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 the operator of a Centennial Wheel that has four gondolas, and each gondola has room for up to four people. You have the ability to rotate the gondolas counterclockwise, which costs you runningCost dollars.
You are given an array customers of length n where customers[i] is the number of new customers arriving just before the ith rotation (0-indexed). This means you must rotate the wheel i times before the customers[i] customers arrive. You cannot make customers wait if there is room in the gondola. Each customer pays boardingCost dollars when they board on the gondola closest to the ground and will exit once that gondola reaches the ground again.
You can stop the wheel at any time, including before serving all customers. If you decide to stop serving customers, all subsequent rotations are free in order to get all the customers down safely. Note that if there are currently more than four customers waiting at the wheel, only four will board the gondola, and the rest will wait for the next rotation.
Return the minimum number of rotations you need to perform to maximize your profit. If there is no scenario where the profit is positive, return -1.
Example 1:
Input: customers = [8,3], boardingCost = 5, runningCost = 6 Output: 3 Explanation: The numbers written on the gondolas are the number of people currently there. 1. 8 customers arrive, 4 board and 4 wait for the next gondola, the wheel rotates. Current profit is 4 * $5 - 1 * $6 = $14. 2. 3 customers arrive, the 4 waiting board the wheel and the other 3 wait, the wheel rotates. Current profit is 8 * $5 - 2 * $6 = $28. 3. The final 3 customers board the gondola, the wheel rotates. Current profit is 11 * $5 - 3 * $6 = $37. The highest profit was $37 after rotating the wheel 3 times.
Example 2:
Input: customers = [10,9,6], boardingCost = 6, runningCost = 4 Output: 7 Explanation: 1. 10 customers arrive, 4 board and 6 wait for the next gondola, the wheel rotates. Current profit is 4 * $6 - 1 * $4 = $20. 2. 9 customers arrive, 4 board and 11 wait (2 originally waiting, 9 newly waiting), the wheel rotates. Current profit is 8 * $6 - 2 * $4 = $40. 3. The final 6 customers arrive, 4 board and 13 wait, the wheel rotates. Current profit is 12 * $6 - 3 * $4 = $60. 4. 4 board and 9 wait, the wheel rotates. Current profit is 16 * $6 - 4 * $4 = $80. 5. 4 board and 5 wait, the wheel rotates. Current profit is 20 * $6 - 5 * $4 = $100. 6. 4 board and 1 waits, the wheel rotates. Current profit is 24 * $6 - 6 * $4 = $120. 7. 1 boards, the wheel rotates. Current profit is 25 * $6 - 7 * $4 = $122. The highest profit was $122 after rotating the wheel 7 times.
Example 3:
Input: customers = [3,4,0,5,1], boardingCost = 1, runningCost = 92 Output: -1 Explanation: 1. 3 customers arrive, 3 board and 0 wait, the wheel rotates. Current profit is 3 * $1 - 1 * $92 = -$89. 2. 4 customers arrive, 4 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 2 * $92 = -$177. 3. 0 customers arrive, 0 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 3 * $92 = -$269. 4. 5 customers arrive, 4 board and 1 waits, the wheel rotates. Current profit is 11 * $1 - 4 * $92 = -$357. 5. 1 customer arrives, 2 board and 0 wait, the wheel rotates. Current profit is 13 * $1 - 5 * $92 = -$447. The profit was never positive, so return -1.
Constraints:
n == customers.length1 <= n <= 1050 <= customers[i] <= 501 <= boardingCost, runningCost <= 100Problem summary: You are the operator of a Centennial Wheel that has four gondolas, and each gondola has room for up to four people. You have the ability to rotate the gondolas counterclockwise, which costs you runningCost dollars. You are given an array customers of length n where customers[i] is the number of new customers arriving just before the ith rotation (0-indexed). This means you must rotate the wheel i times before the customers[i] customers arrive. You cannot make customers wait if there is room in the gondola. Each customer pays boardingCost dollars when they board on the gondola closest to the ground and will exit once that gondola reaches the ground again. You can stop the wheel at any time, including before serving all customers. If you decide to stop serving customers, all subsequent rotations are free in order to get all the customers down safely. Note that if there are currently more
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Array
[8,3] 5 6
[10,9,6] 6 4
[3,4,0,5,1] 1 92
Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #1599: Maximum Profit of Operating a Centennial Wheel
class Solution {
public int minOperationsMaxProfit(int[] customers, int boardingCost, int runningCost) {
int ans = -1;
int mx = 0, t = 0;
int wait = 0, i = 0;
while (wait > 0 || i < customers.length) {
wait += i < customers.length ? customers[i] : 0;
int up = Math.min(4, wait);
wait -= up;
++i;
t += up * boardingCost - runningCost;
if (t > mx) {
mx = t;
ans = i;
}
}
return ans;
}
}
// Accepted solution for LeetCode #1599: Maximum Profit of Operating a Centennial Wheel
func minOperationsMaxProfit(customers []int, boardingCost int, runningCost int) int {
ans := -1
t, mx := 0, 0
wait, i := 0, 0
for wait > 0 || i < len(customers) {
if i < len(customers) {
wait += customers[i]
}
up := min(4, wait)
wait -= up
t += up*boardingCost - runningCost
i++
if t > mx {
mx = t
ans = i
}
}
return ans
}
# Accepted solution for LeetCode #1599: Maximum Profit of Operating a Centennial Wheel
class Solution:
def minOperationsMaxProfit(
self, customers: List[int], boardingCost: int, runningCost: int
) -> int:
ans = -1
mx = t = 0
wait = 0
i = 0
while wait or i < len(customers):
wait += customers[i] if i < len(customers) else 0
up = wait if wait < 4 else 4
wait -= up
t += up * boardingCost - runningCost
i += 1
if t > mx:
mx = t
ans = i
return ans
// Accepted solution for LeetCode #1599: Maximum Profit of Operating a Centennial Wheel
impl Solution {
pub fn min_operations_max_profit(
customers: Vec<i32>,
boarding_cost: i32,
running_cost: i32,
) -> i32 {
let mut ans = -1;
let mut mx = 0;
let mut t = 0;
let mut wait = 0;
let mut i = 0;
while wait > 0 || i < customers.len() {
wait += if i < customers.len() { customers[i] } else { 0 };
let up = std::cmp::min(4, wait);
wait -= up;
i += 1;
t += up * boarding_cost - running_cost;
if t > mx {
mx = t;
ans = i as i32;
}
}
ans
}
}
// Accepted solution for LeetCode #1599: Maximum Profit of Operating a Centennial Wheel
function minOperationsMaxProfit(
customers: number[],
boardingCost: number,
runningCost: number,
): number {
let ans: number = -1;
let [mx, t, wait, i] = [0, 0, 0, 0];
while (wait > 0 || i < customers.length) {
wait += i < customers.length ? customers[i] : 0;
let up: number = Math.min(4, wait);
wait -= up;
++i;
t += up * boardingCost - runningCost;
if (t > mx) {
mx = t;
ans = i;
}
}
return ans;
}
Use this to step through a reusable interview workflow for this problem.
Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.
Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.
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.