ratio

Incorrect calculation of minimum cost, possibly overpaying

✅ Always sort workers by ratio ascending before processing

Using min heap instead of max heap for qualities

Keeps smallest qualities instead of removing largest, leading to incorrect sum

✅ Use max heap to remove largest quality when heap size exceeds K

Calculating cost as sum of wages instead of ratio * sum of qualities

Incorrect total cost calculation, fails fairness condition

✅ Calculate cost as current ratio multiplied by sum of qualities

Not updating minimum cost when heap size equals K

Misses potential minimum cost, returns wrong answer

✅ Update minimum cost only when heap size is exactly K

Integer division instead of floating point for ratio

Incorrect ratio calculation, leading to wrong cost

✅ Use floating point division for wage/quality ratio

🧠

Brute Force (Check All Subsets)

💡 This approach exists to understand the problem's core constraints and why naive enumeration is infeasible. It helps grasp the fairness condition and the ratio concept.

Intuition

Try every combination of K workers, calculate the pay ratio needed to satisfy all, and pick the minimum total cost.

Algorithm

1. Generate all subsets of workers of size K.
2. For each subset, find the maximum wage-to-quality ratio among its workers.
3. Calculate total cost as this ratio multiplied by sum of qualities in the subset.
4. Return the minimum total cost found.

💡 The challenge is understanding the fairness ratio and how it determines the total cost for each subset.

</>

Code

from itertools import combinations

def mincostToHireWorkers(quality, wage, K):
    n = len(quality)
    min_cost = float('inf')
    for group in combinations(range(n), K):
        max_ratio = 0
        total_quality = 0
        for i in group:
            ratio = wage[i] / quality[i]
            if ratio > max_ratio:
                max_ratio = ratio
            total_quality += quality[i]
        cost = max_ratio * total_quality
        if cost < min_cost:
            min_cost = cost
    return min_cost

# Driver code
if __name__ == '__main__':
    quality = [10, 20, 5]
    wage = [70, 50, 30]
    K = 2
    print(mincostToHireWorkers(quality, wage, K))  # Expected: 105.0

Line Notes

from itertools import combinationsImport combinations to generate all subsets of size K, essential for brute force enumeration.

for group in combinations(range(n), K):Enumerate every possible group of K workers to check all combinations.

max_ratio = 0Initialize max ratio for current group to find the fairness ratio that satisfies all workers.

ratio = wage[i] / quality[i]Calculate wage-to-quality ratio for each worker to determine minimum pay ratio.

cost = max_ratio * total_qualityCalculate total cost based on max ratio and sum of qualities, ensuring fairness.

if cost < min_cost:Update minimum cost if current group is cheaper, tracking the best solution.

import java.util.*;

public class Solution {
    public static double mincostToHireWorkers(int[] quality, int[] wage, int K) {
        int n = quality.length;
        double minCost = Double.MAX_VALUE;
        List<Integer> workers = new ArrayList<>();
        for (int i = 0; i < n; i++) workers.add(i);
        List<List<Integer>> combinations = new ArrayList<>();
        combine(workers, K, 0, new ArrayList<>(), combinations);
        for (List<Integer> group : combinations) {
            double maxRatio = 0;
            int totalQuality = 0;
            for (int i : group) {
                double ratio = (double) wage[i] / quality[i];
                if (ratio > maxRatio) maxRatio = ratio;
                totalQuality += quality[i];
            }
            double cost = maxRatio * totalQuality;
            if (cost < minCost) minCost = cost;
        }
        return minCost;
    }

    private static void combine(List<Integer> arr, int k, int start, List<Integer> temp, List<List<Integer>> res) {
        if (temp.size() == k) {
            res.add(new ArrayList<>(temp));
            return;
        }
        for (int i = start; i < arr.size(); i++) {
            temp.add(arr.get(i));
            combine(arr, k, i + 1, temp, res);
            temp.remove(temp.size() - 1);
        }
    }

    public static void main(String[] args) {
        int[] quality = {10, 20, 5};
        int[] wage = {70, 50, 30};
        int K = 2;
        System.out.println(mincostToHireWorkers(quality, wage, K)); // Expected: 105.0
    }
}

Line Notes

List<Integer> workers = new ArrayList<>();Create list of worker indices for generating combinations.

combine(workers, K, 0, new ArrayList<>(), combinations);Generate all subsets of size K recursively to enumerate all groups.

double ratio = (double) wage[i] / quality[i];Calculate wage-to-quality ratio for fairness condition.

double cost = maxRatio * totalQuality;Compute total cost for current group based on max ratio and total quality.

