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Commitb0e7b54

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Sri Hari: Batch-5/Neetcode-ALL/Added-articles (neetcode-gh#3841)
* Batch-5/Neetcode-ALL/Added-articles* Batch-5/Neetcode-ALL/Added-articles* Batch-5/Neetcode-ALL/Added-articles* Batch-5/Neetcode-ALL/Added-articles* Batch-5/Neetcode-ALL/Added-articles* Batch-5/Neetcode-ALL/Added-articles* Batch-5/Neetcode-ALL/Added-articles* Batch-5/Neetcode-ALL/Added-articles* Batch-5/Neetcode-ALL/Added-articles* Batch-5/Neetcode-ALL/Added-articles* Batch-5/Neetcode-ALL/Added-articles* Batch-5/Neetcode-ALL/Added-articles* Batch-5/Neetcode-ALL/Added-articles
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‎articles/all-possible-full-binary-trees.md

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‎articles/binary-search-tree-iterator.md

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‎articles/concatenated-words.md

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‎articles/count-all-valid-pickup-and-delivery-options.md

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‎articles/data-stream-as-disjoint-intervals.md

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‎articles/design-browser-history.md

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‎articles/distribute-coins-in-binary-tree.md

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##1. Sorting
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::tabs-start
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```python
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classSolution:
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defeliminateMaximum(self,dist: List[int],speed: List[int]) ->int:
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minReach= [math.ceil(d/ s)for d, sinzip(dist, speed)]
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minReach.sort()
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res=0
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for minuteinrange(len(minReach)):
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if minute>= minReach[minute]:
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return res
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res+=1
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return res
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```
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```java
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publicclassSolution {
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publicinteliminateMaximum(int[]dist,int[]speed) {
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int n= dist.length;
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int[] minReach=newint[n];
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for (int i=0; i< n; i++) {
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minReach[i]= (int)Math.ceil((double) dist[i]/ speed[i]);
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}
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Arrays.sort(minReach);
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int res=0;
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for (int minute=0; minute< n; minute++) {
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if (minute>= minReach[minute]) {
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return res;
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}
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res++;
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}
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return res;
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}
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}
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```
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```cpp
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classSolution {
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public:
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int eliminateMaximum(vector<int>& dist, vector<int>& speed) {
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int n = dist.size();
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vector<int> minReach(n);
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for (int i = 0; i < n; i++) {
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minReach[i] = ceil((double)dist[i] / speed[i]);
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}
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sort(minReach.begin(), minReach.end());
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int res = 0;
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for (int minute = 0; minute < n; minute++) {
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if (minute >= minReach[minute]) {
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return res;
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}
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res++;
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}
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return res;
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}
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};
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```
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```javascript
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class Solution {
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/**
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* @param {number[]} dist
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* @param {number[]} speed
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* @return {number}
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*/
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eliminateMaximum(dist, speed) {
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let n = dist.length;
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let minReach = new Array(n);
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for (let i = 0; i < n; i++) {
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minReach[i] = Math.ceil(dist[i] / speed[i]);
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}
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minReach.sort((a, b) => a - b);
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let res = 0;
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for (let minute = 0; minute < n; minute++) {
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if (minute >= minReach[minute]) {
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return res;
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}
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res++;
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}
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return res;
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}
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}
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```
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::tabs-end
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###Time & Space Complexity
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* Time complexity: $O(n \log n)$
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* Space complexity: $O(n)$
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---
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##2. Sorting (Overwrting Input Array)
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::tabs-start
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```python
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classSolution:
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defeliminateMaximum(self,dist: List[int],speed: List[int]) ->int:
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for iinrange(len(dist)):
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dist[i]= math.ceil(dist[i]/ speed[i])
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dist.sort()
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for minuteinrange(len(dist)):
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if minute>= dist[minute]:
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return minute
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returnlen(dist)
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```
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```java
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publicclassSolution {
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publicinteliminateMaximum(int[]dist,int[]speed) {
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int n= dist.length;
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for (int i=0; i< n; i++) {
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dist[i]= (int)Math.ceil((double) dist[i]/ speed[i]);
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}
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Arrays.sort(dist);
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for (int minute=0; minute< n; minute++) {
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if (minute>= dist[minute]) {
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return minute;
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}
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}
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return n;
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}
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}
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```
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```cpp
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classSolution {
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public:
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int eliminateMaximum(vector<int>& dist, vector<int>& speed) {
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int n = dist.size();
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for (int i = 0; i < n; i++) {
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dist[i] = ceil((double)dist[i] / speed[i]);
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}
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sort(dist.begin(), dist.end());
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for (int minute = 0; minute < n; minute++) {
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if (minute >= dist[minute]) {
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return minute;
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}
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}
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return n;
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}
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};
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```
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```javascript
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classSolution {
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/**
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*@param{number[]}dist
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*@param{number[]}speed
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*@return{number}
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*/
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eliminateMaximum(dist,speed) {
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let n=dist.length;
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for (let i=0; i< n; i++) {
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dist[i]=Math.ceil(dist[i]/ speed[i]);
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}
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dist.sort((a,b)=> a- b);
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for (let minute=0; minute< n; minute++) {
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if (minute>= dist[minute]) {
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return minute;
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}
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}
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return n;
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}
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}
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```
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::tabs-end
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###Time & Space Complexity
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* Time complexity: $O(n \log n)$
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* Space complexity: $O(1)$ or $O(n)$ depending on the sorting algorithm.
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---
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##3. Min-Heap
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::tabs-start
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```python
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classSolution:
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defeliminateMaximum(self,dist: List[int],speed: List[int]) ->int:
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minHeap= []
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for iinrange(len(dist)):
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heapq.heappush(minHeap, dist[i]/ speed[i])
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res=0
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while minHeap:
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if res>= heapq.heappop(minHeap):
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return res
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res+=1
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return res
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```
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```java
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publicclassSolution {
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publicinteliminateMaximum(int[]dist,int[]speed) {
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PriorityQueue<Double> minHeap=newPriorityQueue<>();
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for (int i=0; i< dist.length; i++) {
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minHeap.add((double) dist[i]/ speed[i]);
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}
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int res=0;
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while (!minHeap.isEmpty()) {
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if (res>= minHeap.poll()) {
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return res;
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}
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res++;
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}
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return res;
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}
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}
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```
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```cpp
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classSolution {
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public:
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int eliminateMaximum(vector<int>& dist, vector<int>& speed) {
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priority_queue<double, vector<double>, greater<double>> minHeap;
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for (int i = 0; i < dist.size(); i++) {
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minHeap.push((double)dist[i] / speed[i]);
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}
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int res = 0;
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while (!minHeap.empty()) {
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if (res >= minHeap.top()) {
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return res;
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}
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minHeap.pop();
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res++;
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}
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return res;
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}
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};
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```
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```javascript
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classSolution {
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/**
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*@param{number[]}dist
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*@param{number[]}speed
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*@return{number}
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*/
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eliminateMaximum(dist,speed) {
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constminHeap=newMinPriorityQueue();
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for (let i=0; i<dist.length; i++) {
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minHeap.enqueue(dist[i]/ speed[i]);
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}
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let res=0;
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while (!minHeap.isEmpty()) {
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if (res>=minHeap.dequeue().element) {
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return res;
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}
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res++;
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}
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return res;
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}
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}
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```
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::tabs-end
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###Time & Space Complexity
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* Time complexity: $O(n \log n)$
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* Space complexity: $O(n)$

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