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feat: add solutions to lc problem: No.1942 (#4035)
No.1942.The Number of the Smallest Unoccupied Chair
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solution/1900-1999/1942.The Number of the Smallest Unoccupied Chair/README.md

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@@ -80,7 +80,15 @@ tags:
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<!-- solution:start -->
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### 方法一:优先队列(最小堆)
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### 方法一:优先队列(小根堆)
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我们首先将每个朋友的到达时间、离开时间和编号组成一个三元组,然后按到达时间排序。
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我们使用一个小根堆 $\textit{idle}$ 来存储当前空闲的椅子编号,初始时,我们将 $0, 1, \ldots, n-1$ 加入 $\textit{idle}$ 中。使用一个小根堆 $\textit{busy}$ 存储二元组 $(\textit{leaving}, \textit{chair})$,其中 $\textit{leaving}$ 表示离开时间,而 $\textit{chair}$ 表示椅子编号。
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遍历每个朋友的到达时间、离开时间和编号,对于每个朋友,我们首先将所有离开时间小于等于当前朋友到达时间的朋友从 $\textit{busy}$ 中弹出,将他们占据的椅子编号加入 $\textit{idle}$ 中。然后我们从 $\textit{idle}$ 中弹出一个椅子编号,将其分配给当前朋友,将 $(\textit{leaving}, \textit{chair})$ 加入 $\textit{busy}$ 中。如果当前朋友的编号等于 $\textit{targetFriend}$,我们返回当前分配的椅子编号。
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时间复杂度 $O(n \times \log n)$,空间复杂度 $O(n)$。其中 $n$ 为朋友的个数。
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<!-- tabs:start -->
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@@ -90,20 +98,19 @@ tags:
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class Solution:
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def smallestChair(self, times: List[List[int]], targetFriend: int) -> int:
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n = len(times)
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h = list(range(n))
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heapify(h)
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for i in range(n):
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times[i].append(i)
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times.sort()
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idle = list(range(n))
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heapify(idle)
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busy = []
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for a, b, i in times:
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while busy and busy[0][0] <= a:
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heappush(h, heappop(busy)[1])
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c = heappop(h)
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for arrival, leaving, i in times:
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while busy and busy[0][0] <= arrival:
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heappush(idle, heappop(busy)[1])
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j = heappop(idle)
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if i == targetFriend:
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return c
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heappush(busy, (b, c))
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return -1
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return j
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heappush(busy, (leaving, j))
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```
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#### Java
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class Solution {
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public int smallestChair(int[][] times, int targetFriend) {
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int n = times.length;
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int[][] ts = new int[n][3];
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PriorityQueue<Integer> q = new PriorityQueue<>();
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PriorityQueue<int[]> busy = new PriorityQueue<>((a, b) -> a[0] - b[0]);
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PriorityQueue<Integer> idle = new PriorityQueue<>();
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PriorityQueue<int[]> busy = new PriorityQueue<>(Comparator.comparingInt(a -> a[0]));
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for (int i = 0; i < n; ++i) {
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ts[i] = new int[] {times[i][0], times[i][1], i};
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q.offer(i);
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times[i] = new int[] {times[i][0], times[i][1], i};
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idle.offer(i);
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}
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Arrays.sort(ts, (a, b) -> a[0] - b[0]);
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for (int[] t : ts) {
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int a = t[0], b = t[1], i = t[2];
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while (!busy.isEmpty() && busy.peek()[0] <= a) {
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q.offer(busy.poll()[1]);
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Arrays.sort(times, Comparator.comparingInt(a -> a[0]));
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for (var e : times) {
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int arrival = e[0], leaving = e[1], i = e[2];
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while (!busy.isEmpty() && busy.peek()[0] <= arrival) {
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idle.offer(busy.poll()[1]);
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}
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int c = q.poll();
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int j = idle.poll();
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if (i == targetFriend) {
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return c;
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return j;
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}
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busy.offer(new int[] {b, c});
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busy.offer(new int[] {leaving, j});
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}
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return -1;
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}
@@ -139,35 +145,138 @@ class Solution {
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#### C++
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```cpp
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using pii = pair<int, int>;
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class Solution {
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public:
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int smallestChair(vector<vector<int>>& times, int targetFriend) {
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priority_queue<int, vector<int>, greater<int>> q;
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using pii = pair<int, int>;
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priority_queue<pii, vector<pii>, greater<pii>> busy;
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priority_queue<int, vector<int>, greater<int>> idle;
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int n = times.size();
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for (int i = 0; i < n; ++i) {
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times[i].push_back(i);
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q.push(i);
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idle.push(i);
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}
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sort(times.begin(), times.end());
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for (auto& t : times) {
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int a = t[0], b = t[1], i = t[2];
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while (!busy.empty() && busy.top().first <= a) {
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q.push(busy.top().second);
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ranges::sort(times);
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for (const auto& e : times) {
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int arrival = e[0], leaving = e[1], i = e[2];
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while (!busy.empty() && busy.top().first <= arrival) {