#include <bits/stdc++.h>
using namespace std;

void combine(vector<int>& arr, int k, int start, vector<int>& temp, vector<vector<int>>& res) {
    if ((int)temp.size() == k) {
        res.push_back(temp);
        return;
    }
    for (int i = start; i < (int)arr.size(); i++) {
        temp.push_back(arr[i]);
        combine(arr, k, i + 1, temp, res);
        temp.pop_back();
    }
}

double mincostToHireWorkers(vector<int>& quality, vector<int>& wage, int K) {
    int n = (int)quality.size();
    vector<int> workers(n);
    iota(workers.begin(), workers.end(), 0);
    vector<vector<int>> combinations;
    vector<int> temp;
    combine(workers, K, 0, temp, combinations);
    double minCost = 1e15;
    for (auto& group : combinations) {
        double maxRatio = 0;
        int totalQuality = 0;
        for (int i : group) {
            double ratio = (double)wage[i] / quality[i];
            maxRatio = max(maxRatio, ratio);
            totalQuality += quality[i];
        }
        double cost = maxRatio * totalQuality;
        minCost = min(minCost, cost);
    }
    return minCost;
}

int main() {
    vector<int> quality = {10, 20, 5};
    vector<int> wage = {70, 50, 30};
    int K = 2;
    cout << mincostToHireWorkers(quality, wage, K) << "\n"; // Expected: 105.0
    return 0;
}

Line Notes

iota(workers.begin(), workers.end(), 0);Fill workers vector with indices 0 to n-1 for combination generation.

combine(workers, K, 0, temp, combinations);Generate all combinations of size K recursively.

double ratio = (double)wage[i] / quality[i];Calculate wage-to-quality ratio for fairness.

minCost = min(minCost, cost);Update minimum cost if current group is cheaper.

function combine(arr, k, start, temp, res) {
    if (temp.length === k) {
        res.push([...temp]);
        return;
    }
    for (let i = start; i < arr.length; i++) {
        temp.push(arr[i]);
        combine(arr, k, i + 1, temp, res);
        temp.pop();
    }
}

function mincostToHireWorkers(quality, wage, K) {
    const n = quality.length;
    const workers = Array.from({length: n}, (_, i) => i);
    const combinations = [];
    combine(workers, K, 0, [], combinations);
    let minCost = Infinity;
    for (const group of combinations) {
        let maxRatio = 0;
        let totalQuality = 0;
        for (const i of group) {
            const ratio = wage[i] / quality[i];
            if (ratio > maxRatio) maxRatio = ratio;
            totalQuality += quality[i];
        }
        const cost = maxRatio * totalQuality;
        if (cost < minCost) minCost = cost;
    }
    return minCost;
}

// Driver code
const quality = [10, 20, 5];
const wage = [70, 50, 30];
const K = 2;
console.log(mincostToHireWorkers(quality, wage, K)); // Expected: 105.0

Line Notes

const workers = Array.from({length: n}, (_, i) => i);Create array of worker indices for combination generation.

combine(workers, K, 0, [], combinations);Generate all subsets of size K recursively.

const ratio = wage[i] / quality[i];Calculate wage-to-quality ratio for fairness.

if (cost < minCost) minCost = cost;Update minimum cost if current group is cheaper.

⏱

Complexity

TimeO(n choose K * K)

SpaceO(n choose K * K)

We generate all subsets of size K (n choose K), and for each subset, we compute max ratio and sum quality in O(K). This is exponential and infeasible for large n.

💡 For n=20 and K=10, this means over 184,000 subsets, each requiring work, which is too slow for interviews.

