LeetCode #857 — HARD

Minimum Cost to Hire K Workers

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

The Problem

Problem Statement

There are n workers. You are given two integer arrays quality and wage where quality[i] is the quality of the i^th worker and wage[i] is the minimum wage expectation for the i^th worker.

We want to hire exactly k workers to form a paid group. To hire a group of k workers, we must pay them according to the following rules:

Every worker in the paid group must be paid at least their minimum wage expectation.
In the group, each worker's pay must be directly proportional to their quality. This means if a worker’s quality is double that of another worker in the group, then they must be paid twice as much as the other worker.

Given the integer k, return the least amount of money needed to form a paid group satisfying the above conditions. Answers within 10^-5 of the actual answer will be accepted.

Example 1:

Input: quality = [10,20,5], wage = [70,50,30], k = 2
Output: 105.00000
Explanation: We pay 70 to 0^th worker and 35 to 2^nd worker.

Example 2:

Input: quality = [3,1,10,10,1], wage = [4,8,2,2,7], k = 3
Output: 30.66667
Explanation: We pay 4 to 0^th worker, 13.33333 to 2^nd and 3^rd workers separately.

Constraints:

n == quality.length == wage.length
1 <= k <= n <= 10⁴
1 <= quality[i], wage[i] <= 10⁴

Step 01

Brute Force Baseline

Problem summary: There are n workers. You are given two integer arrays quality and wage where quality[i] is the quality of the ith worker and wage[i] is the minimum wage expectation for the ith worker. We want to hire exactly k workers to form a paid group. To hire a group of k workers, we must pay them according to the following rules: Every worker in the paid group must be paid at least their minimum wage expectation. In the group, each worker's pay must be directly proportional to their quality. This means if a worker’s quality is double that of another worker in the group, then they must be paid twice as much as the other worker. Given the integer k, return the least amount of money needed to form a paid group satisfying the above conditions. Answers within 10-5 of the actual answer will be accepted.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Greedy

Example 1

[10,20,5]
[70,50,30]
2

Example 2

[3,1,10,10,1]
[4,8,2,2,7]
3

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Largest constraint values

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #857: Minimum Cost to Hire K Workers
class Solution {
    public double mincostToHireWorkers(int[] quality, int[] wage, int k) {
        int n = quality.length;
        Pair<Double, Integer>[] t = new Pair[n];
        for (int i = 0; i < n; ++i) {
            t[i] = new Pair<>((double) wage[i] / quality[i], quality[i]);
        }
        Arrays.sort(t, (a, b) -> Double.compare(a.getKey(), b.getKey()));
        PriorityQueue<Integer> pq = new PriorityQueue<>((a, b) -> b - a);
        double ans = 1e18;
        int tot = 0;
        for (var e : t) {
            tot += e.getValue();
            pq.offer(e.getValue());
            if (pq.size() == k) {
                ans = Math.min(ans, tot * e.getKey());
                tot -= pq.poll();
            }
        }
        return ans;
    }
}

// Accepted solution for LeetCode #857: Minimum Cost to Hire K Workers
func mincostToHireWorkers(quality []int, wage []int, k int) float64 {
	t := []pair{}
	for i, q := range quality {
		t = append(t, pair{float64(wage[i]) / float64(q), q})
	}
	sort.Slice(t, func(i, j int) bool { return t[i].x < t[j].x })
	tot := 0
	var ans float64 = 1e18
	pq := hp{}
	for _, e := range t {
		tot += e.q
		heap.Push(&pq, e.q)
		if pq.Len() == k {
			ans = min(ans, float64(tot)*e.x)
			tot -= heap.Pop(&pq).(int)
		}
	}
	return ans
}

type pair struct {
	x float64
	q int
}

type hp struct{ sort.IntSlice }

func (h *hp) Push(v any) { h.IntSlice = append(h.IntSlice, v.(int)) }
func (h *hp) Pop() any {
	a := h.IntSlice
	v := a[len(a)-1]
	h.IntSlice = a[:len(a)-1]
	return v
}
func (h *hp) Less(i, j int) bool { return h.IntSlice[i] > h.IntSlice[j] }