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idle.push(busy.top().second);
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busy.pop();
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}
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int c = q.top();
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q.pop();
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if (i == targetFriend) return c;
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busy.push({b, c});
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int j = idle.top();
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if (i == targetFriend) {
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return j;
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}
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idle.pop();
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busy.emplace(leaving, j);
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}
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return -1;
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}
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};
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```
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#### Go
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```go
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func smallestChair(times [][]int, targetFriend int) int {
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idle := hp{}
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busy := hp2{}
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for i := range times {
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times[i] = append(times[i], i)
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heap.Push(&idle, i)
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}
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sort.Slice(times, func(i, j int) bool { return times[i][0] < times[j][0] })
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for _, e := range times {
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arrival, leaving, i := e[0], e[1], e[2]
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for len(busy) > 0 && busy[0].t <= arrival {
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heap.Push(&idle, heap.Pop(&busy).(pair).i)
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}
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j := heap.Pop(&idle).(int)
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if i == targetFriend {
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return j
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}
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heap.Push(&busy, pair{leaving, j})
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}
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return -1
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}
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type hp struct{ sort.IntSlice }
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func (h hp) Less(i, j int) bool { return h.IntSlice[i] < h.IntSlice[j] }
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func (h *hp) Push(v any) { h.IntSlice = append(h.IntSlice, v.(int)) }
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func (h *hp) Pop() any {
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a := h.IntSlice
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v := a[len(a)-1]
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h.IntSlice = a[:len(a)-1]
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return v
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}
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type pair struct{ t, i int }
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type hp2 []pair
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func (h hp2) Len() int { return len(h) }
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func (h hp2) Less(i, j int) bool { return h[i].t < h[j].t || (h[i].t == h[j].t && h[i].i < h[j].i) }
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func (h hp2) Swap(i, j int) { h[i], h[j] = h[j], h[i] }
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func (h *hp2) Push(v any) { *h = append(*h, v.(pair)) }
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func (h *hp2) Pop() any { a := *h; v := a[len(a)-1]; *h = a[:len(a)-1]; return v }
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```
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#### TypeScript
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```ts
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function smallestChair(times: number[][], targetFriend: number): number {
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const n = times.length;
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const idle = new MinPriorityQueue();
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const busy = new MinPriorityQueue({ priority: v => v[0] });
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for (let i = 0; i < n; ++i) {
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times[i].push(i);
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idle.enqueue(i);
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}
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times.sort((a, b) => a[0] - b[0]);
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for (const [arrival, leaving, i] of times) {
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while (busy.size() > 0 && busy.front().element[0] <= arrival) {
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idle.enqueue(busy.dequeue().element[1]);
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}
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const j = idle.dequeue().element;
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if (i === targetFriend) {
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return j;
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}
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busy.enqueue([leaving, j]);
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}
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return -1;
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}
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```
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#### JavaScript
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```js
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/**
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* @param {number[][]} times
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* @param {number} targetFriend
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* @return {number}
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*/
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var smallestChair = function (times, targetFriend) {
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const n = times.length;
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const idle = new MinPriorityQueue();
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const busy = new MinPriorityQueue({ priority: v => v[0] });
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for (let i = 0; i < n; ++i) {
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times[i].push(i);
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idle.enqueue(i);
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}
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times.sort((a, b) => a[0] - b[0]);
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for (const [arrival, leaving, i] of times) {
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while (busy.size() > 0 && busy.front().element[0] <= arrival) {
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idle.enqueue(busy.dequeue().element[1]);
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}
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const j = idle.dequeue().element;
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if (i === targetFriend) {
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return j;
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}
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busy.enqueue([leaving, j]);
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}
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};
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```
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<!-- tabs:end -->
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<!-- solution:end -->

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