Interview Verdict: TLE

This approach is too slow for large inputs but helps understand the problem and fairness condition.

🧠

Sorting by Ratio + Heap to Maintain K Workers

💡 This approach introduces sorting by wage-to-quality ratio and uses a max heap to efficiently maintain the group with minimum total quality, which reduces cost.

Intuition

Sort workers by their wage-to-quality ratio. For each ratio, consider hiring workers with ratio less or equal to it. Use a max heap to keep track of the smallest total quality sum for K workers to minimize cost.

Algorithm

1. Calculate wage-to-quality ratio for each worker and sort workers by this ratio ascending.
2. Iterate over sorted workers, adding their quality to a max heap and maintaining sum of qualities.
3. If heap size exceeds K, remove the worker with highest quality to minimize total quality sum.
4. When heap size is exactly K, calculate cost as current ratio * sum of qualities and update minimum cost.

💡 The key insight is that the ratio of the current worker sets the pay ratio for the group, and we want to minimize sum of qualities for that ratio.

</>

Code

import heapq

def mincostToHireWorkers(quality, wage, K):
    workers = sorted([(w/q, q) for q, w in zip(quality, wage)])
    heap = []
    sum_quality = 0
    min_cost = float('inf')
    for ratio, q in workers:
        heapq.heappush(heap, -q)  # max heap by pushing negative
        sum_quality += q
        if len(heap) > K:
            removed = -heapq.heappop(heap)
            sum_quality -= removed
        if len(heap) == K:
            cost = ratio * sum_quality
            if cost < min_cost:
                min_cost = cost
    return min_cost

# Driver code
if __name__ == '__main__':
    quality = [10, 20, 5]
    wage = [70, 50, 30]
    K = 2
    print(mincostToHireWorkers(quality, wage, K))  # Expected: 105.0

Line Notes

workers = sorted([(w/q, q) for q, w in zip(quality, wage)])Sort workers by wage-to-quality ratio ascending to consider pay thresholds in order.

heapq.heappush(heap, -q)Push negative quality to simulate max heap, so largest quality can be removed efficiently.

if len(heap) > K:Remove largest quality worker to keep only K workers with smallest total quality.

cost = ratio * sum_qualityCalculate total cost with current ratio and sum of qualities, ensuring fairness.

if cost < min_cost:Update minimum cost if current group is cheaper, tracking best solution.

import java.util.*;

public class Solution {
    public static double mincostToHireWorkers(int[] quality, int[] wage, int K) {
        int n = quality.length;
        double[][] workers = new double[n][2];
        for (int i = 0; i < n; i++) {
            workers[i][0] = (double) wage[i] / quality[i];
            workers[i][1] = quality[i];
        }
        Arrays.sort(workers, Comparator.comparingDouble(a -> a[0]));
        PriorityQueue<Integer> maxHeap = new PriorityQueue<>(Collections.reverseOrder());
        int sumQuality = 0;
        double minCost = Double.MAX_VALUE;
        for (double[] worker : workers) {
            maxHeap.offer((int) worker[1]);
            sumQuality += (int) worker[1];
            if (maxHeap.size() > K) {
                sumQuality -= maxHeap.poll();
            }
            if (maxHeap.size() == K) {
                double cost = worker[0] * sumQuality;
                if (cost < minCost) minCost = cost;
            }
        }
        return minCost;
    }

    public static void main(String[] args) {
        int[] quality = {10, 20, 5};
        int[] wage = {70, 50, 30};
        int K = 2;
        System.out.println(mincostToHireWorkers(quality, wage, K)); // Expected: 105.0
    }
}

Line Notes

workers[i][0] = (double) wage[i] / quality[i];Calculate ratio for each worker to determine pay threshold.

Arrays.sort(workers, Comparator.comparingDouble(a -> a[0]));Sort workers by ratio ascending to process in order of increasing pay ratio.

PriorityQueue<Integer> maxHeap = new PriorityQueue<>(Collections.reverseOrder());Max heap to track largest quality for efficient removal.

if (maxHeap.size() > K) { sumQuality -= maxHeap.poll(); }Remove largest quality to keep only K workers with smallest total quality.

if (maxHeap.size() == K) {Calculate cost only when exactly K workers are selected.

