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Commit fcecf02

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feat: add js/ts solutions to lc problem: No.2332 (#2223)
No.2332.The Latest Time to Catch a Bus
1 parent 0be93a9 commit fcecf02

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‎solution/2300-2399/2332.The Latest Time to Catch a Bus/README.md‎

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@@ -56,14 +56,14 @@
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**方法一:模拟**
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先排序,然后用双指针模拟乘客上车的过程:遍历公交车 $bus,ドル乘客遵循"先到先上车"的原则。
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我们先排序,然后用双指针模拟乘客上车的过程:遍历公交车 $bus,ドル乘客遵循"先到先上车"的原则。
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模拟结束后,判断最后一班公交车是否还有空位:
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- 若有空位,我们可以在公交车发车时 $bus[bus.length-1]$ 到达公交站;若此时有人,可以顺着往前找到没人到达的时刻;
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- 若有空位,我们可以在公交车发车时 $bus[|bus|-1]$ 到达公交站;若此时有人,可以顺着往前找到没人到达的时刻;
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- 若无空位,我们可以找到上一个上车的乘客,顺着他往前找到没人到达的时刻。
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时间复杂度 $O(nlogn+mlogm)$。
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时间复杂度 $O(n \times \log n + m \times \log m),ドル空间复杂度 $O(\log n + \log m)$。其中 $n$ 和 $m$ 分别是公交车和乘客的数量。
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<!-- tabs:start -->
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@@ -169,7 +169,25 @@ func latestTimeCatchTheBus(buses []int, passengers []int, capacity int) int {
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### **TypeScript**
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```ts
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function latestTimeCatchTheBus(buses: number[], passengers: number[], capacity: number): number {
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buses.sort((a, b) => a - b);
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passengers.sort((a, b) => a - b);
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let [j, c] = [0, 0];
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for (const t of buses) {
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c = capacity;
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while (c && j < passengers.length && passengers[j] <= t) {
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--c;
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++j;
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}
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}
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--j;
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let ans = c > 0 ? buses.at(-1)! : passengers[j];
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while (j >= 0 && passengers[j] === ans) {
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--ans;
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--j;
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}
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return ans;
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}
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```
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### **JavaScript**
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var latestTimeCatchTheBus = function (buses, passengers, capacity) {
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buses.sort((a, b) => a - b);
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passengers.sort((a, b) => a - b);
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let j = 0,
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c;
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let [j, c] = [0, 0];
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for (const t of buses) {
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c = capacity;
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while (c && j < passengers.length && passengers[j] <= t) {
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}
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}
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--j;
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let ans = c > 0 ? buses[buses.length-1] : passengers[j];
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let ans = c > 0 ? buses.at(-1) : passengers[j];
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while (j >= 0 && passengers[j] === ans) {
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--ans;
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--j;

‎solution/2300-2399/2332.The Latest Time to Catch a Bus/README_EN.md‎

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@@ -57,6 +57,17 @@ Notice if you had arrived any later, then the 6<sup>th</sup> passenger would hav
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## Solutions
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**Solution 1: Simulation**
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First, we sort, and then use double pointers to simulate the process of passengers getting on the bus: traverse the bus $bus,ドル passengers follow the principle of "first come, first served".
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After the simulation ends, judge whether the last bus still has seats:
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- If there are seats, we can arrive at the bus station when the bus departs at $bus[|bus|-1]$; if there are people at this time, we can find the time when no one arrives by going forward.
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- If there are no seats, we can find the last passenger who got on the bus, and find the time when no one arrives by going forward from him.
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The time complexity is $O(n \times \log n + m \times \log m),ドル and the space complexity is $O(\log n + \log m)$. Where $n$ and $m$ are the numbers of buses and passengers respectively.
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<!-- tabs:start -->
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### **Python3**
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### **TypeScript**
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```ts
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function latestTimeCatchTheBus(buses: number[], passengers: number[], capacity: number): number {
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buses.sort((a, b) => a - b);
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passengers.sort((a, b) => a - b);
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let [j, c] = [0, 0];
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for (const t of buses) {
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c = capacity;
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while (c && j < passengers.length && passengers[j] <= t) {
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--c;
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++j;
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}
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}
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--j;
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let ans = c > 0 ? buses.at(-1)! : passengers[j];
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while (j >= 0 && passengers[j] === ans) {
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--ans;
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--j;
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}
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return ans;
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}
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```
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### **JavaScript**
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var latestTimeCatchTheBus = function (buses, passengers, capacity) {
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buses.sort((a, b) => a - b);
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passengers.sort((a, b) => a - b);
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let j = 0,
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c;
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let [j, c] = [0, 0];
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for (const t of buses) {
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c = capacity;
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while (c && j < passengers.length && passengers[j] <= t) {
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}
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}
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--j;
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let ans = c > 0 ? buses[buses.length-1] : passengers[j];
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let ans = c > 0 ? buses.at(-1) : passengers[j];
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while (j >= 0 && passengers[j] === ans) {
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--ans;
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--j;
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/**
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* @param {number[]} buses
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* @param {number[]} passengers
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* @param {number} capacity
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* @return {number}
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*/
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var latestTimeCatchTheBus = function (buses, passengers, capacity) {
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buses.sort((a, b) => a - b);
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passengers.sort((a, b) => a - b);
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let [j, c] = [0, 0];
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for (const t of buses) {
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c = capacity;
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while (c && j < passengers.length && passengers[j] <= t) {
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--c;
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++j;
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}
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}
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--j;
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let ans = c > 0 ? buses.at(-1) : passengers[j];
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while (j >= 0 && passengers[j] === ans) {
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--ans;
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--j;
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}
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return ans;
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};
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function latestTimeCatchTheBus(buses: number[], passengers: number[], capacity: number): number {
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buses.sort((a, b) => a - b);
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passengers.sort((a, b) => a - b);
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let [j, c] = [0, 0];
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for (const t of buses) {
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c = capacity;
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while (c && j < passengers.length && passengers[j] <= t) {
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--c;
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++j;
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}
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}
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--j;
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let ans = c > 0 ? buses.at(-1)! : passengers[j];
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while (j >= 0 && passengers[j] === ans) {
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--ans;
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--j;
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}
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return ans;
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}

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