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Batch-4/Neetcode-All/Added-articles (neetcode-gh#3766)
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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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class Solution:
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def findContentChildren(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 i in g:
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minIdx = -1
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for j in range(len(s)):
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if s[j] < i:
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continue
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if minIdx == -1 or 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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public class Solution {
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public int findContentChildren(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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class Solution {
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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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class Solution {
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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 i of 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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class Solution:
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def findContentChildren(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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public class Solution {
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public int findContentChildren(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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class Solution {
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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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class Solution {
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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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class Solution:
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def findContentChildren(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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public class Solution {
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public int findContentChildren(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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class Solution {
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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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class Solution {
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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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