#include <bits/stdc++.h>
using namespace std;

double mincostToHireWorkers(vector<int>& quality, vector<int>& wage, int K) {
    int n = (int)quality.size();
    vector<pair<double, int>> workers(n);
    for (int i = 0; i < n; i++) {
        workers[i] = {(double)wage[i] / quality[i], quality[i]};
    }
    sort(workers.begin(), workers.end());
    priority_queue<int> maxHeap;
    int sumQuality = 0;
    double minCost = 1e15;
    for (auto& w : workers) {
        maxHeap.push(w.second);
        sumQuality += w.second;
        if ((int)maxHeap.size() > K) {
            sumQuality -= maxHeap.top();
            maxHeap.pop();
        }
        if ((int)maxHeap.size() == K) {
            double cost = w.first * sumQuality;
            minCost = min(minCost, cost);
        }
    }
    return minCost;
}

int main() {
    vector<int> quality = {10, 20, 5};
    vector<int> wage = {70, 50, 30};
    int K = 2;
    cout << mincostToHireWorkers(quality, wage, K) << "\n"; // Expected: 105.0
    return 0;
}

Line Notes

workers[i] = {(double)wage[i] / quality[i], quality[i]};Store ratio and quality for sorting and processing.

sort(workers.begin(), workers.end());Sort workers by ratio ascending to consider pay thresholds in order.

priority_queue<int> maxHeap;Max heap to track largest quality for efficient removal.

if ((int)maxHeap.size() > K) { sumQuality -= maxHeap.top(); maxHeap.pop(); }Remove largest quality to maintain K workers with smallest total quality.

if ((int)maxHeap.size() == K) {Calculate cost only when exactly K workers are selected.

function mincostToHireWorkers(quality, wage, K) {
    const workers = quality.map((q, i) => [wage[i] / q, q]);
    workers.sort((a, b) => a[0] - b[0]);
    const maxHeap = [];
    let sumQuality = 0;
    let minCost = Infinity;

    function heapPush(val) {
        maxHeap.push(val);
        let i = maxHeap.length - 1;
        while (i > 0) {
            let parent = Math.floor((i - 1) / 2);
            if (maxHeap[parent] >= maxHeap[i]) break;
            [maxHeap[parent], maxHeap[i]] = [maxHeap[i], maxHeap[parent]];
            i = parent;
        }
    }

    function heapPop() {
        const top = maxHeap[0];
        const end = maxHeap.pop();
        if (maxHeap.length > 0) {
            maxHeap[0] = end;
            let i = 0;
            while (true) {
                let left = 2 * i + 1, right = 2 * i + 2, largest = i;
                if (left < maxHeap.length && maxHeap[left] > maxHeap[largest]) largest = left;
                if (right < maxHeap.length && maxHeap[right] > maxHeap[largest]) largest = right;
                if (largest === i) break;
                [maxHeap[i], maxHeap[largest]] = [maxHeap[largest], maxHeap[i]];
                i = largest;
            }
        }
        return top;
    }

    for (const [ratio, q] of workers) {
        heapPush(q);
        sumQuality += q;
        if (maxHeap.length > K) {
            sumQuality -= heapPop();
        }
        if (maxHeap.length === K) {
            const cost = ratio * sumQuality;
            if (cost < minCost) minCost = cost;
        }
    }
    return minCost;
}

// Driver code
const quality = [10, 20, 5];
const wage = [70, 50, 30];
const K = 2;
console.log(mincostToHireWorkers(quality, wage, K)); // Expected: 105.0

Line Notes

const workers = quality.map((q, i) => [wage[i] / q, q]);Calculate ratio and quality pairs for sorting and processing.

workers.sort((a, b) => a[0] - b[0]);Sort workers by ratio ascending to consider pay thresholds in order.

function heapPush(val)Insert quality into max heap to track largest quality for efficient removal.

if (maxHeap.length > K) { sumQuality -= heapPop(); }Remove largest quality to keep heap size K and minimize total quality.

if (maxHeap.length === K) {Calculate cost only when exactly K workers are selected.

