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.
Move from brute-force thinking to an efficient approach using array strategy.
You are given a 0-indexed integer array costs where costs[i] is the cost of hiring the ith worker.
You are also given two integers k and candidates. We want to hire exactly k workers according to the following rules:
k sessions and hire exactly one worker in each session.candidates workers or the last candidates workers. Break the tie by the smallest index.
costs = [3,2,7,7,1,2] and candidates = 2, then in the first hiring session, we will choose the 4th worker because they have the lowest cost [3,2,7,7,1,2].1st worker because they have the same lowest cost as 4th worker but they have the smallest index [3,2,7,7,2]. Please note that the indexing may be changed in the process.Return the total cost to hire exactly k workers.
Example 1:
Input: costs = [17,12,10,2,7,2,11,20,8], k = 3, candidates = 4 Output: 11 Explanation: We hire 3 workers in total. The total cost is initially 0. - In the first hiring round we choose the worker from [17,12,10,2,7,2,11,20,8]. The lowest cost is 2, and we break the tie by the smallest index, which is 3. The total cost = 0 + 2 = 2. - In the second hiring round we choose the worker from [17,12,10,7,2,11,20,8]. The lowest cost is 2 (index 4). The total cost = 2 + 2 = 4. - In the third hiring round we choose the worker from [17,12,10,7,11,20,8]. The lowest cost is 7 (index 3). The total cost = 4 + 7 = 11. Notice that the worker with index 3 was common in the first and last four workers. The total hiring cost is 11.
Example 2:
Input: costs = [1,2,4,1], k = 3, candidates = 3 Output: 4 Explanation: We hire 3 workers in total. The total cost is initially 0. - In the first hiring round we choose the worker from [1,2,4,1]. The lowest cost is 1, and we break the tie by the smallest index, which is 0. The total cost = 0 + 1 = 1. Notice that workers with index 1 and 2 are common in the first and last 3 workers. - In the second hiring round we choose the worker from [2,4,1]. The lowest cost is 1 (index 2). The total cost = 1 + 1 = 2. - In the third hiring round there are less than three candidates. We choose the worker from the remaining workers [2,4]. The lowest cost is 2 (index 0). The total cost = 2 + 2 = 4. The total hiring cost is 4.
Constraints:
1 <= costs.length <= 105 1 <= costs[i] <= 1051 <= k, candidates <= costs.lengthProblem summary: You are given a 0-indexed integer array costs where costs[i] is the cost of hiring the ith 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] and candidates = 2, then in the first hiring session, we will choose the 4th worker because they have the lowest cost [3,2,7,7,1,2]. In the second hiring session, we will choose 1st worker because they have the same lowest cost as 4th worker but they have the smallest index [3,2,7,7,2]. Please note that the indexing may be changed in the process. If there are fewer than candidates workers remaining,
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Array · Two Pointers
[17,12,10,2,7,2,11,20,8] 3 4
[1,2,4,1] 3 3
meeting-rooms-ii)time-to-cross-a-bridge)Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #2462: Total Cost to Hire K Workers
class Solution {
public long totalCost(int[] costs, int k, int candidates) {
int n = costs.length;
long ans = 0;
if (candidates * 2 >= n) {
Arrays.sort(costs);
for (int i = 0; i < k; ++i) {
ans += costs[i];
}
return ans;
}
PriorityQueue<int[]> pq
= new PriorityQueue<>((a, b) -> a[0] == b[0] ? a[1] - b[1] : a[0] - b[0]);
for (int i = 0; i < candidates; ++i) {
pq.offer(new int[] {costs[i], i});
pq.offer(new int[] {costs[n - i - 1], n - i - 1});
}
int l = candidates, r = n - candidates - 1;
while (k-- > 0) {
var p = pq.poll();
ans += p[0];
if (l > r) {
continue;
}
if (p[1] < l) {
pq.offer(new int[] {costs[l], l++});
} else {
pq.offer(new int[] {costs[r], r--});
}
}
return ans;
}
}
// Accepted solution for LeetCode #2462: Total Cost to Hire K Workers
func totalCost(costs []int, k int, candidates int) (ans int64) {
n := len(costs)
if candidates*2 > n {
sort.Ints(costs)
