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Commitff45c13

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Batch-4/Neetcode-All/Added-articles (neetcode-gh#3766)
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26 files changed

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26 files changed

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‎articles/arranging-coins.md

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‎articles/assign-cookies.md

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##1. Brute Force
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::tabs-start
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```python
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classSolution:
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deffindContentChildren(self,g: List[int],s: List[int]) ->int:
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s.sort()
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res=0
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for iin g:
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minIdx=-1
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for jinrange(len(s)):
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if s[j]< i:
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continue
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if minIdx==-1or s[minIdx]> s[j]:
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minIdx= j
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if minIdx!=-1:
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s[minIdx]=-1
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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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publicintfindContentChildren(int[]g,int[]s) {
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Arrays.sort(s);
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int res=0;
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for (int i: g) {
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int minIdx=-1;
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for (int j=0; j< s.length; j++) {
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if (s[j]< i)continue;
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if (minIdx==-1|| s[minIdx]> s[j]) {
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minIdx= j;
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}
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}
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if (minIdx!=-1) {
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s[minIdx]=-1;
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res++;
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}
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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 findContentChildren(vector<int>& g, vector<int>& s) {
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sort(s.begin(), s.end());
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int res = 0;
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for (int i : g) {
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int minIdx = -1;
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for (int j = 0; j < s.size(); j++) {
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if (s[j] < i) continue;
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if (minIdx == -1 || s[minIdx] > s[j]) {
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minIdx = j;
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}
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}
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if (minIdx != -1) {
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s[minIdx] = -1;
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res++;
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}
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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[]}g
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*@param{number[]}s
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*@return{number}
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*/
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findContentChildren(g,s) {
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s.sort((a,b)=> a- b);
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let res=0;
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for (let iof g) {
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let minIdx=-1;
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for (let j=0; j<s.length; j++) {
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if (s[j]< i)continue;
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if (minIdx===-1|| s[minIdx]> s[j]) {
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minIdx= j;
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}
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}
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if (minIdx!==-1) {
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s[minIdx]=-1;
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res++;
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}
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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: $O(n * m + m \log m)$
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* Space complexity: $O(1)$ or $O(m)$ depending on the sorting algorithm.
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>Where $n$ is the size of the array $g$ and $m$ is the size of the array $s$.
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---
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##2. Two Pointers - I
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::tabs-start
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```python
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classSolution:
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deffindContentChildren(self,g: List[int],s: List[int]) ->int:
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g.sort()
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s.sort()
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i= j=0
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while i<len(g):
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while j<len(s)and g[i]> s[j]:
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j+=1
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if j==len(s):
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break
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i+=1
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j+=1
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return i
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```
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```java
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publicclassSolution {
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publicintfindContentChildren(int[]g,int[]s) {
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Arrays.sort(g);
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Arrays.sort(s);
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int i=0, j=0;
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while (i< g.length) {
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while (j< s.length&& g[i]> s[j]) {
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j++;
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}
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if (j== s.length)break;
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i++;
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j++;
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}
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return i;
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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 findContentChildren(vector<int>& g, vector<int>& s) {
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sort(g.begin(), g.end());
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sort(s.begin(), s.end());
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int i = 0, j = 0;
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while (i < g.size()) {
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while (j < s.size() && g[i] > s[j]) {
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j++;
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}
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if (j == s.size())break;
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i++;
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j++;
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}
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return i;
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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[]}g
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*@param{number[]}s
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*@return{number}
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*/
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findContentChildren(g,s) {
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g.sort((a,b)=> a- b);
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s.sort((a,b)=> a- b);
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let i=0, j=0;
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while (i<g.length) {
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while (j<s.length&& g[i]> s[j]) {
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j++;
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}
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if (j===s.length)break;
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i++;
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j++;
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}
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return i;
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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: $O(n \log n + m \log m)$
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* Space complexity: $O(1)$ or $O(n + m)$ depending on the sorting algorithm.
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>Where $n$ is the size of the array $g$ and $m$ is the size of the array $s$.
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---
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##3. Two Pointers - II
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::tabs-start
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```python
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classSolution:
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deffindContentChildren(self,g: List[int],s: List[int]) ->int:
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g.sort()
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s.sort()
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i= j=0
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while i<len(g)and j<len(s):
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if g[i]<= s[j]:
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i+=1
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j+=1
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return i
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```
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```java
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publicclassSolution {
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publicintfindContentChildren(int[]g,int[]s) {
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Arrays.sort(g);
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Arrays.sort(s);
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int i=0;
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for (int j=0; i< g.length&& j< s.length; j++) {
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if (g[i]<= s[j]) i++;
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}
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return i;
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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 findContentChildren(vector<int>& g, vector<int>& s) {
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sort(g.begin(), g.end());
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sort(s.begin(), s.end());
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int i = 0;
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for (int j = 0; i < g.size() && j < s.size(); j++) {
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if (g[i] <= s[j]) i++;
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}
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return i;
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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[]}g
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*@param{number[]}s
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*@return{number}
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*/
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findContentChildren(g,s) {
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g.sort((a,b)=> a- b);
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s.sort((a,b)=> a- b);
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let i=0;
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for (let j=0; i<g.length&& j<s.length; j++) {
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if (g[i]<= s[j]) i++;
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}
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return i;
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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: $O(n \log n + m \log m)$
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* Space complexity: $O(1)$ or $O(n + m)$ depending on the sorting algorithm.
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>Where $n$ is the size of the array $g$ and $m$ is the size of the array $s$.

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