⏱

Complexity

TimeO(n log n)

SpaceO(n)

Sorting takes O(n log n). Each worker is pushed and popped at most once from the heap, so heap operations total O(n log K). Overall O(n log n).

💡 For n=10^5, this approach is efficient and suitable for interviews.

Interview Verdict: Accepted

This is the optimal approach commonly expected in interviews for this problem.

🧠

Sorting by Ratio + Balanced Tree / Multiset Instead of Heap

💡 This approach is a variant using balanced trees or multisets to maintain the K smallest qualities instead of a heap, useful in languages without built-in heaps or when needing ordered data.

Intuition

Similar to the heap approach, but use a balanced tree to insert and remove qualities efficiently, maintaining the sum and size constraints.

Algorithm

1. Calculate ratio and quality pairs and sort by ratio ascending.
2. Iterate over workers, insert quality into balanced tree and update sum.
3. If size exceeds K, remove largest quality from tree and update sum.
4. When size is K, calculate cost and update minimum.

💡 The logic is the same as heap but uses a different data structure to maintain order.

</>

Code

import bisect

def mincostToHireWorkers(quality, wage, K):
    workers = sorted([(w/q, q) for q, w in zip(quality, wage)])
    qualities = []
    sum_quality = 0
    min_cost = float('inf')
    for ratio, q in workers:
        bisect.insort(qualities, q)
        sum_quality += q
        if len(qualities) > K:
            removed = qualities.pop()  # largest quality
            sum_quality -= removed
        if len(qualities) == K:
            cost = ratio * sum_quality
            if cost < min_cost:
                min_cost = cost
    return min_cost

# Driver code
if __name__ == '__main__':
    quality = [10, 20, 5]
    wage = [70, 50, 30]
    K = 2
    print(mincostToHireWorkers(quality, wage, K))  # Expected: 105.0

Line Notes

bisect.insort(qualities, q)Insert quality in sorted order to maintain balanced tree behavior for efficient largest removal.

removed = qualities.pop()Remove largest quality to keep size K and minimize total quality sum.

cost = ratio * sum_qualityCalculate total cost for current ratio and sum of qualities.

if cost < min_cost:Update minimum cost if current group is cheaper.

import java.util.*;

public class Solution {
    public static double mincostToHireWorkers(int[] quality, int[] wage, int K) {
        int n = quality.length;
        double[][] workers = new double[n][2];
        for (int i = 0; i < n; i++) {
            workers[i][0] = (double) wage[i] / quality[i];
            workers[i][1] = quality[i];
        }
        Arrays.sort(workers, Comparator.comparingDouble(a -> a[0]));
        TreeSet<Integer> qualities = new TreeSet<>(Collections.reverseOrder());
        int sumQuality = 0;
        double minCost = Double.MAX_VALUE;
        for (double[] worker : workers) {
            qualities.add((int) worker[1]);
            sumQuality += (int) worker[1];
            if (qualities.size() > K) {
                int largest = qualities.first();
                qualities.remove(largest);
                sumQuality -= largest;
            }
            if (qualities.size() == K) {
                double cost = worker[0] * sumQuality;
                if (cost < minCost) minCost = cost;
            }
        }
        return minCost;
    }

    public static void main(String[] args) {
        int[] quality = {10, 20, 5};
        int[] wage = {70, 50, 30};
        int K = 2;
        System.out.println(mincostToHireWorkers(quality, wage, K)); // Expected: 105.0
    }
}

Line Notes

TreeSet<Integer> qualities = new TreeSet<>(Collections.reverseOrder());Use TreeSet as balanced tree to maintain qualities in descending order for efficient largest removal.

int largest = qualities.first();Get largest quality to remove when size exceeds K.

qualities.remove(largest);Remove largest quality to maintain size K and minimize total quality.

double cost = worker[0] * sumQuality;Calculate total cost for current ratio and sum of qualities.