for _, x := range costs[:k] {
ans += int64(x)
}
return
}
pq := hp{}
for i, x := range costs[:candidates] {
heap.Push(&pq, pair{x, i})
heap.Push(&pq, pair{costs[n-i-1], n - i - 1})
}
l, r := candidates, n-candidates-1
for ; k > 0; k-- {
p := heap.Pop(&pq).(pair)
ans += int64(p.cost)
if l > r {
continue
}
if p.i < l {
heap.Push(&pq, pair{costs[l], l})
l++
} else {
heap.Push(&pq, pair{costs[r], r})
r--
}
}
return
}
type pair struct{ cost, i int }
type hp []pair
func (h hp) Len() int { return len(h) }
func (h hp) Less(i, j int) bool {
return h[i].cost < h[j].cost || (h[i].cost == h[j].cost && h[i].i < h[j].i)
}
func (h hp) Swap(i, j int) { h[i], h[j] = h[j], h[i] }
func (h *hp) Push(v any) { *h = append(*h, v.(pair)) }
func (h *hp) Pop() any { a := *h; v := a[len(a)-1]; *h = a[:len(a)-1]; return v }
# Accepted solution for LeetCode #2462: Total Cost to Hire K Workers
class Solution:
def totalCost(self, costs: List[int], k: int, candidates: int) -> int:
n = len(costs)
if candidates * 2 >= n:
return sum(sorted(costs)[:k])
pq = []
for i, c in enumerate(costs[:candidates]):
heappush(pq, (c, i))
for i in range(n - candidates, n):
heappush(pq, (costs[i], i))
heapify(pq)
l, r = candidates, n - candidates - 1
ans = 0
for _ in range(k):
c, i = heappop(pq)
ans += c
if l > r:
continue
if i < l:
heappush(pq, (costs[l], l))
l += 1
else:
heappush(pq, (costs[r], r))
r -= 1
return ans
// Accepted solution for LeetCode #2462: Total Cost to Hire K Workers
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
pub fn rust_example() {
// Port the logic from the reference block below.
}
}
// Reference (java):
// // Accepted solution for LeetCode #2462: Total Cost to Hire K Workers
// class Solution {
// public long totalCost(int[] costs, int k, int candidates) {
// int n = costs.length;
// long ans = 0;
// if (candidates * 2 >= n) {
// Arrays.sort(costs);
// for (int i = 0; i < k; ++i) {
// ans += costs[i];
// }
// return ans;
// }
// PriorityQueue<int[]> pq
// = new PriorityQueue<>((a, b) -> a[0] == b[0] ? a[1] - b[1] : a[0] - b[0]);
// for (int i = 0; i < candidates; ++i) {
// pq.offer(new int[] {costs[i], i});
// pq.offer(new int[] {costs[n - i - 1], n - i - 1});
// }
// int l = candidates, r = n - candidates - 1;
// while (k-- > 0) {
// var p = pq.poll();
// ans += p[0];
// if (l > r) {
// continue;
// }
// if (p[1] < l) {
// pq.offer(new int[] {costs[l], l++});
// } else {
// pq.offer(new int[] {costs[r], r--});
// }
// }
// return ans;
// }
// }
// Accepted solution for LeetCode #2462: Total Cost to Hire K Workers
function totalCost(costs: number[], k: number, candidates: number): number {
const n = costs.length;
if (candidates * 2 >= n) {
costs.sort((a, b) => a - b);
return costs.slice(0, k).reduce((acc, x) => acc + x, 0);
}
const pq = new PriorityQueue<number[]>((a, b) => (a[0] === b[0] ? a[1] - b[1] : a[0] - b[0]));
for (let i = 0; i < candidates; ++i) {
pq.enqueue([costs[i], i]);
pq.enqueue([costs[n - i - 1], n - i - 1]);
}
let [l, r] = [candidates, n - candidates - 1];
let ans = 0;
while (k--) {
const [cost, i] = pq.dequeue()!;
ans += cost;
if (l > r) {
continue;
}
if (i < l) {
pq.enqueue([costs[l], l++]);
} else {
pq.enqueue([costs[r], r--]);
}
}
return ans;
}
Use this to step through a reusable interview workflow for this problem.
Two nested loops check every pair of elements. The outer loop picks one element, the inner loop scans the rest. For n elements that is n × (n−1)/2 comparisons = O(n²). No extra memory — just two loop variables.
Each pointer traverses the array at most once. With two pointers moving inward (or both moving right), the total number of steps is bounded by n. Each comparison is O(1), giving O(n) overall. No auxiliary data structures are needed — just two index variables.
Review these before coding to avoid predictable interview regressions.
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.
Wrong move: Advancing both pointers shrinks the search space too aggressively and skips candidates.
Usually fails on: A valid pair can be skipped when only one side should move.
Fix: Move exactly one pointer per decision branch based on invariant.