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Commit f10e4c9

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feat: add solutions to lc problem: No.1599 (doocs#2170)
No.1599.Maximum Profit of Operating a Centennial Wheel
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‎solution/1500-1599/1599.Maximum Profit of Operating a Centennial Wheel/README.md‎

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我们直接模拟摩天轮的轮转过程,每次轮转时,累加等待的游客以及新到达的游客,然后最多 4ドル$ 个人上船,更新等待的游客数和利润,记录最大利润与其对应的轮转次数。
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时间复杂度 $O(\frac{S}{4}),ドル其中 $S$ 为数组 `customers` 的元素和,即游客总数。空间复杂度 $O(1)$。
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时间复杂度 $O(n),ドル其中 $n$ 为数组 `customers` 的长度。空间复杂度 $O(1)$。
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<!-- tabs:start -->
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}
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```
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### **TypeScript**
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```ts
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function minOperationsMaxProfit(
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customers: number[],
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boardingCost: number,
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runningCost: number,
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): number {
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let ans: number = -1;
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let [mx, t, wait, i] = [0, 0, 0, 0];
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while (wait > 0 || i < customers.length) {
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wait += i < customers.length ? customers[i] : 0;
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let up: number = Math.min(4, wait);
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wait -= up;
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++i;
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t += up * boardingCost - runningCost;
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if (t > mx) {
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mx = t;
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ans = i;
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}
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}
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return ans;
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}
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```
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### **Rust**
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```rust
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impl Solution {
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pub fn min_operations_max_profit(
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customers: Vec<i32>,
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boarding_cost: i32,
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running_cost: i32
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) -> i32 {
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let mut ans = -1;
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let mut mx = 0;
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let mut t = 0;
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let mut wait = 0;
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let mut i = 0;
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while wait > 0 || i < customers.len() {
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wait += if i < customers.len() { customers[i] } else { 0 };
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let up = std::cmp::min(4, wait);
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wait -= up;
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i += 1;
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t += up * boarding_cost - running_cost;
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if t > mx {
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mx = t;
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ans = i as i32;
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}
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}
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ans
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}
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}
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```
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### **...**
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```

‎solution/1500-1599/1599.Maximum Profit of Operating a Centennial Wheel/README_EN.md‎

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## Solutions
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**Solution 1: Simulation**
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We directly simulate the rotation process of the Ferris wheel. Each time it rotates, we add up the waiting customers and the newly arrived customers, then at most 4ドル$ people get on the ride, update the number of waiting customers and profit, and record the maximum profit and its corresponding number of rotations.
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The time complexity is $O(n),ドル where $n$ is the length of the `customers` array. The space complexity is $O(1)$.
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<!-- tabs:start -->
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### **Python3**
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}
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```
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### **TypeScript**
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```ts
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function minOperationsMaxProfit(
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customers: number[],
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boardingCost: number,
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runningCost: number,
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): number {
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let ans: number = -1;
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let [mx, t, wait, i] = [0, 0, 0, 0];
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while (wait > 0 || i < customers.length) {
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wait += i < customers.length ? customers[i] : 0;
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let up: number = Math.min(4, wait);
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wait -= up;
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++i;
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t += up * boardingCost - runningCost;
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if (t > mx) {
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mx = t;
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ans = i;
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}
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}
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return ans;
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}
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```
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### **Rust**
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```rust
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impl Solution {
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pub fn min_operations_max_profit(
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customers: Vec<i32>,
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boarding_cost: i32,
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running_cost: i32
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) -> i32 {
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let mut ans = -1;
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let mut mx = 0;
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let mut t = 0;
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let mut wait = 0;
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let mut i = 0;
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while wait > 0 || i < customers.len() {
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wait += if i < customers.len() { customers[i] } else { 0 };
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let up = std::cmp::min(4, wait);
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wait -= up;
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i += 1;
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t += up * boarding_cost - running_cost;
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if t > mx {
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mx = t;
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ans = i as i32;
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}
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}
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ans
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}
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}
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```
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### **...**
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```
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impl Solution {
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pub fn min_operations_max_profit(
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customers: Vec<i32>,
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boarding_cost: i32,
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running_cost: i32
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) -> i32 {
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let mut ans = -1;
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let mut mx = 0;
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let mut t = 0;
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let mut wait = 0;
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let mut i = 0;
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while wait > 0 || i < customers.len() {
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wait += if i < customers.len() { customers[i] } else { 0 };
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let up = std::cmp::min(4, wait);
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wait -= up;
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i += 1;
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t += up * boarding_cost - running_cost;
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if t > mx {
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mx = t;
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ans = i as i32;
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}
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}
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ans
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}
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}
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function minOperationsMaxProfit(
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customers: number[],
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boardingCost: number,
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runningCost: number,
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): number {
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let ans: number = -1;
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let [mx, t, wait, i] = [0, 0, 0, 0];
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while (wait > 0 || i < customers.length) {
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wait += i < customers.length ? customers[i] : 0;
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let up: number = Math.min(4, wait);
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wait -= up;
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++i;
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t += up * boardingCost - runningCost;
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if (t > mx) {
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mx = t;
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ans = i;
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}
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}
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return ans;
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}

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