#include <bits/stdc++.h>
using namespace std;

double mincostToHireWorkers(vector<int>& quality, vector<int>& wage, int K) {
    int n = (int)quality.size();
    vector<pair<double, int>> workers(n);
    for (int i = 0; i < n; i++) {
        workers[i] = {(double)wage[i] / quality[i], quality[i]};
    }
    sort(workers.begin(), workers.end());
    multiset<int, greater<int>> qualities;
    int sumQuality = 0;
    double minCost = 1e15;
    for (auto& w : workers) {
        qualities.insert(w.second);
        sumQuality += w.second;
        if ((int)qualities.size() > K) {
            auto it = qualities.begin();
            sumQuality -= *it;
            qualities.erase(it);
        }
        if ((int)qualities.size() == K) {
            double cost = w.first * sumQuality;
            minCost = min(minCost, cost);
        }
    }
    return minCost;
}

int main() {
    vector<int> quality = {10, 20, 5};
    vector<int> wage = {70, 50, 30};
    int K = 2;
    cout << mincostToHireWorkers(quality, wage, K) << "\n"; // Expected: 105.0
    return 0;
}

Line Notes

multiset<int, greater<int>> qualities;Use multiset as balanced tree sorted descending to maintain qualities.

auto it = qualities.begin();Iterator to largest quality to remove when size exceeds K.

sumQuality -= *it;Update sum after removing largest quality to minimize total cost.

minCost = min(minCost, cost);Update minimum cost if current group is cheaper.

// JavaScript does not have built-in balanced tree, so this approach is less common
// but can be simulated with sorted array and binary search (inefficient for large n)

function bisectInsert(arr, val) {
    let left = 0, right = arr.length;
    while (left < right) {
        let mid = Math.floor((left + right) / 2);
        if (arr[mid] < val) left = mid + 1;
        else right = mid;
    }
    arr.splice(left, 0, val);
}

function mincostToHireWorkers(quality, wage, K) {
    const workers = quality.map((q, i) => [wage[i] / q, q]);
    workers.sort((a, b) => a[0] - b[0]);
    const qualities = [];
    let sumQuality = 0;
    let minCost = Infinity;
    for (const [ratio, q] of workers) {
        bisectInsert(qualities, q);
        sumQuality += q;
        if (qualities.length > K) {
            const removed = qualities.pop();
            sumQuality -= removed;
        }
        if (qualities.length === K) {
            const cost = ratio * sumQuality;
            if (cost < minCost) minCost = cost;
        }
    }
    return minCost;
}

// Driver code
const quality = [10, 20, 5];
const wage = [70, 50, 30];
const K = 2;
console.log(mincostToHireWorkers(quality, wage, K)); // Expected: 105.0

Line Notes

function bisectInsert(arr, val)Insert value into sorted array to simulate balanced tree insertion for order maintenance.

qualities.pop()Remove largest quality (last element) to maintain size K and minimize total quality.

const cost = ratio * sumQuality;Calculate total cost for current ratio and sum of qualities.

if (cost < minCost) minCost = cost;Update minimum cost if current group is cheaper.

⏱

Complexity

TimeO(n log n) for sorting + O(n log K) for balanced tree operations

SpaceO(n)

Balanced tree insertions and removals take O(log K) each, done n times after sorting.

💡 This approach is similar in complexity to the heap approach but uses different data structures.

Interview Verdict: Accepted

This is a valid alternative approach, especially in languages with balanced tree support.

📊

All Approaches - One-Glance Tradeoffs

💡 Use the heap-based sorting approach in 95% of interviews for efficiency and clarity.

Approach	Time	Space	Stack Risk	Reconstruct	Use In Interview
1. Brute Force	O(n choose K * K)	O(n choose K * K)	Yes (deep recursion in combinations)	Yes	Mention only - never code due to inefficiency
2. Sorting + Max Heap	O(n log n)	O(n)	No	Yes	Optimal approach to code in interviews
3. Sorting + Balanced Tree / Multiset	O(n log n)	O(n)	No	Yes	Alternative approach if language supports balanced trees

💼

Interview Strategy

💡 Use this guide to understand the problem deeply, practice the brute force to grasp the fairness condition, then master the heap-based optimal approach for interviews.