# Accepted solution for LeetCode #857: Minimum Cost to Hire K Workers
class Solution:
    def mincostToHireWorkers(
        self, quality: List[int], wage: List[int], k: int
    ) -> float:
        t = sorted(zip(quality, wage), key=lambda x: x[1] / x[0])
        ans, tot = inf, 0
        h = []
        for q, w in t:
            tot += q
            heappush(h, -q)
            if len(h) == k:
                ans = min(ans, w / q * tot)
                tot += heappop(h)
        return ans

// Accepted solution for LeetCode #857: Minimum Cost to Hire K Workers
struct Solution;

use std::collections::BinaryHeap;

impl Solution {
    fn mincost_to_hire_workers(quality: Vec<i32>, wage: Vec<i32>, k: i32) -> f64 {
        let k = k as usize;
        let n = quality.len();
        let mut workers: Vec<usize> = (0..n).collect();
        let rate: Vec<f64> = (0..n).map(|i| wage[i] as f64 / quality[i] as f64).collect();
        workers.sort_by(|&i, &j| rate[i].partial_cmp(&rate[j]).unwrap());
        let mut res = std::f64::MAX;
        let mut queue: BinaryHeap<i32> = BinaryHeap::new();
        let mut sum = 0;
        for i in workers {
            let r = rate[i];
            let q = quality[i];
            queue.push(q);
            sum += q;
            if queue.len() > k {
                sum -= queue.pop().unwrap();
            }
            if queue.len() == k {
                res = res.min(r * sum as f64);
            }
        }
        res
    }
}

#[test]
fn test() {
    use assert_approx_eq::assert_approx_eq;
    let quality = vec![10, 20, 5];
    let wage = vec![70, 50, 30];
    let k = 2;
    let res = 105.00000;
    assert_approx_eq!(Solution::mincost_to_hire_workers(quality, wage, k), res);
    let quality = vec![3, 1, 10, 10, 1];
    let wage = vec![4, 8, 2, 2, 7];
    let k = 3;
    let res = 30.666667;
    assert_approx_eq!(Solution::mincost_to_hire_workers(quality, wage, k), res);
}

// Accepted solution for LeetCode #857: Minimum Cost to Hire K Workers
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #857: Minimum Cost to Hire K Workers
// class Solution {
//     public double mincostToHireWorkers(int[] quality, int[] wage, int k) {
//         int n = quality.length;
//         Pair<Double, Integer>[] t = new Pair[n];
//         for (int i = 0; i < n; ++i) {
//             t[i] = new Pair<>((double) wage[i] / quality[i], quality[i]);
//         }
//         Arrays.sort(t, (a, b) -> Double.compare(a.getKey(), b.getKey()));
//         PriorityQueue<Integer> pq = new PriorityQueue<>((a, b) -> b - a);
//         double ans = 1e18;
//         int tot = 0;
//         for (var e : t) {
//             tot += e.getValue();
//             pq.offer(e.getValue());
//             if (pq.size() == k) {
//                 ans = Math.min(ans, tot * e.getKey());
//                 tot -= pq.poll();
//             }
//         }
//         return ans;
//     }
// }

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n log n)

Space

O(1)

Approach Breakdown

EXHAUSTIVE

O(2ⁿ) time

O(n) space

Try every possible combination of choices. With n items each having two states (include/exclude), the search space is 2ⁿ. Evaluating each combination takes O(n), giving O(n × 2ⁿ). The recursion stack or subset storage uses O(n) space.

GREEDY

O(n log n) time

O(1) space

Greedy algorithms typically sort the input (O(n log n)) then make a single pass (O(n)). The sort dominates. If the input is already sorted or the greedy choice can be computed without sorting, time drops to O(n). Proving greedy correctness (exchange argument) is harder than the implementation.

Shortcut: Sort + single pass → O(n log n). If no sort needed → O(n). The hard part is proving it works.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Using greedy without proof

Wrong move: Locally optimal choices may fail globally.

Usually fails on: Counterexamples appear on crafted input orderings.

Fix: Verify with exchange argument or monotonic objective before committing.