How to Present

Step 1: Clarify the problem and fairness condition (pay proportional to quality, at least minimum wage).Step 2: Explain brute force approach to show understanding of constraints.Step 3: Introduce sorting by ratio and using a heap to maintain K workers with smallest total quality.Step 4: Code the heap-based solution and test with examples.Step 5: Discuss edge cases and possible follow-ups.

Time Allocation

Clarify: 3min → Approach: 5min → Code: 10min → Test: 5min. Total ~23min

What the Interviewer Tests

Interviewer tests your understanding of ratio-based fairness, ability to optimize brute force with sorting and heaps, and handling large inputs efficiently.

Common Follow-ups

• What if we want to hire at least K workers instead of exactly K? → Modify heap logic to consider all sizes ≥ K.
• Can you explain why sorting by ratio works? → Because the highest ratio worker sets the pay ratio for the group.

💡 These follow-ups test deeper understanding of the ratio concept and flexibility in problem constraints.

🔍

Pattern Recognition

When to Use

1) Need to select K elements minimizing cost, 2) Cost depends on ratio of two arrays, 3) Fairness or proportional payment constraints, 4) Efficient selection with heap or sorting.

Signature Phrases

'pay every worker at least their minimum wage''pay is proportional to quality''minimum total wage to hire K workers'

NOT This Pattern When

Problems that only require sorting or simple heaps without ratio or proportional constraints.

Similar Problems

K Closest Points to Origin - uses sorting by ratio and heapTop K Frequent Elements - uses heap to maintain top kTask Scheduler - advanced scheduling with priority queues

Practice

1. Consider the following Python code implementing a frequency stack using a max-heap with timestamp. After executing the sequence of operations below, what is the output of the last pop() call?

fs = FreqStack()
fs.push(1)
fs.push(2)
fs.push(1)
print(fs.pop())
print(fs.pop())

easy

A. 1 and then 1

B. 2 and then 1

C. 1 and then 2

D. 2 and then 2

Solution

1. Step 1: Trace push operations

   Push 1: freq[1]=1, time=1, heap push (-1, -1, 1) Push 2: freq[2]=1, time=2, heap push (-1, -2, 2) Push 1: freq[1]=2, time=3, heap push (-2, -3, 1)

2. Step 2: Trace pop operations

   First pop: heap pops (-2, -3, 1) -> returns 1, freq[1]=1 Second pop: heap now has (-1, -2, 2) and (-1, -1, 1). Max freq is 1, tie broken by time: -2 > -1, so pop (-1, -2, 2) returns 2.

3. Final Answer:

   Option C -> Option C

4. Quick Check:

   Most frequent element 1 popped first, then next most recent with freq 1 is 2 [OK]

Hint: Most frequent and most recent element pops first [OK]

Common Mistakes:

• Ignoring timestamp order in heap
• Mixing up pop order when frequencies tie

2. Given the following Python code implementing Quickselect to find k closest points, what is the final returned list when called with points = [[1,3],[-2,2],[2,-2]] and k = 2?

from typing import List
import random

def kClosest(points: List[List[int]], k: int) -> List[List[int]]:
    def dist(i):
        return points[i][0]*points[i][0] + points[i][1]*points[i][1]

    def partition(left, right, pivot_index):
        pivot_dist = dist(pivot_index)
        points[pivot_index], points[right] = points[right], points[pivot_index]
        store_index = left
        for i in range(left, right):
            if dist(i) < pivot_dist:
                points[store_index], points[i] = points[i], points[store_index]
                store_index += 1
        points[right], points[store_index] = points[store_index], points[right]
        return store_index

    def quickselect(left, right, k_smallest):
        if left == right:
            return
        pivot_index = random.randint(left, right)
        pivot_index = partition(left, right, pivot_index)
        if k_smallest == pivot_index:
            return
        elif k_smallest < pivot_index:
            quickselect(left, pivot_index - 1, k_smallest)
        else:
            quickselect(pivot_index + 1, right, k_smallest)

    quickselect(0, len(points) - 1, k - 1)
    return points[:k]

print(kClosest(points, k))

easy

A. [[2,-2], [1,3]]

B. [[1,3], [-2,2]]

C. [[-2,2], [2,-2]]

D. [[-2,2], [1,3]]

Solution

1. Step 1: Calculate squared distances

   Distances: [1,3] -> 10, [-2,2] -> 8, [2,-2] -> 8.

2. Step 2: Quickselect partitions to find 2 closest points

   Points with distance 8 are closest, so the two closest points are [-2,2] and [2,-2].

3. Final Answer:

   Option C -> Option C

4. Quick Check:

   Returned points have smallest squared distances [OK]

Hint: Squared distances avoid sqrt and preserve order [OK]

Common Mistakes:

• Confusing order of points after partition
• Off-by-one errors in slicing points[:k]

3. Consider the following code snippet for finding the kth largest XOR coordinate value. Given the input matrix = [[5, 2], [1, 6]] and k = 3, what is the value of min_heap[0] returned by the function?

easy

A. 4

B. 5

C. 6

D. 7

Solution

1. Step 1: Compute prefix XOR matrix values

   prefix[1][1] = 5, prefix[1][2] = 7 (5^2), prefix[2][1] = 4 (5^1^0), prefix[2][2] = 0 (prefix[1][2]^prefix[2][1]^prefix[1][1]^matrix[1][1] = 7^4^5^6 = 0)

2. Step 2: Track min_heap of size k=3

   Values inserted: 5, 7, 4. The heap after these is [4,7,5]. The next value 0 is less than min_heap[0]=4, so ignored. The kth largest is min_heap[0] = 4.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Heap after all insertions has 4 as smallest among top 3 [OK]

Hint: Trace prefix XOR and heap insertions carefully [OK]

Common Mistakes:

• Off-by-one in prefix indices
• Miscomputing XOR values
• Confusing heap root with max value

4. You are given a list of words and need to find the top k words that appear most frequently, with ties broken by lexicographical order. Which algorithmic approach guarantees an optimal solution in terms of time complexity?

easy

A. Sort all unique words by frequency descending and lex order ascending, then pick top k

B. Use a dynamic programming approach to count frequencies and select top k

C. Use a greedy approach by repeatedly picking the most frequent word until k are chosen

D. Use bucket sort by frequency, then sort words lexicographically within each bucket, and collect top k

Solution

1. Step 1: Understand frequency distribution

   Counting frequencies is straightforward, but sorting all words fully costs O(n log n).

2. Step 2: Use bucket sort to group words by frequency

   Bucket sort groups words by frequency in O(n), then sorting lex order within buckets is efficient since buckets are smaller.

3. Final Answer:

   Option D -> Option D

4. Quick Check:

   Bucket sort + lex sort within buckets achieves O(n) time [OK]

Hint: Bucket sort by frequency then lex sort is optimal [OK]

Common Mistakes:

• Assuming full sort is optimal
• Using DP for frequency counting
• Greedy without sorting fails tie-break

5. Suppose the problem changes: now elements can be reused infinitely many times, and you want the top k frequent elements considering this reuse. Which modification to the optimal bucket sort approach correctly handles this scenario?

hard

A. Use a min-heap of size k to dynamically track frequencies as elements can be reused infinitely.

B. Modify frequency counting to consider reuse by multiplying counts, then bucket sort as usual.

C. No change needed; bucket sort works the same since frequencies are static counts.

D. Switch to a max-heap approach to handle infinite reuse frequencies efficiently.

Solution

1. Step 1: Understand infinite reuse impact

   Infinite reuse means frequencies are dynamic or unbounded, so static frequency counting and bucket sort no longer apply.

2. Step 2: Choose data structure for dynamic top k

   A min-heap of size k can maintain the top k elements dynamically as frequencies update, suitable for infinite reuse scenarios.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Bucket sort relies on fixed frequencies; min-heap adapts to changing frequencies [OK]

Hint: Bucket sort assumes fixed frequencies; min-heap handles dynamic updates [OK]

Common Mistakes:

• Assuming bucket sort works unchanged
• Using max-heap which grows too large