neetcode-gh · neetcode-gh · Dec 14, 2024 · Dec 13, 2024
diff --git a/articles/arranging-coins.md b/articles/arranging-coins.md
diff --git a/articles/assign-cookies.md b/articles/assign-cookies.md
@@ -0,0 +1,300 @@
+## 1. Brute Force
+
+::tabs-start
+
+```python
+class Solution:
+    def findContentChildren(self, g: List[int], s: List[int]) -> int:
+        s.sort()
+        res = 0
+
+        for i in g:
+            minIdx = -1
+            for j in range(len(s)):
+                if s[j] < i:
+                    continue
+
+                if minIdx == -1 or s[minIdx] > s[j]:
+                    minIdx = j
+
+            if minIdx != -1:
+                s[minIdx] = -1
+                res += 1
+
+        return res
+```
+
+```java
+public class Solution {
+    public int findContentChildren(int[] g, int[] s) {
+        Arrays.sort(s);
+        int res = 0;
+
+        for (int i : g) {
+            int minIdx = -1;
+            for (int j = 0; j < s.length; j++) {
+                if (s[j] < i) continue;
+
+                if (minIdx == -1 || s[minIdx] > s[j]) {
+                    minIdx = j;
+                }
+            }
+
+            if (minIdx != -1) {
+                s[minIdx] = -1;
+                res++;
+            }
+        }
+
+        return res;
+    }
+}
+```
+
+```cpp
+class Solution {
+public:
+    int findContentChildren(vector<int>& g, vector<int>& s) {
+        sort(s.begin(), s.end());
+        int res = 0;
+
+        for (int i : g) {
+            int minIdx = -1;
+            for (int j = 0; j < s.size(); j++) {
+                if (s[j] < i) continue;
+
+                if (minIdx == -1 || s[minIdx] > s[j]) {
+                    minIdx = j;
+                }
+            }
+
+            if (minIdx != -1) {
+                s[minIdx] = -1;
+                res++;
+            }
+        }
+
+        return res;
+    }
+};
+```
+
+```javascript
+class Solution {
+    /**
+     * @param {number[]} g
+     * @param {number[]} s
+     * @return {number}
+     */
+    findContentChildren(g, s) {
+        s.sort((a, b) => a - b);
+        let res = 0;
+
+        for (let i of g) {
+            let minIdx = -1;
+            for (let j = 0; j < s.length; j++) {
+                if (s[j] < i) continue;
+
+                if (minIdx === -1 || s[minIdx] > s[j]) {
+                    minIdx = j;
+                }
+            }
+
+            if (minIdx !== -1) {
+                s[minIdx] = -1;
+                res++;
+            }
+        }
+
+        return res;
+    }
+}
+```
+
+::tabs-end
+
+### Time & Space Complexity
+
+* Time complexity: $O(n * m + m \log m)$
+* Space complexity: $O(1)$ or $O(m)$ depending on the sorting algorithm.
+
+> Where $n$ is the size of the array $g$ and $m$ is the size of the array $s$.
+
+---
+
+## 2. Two Pointers - I
+
+::tabs-start
+
+```python
+class Solution:
+    def findContentChildren(self, g: List[int], s: List[int]) -> int:
+        g.sort()
+        s.sort()
+
+        i = j = 0
+        while i < len(g):
+            while j < len(s) and g[i] > s[j]:
+                j += 1
+            if j == len(s):
+                break
+            i += 1
+            j += 1
+        return i
+```
+
+```java
+public class Solution {
+    public int findContentChildren(int[] g, int[] s) {
+        Arrays.sort(g);
+        Arrays.sort(s);
+
+        int i = 0, j = 0;
+        while (i < g.length) {
+            while (j < s.length && g[i] > s[j]) {
+                j++;
+            }
+            if (j == s.length) break;
+            i++;
+            j++;
+        }
+        return i;
+    }
+}
+```
+
+```cpp
+class Solution {
+public:
+    int findContentChildren(vector<int>& g, vector<int>& s) {
+        sort(g.begin(), g.end());
+        sort(s.begin(), s.end());
+
+        int i = 0, j = 0;
+        while (i < g.size()) {
+            while (j < s.size() && g[i] > s[j]) {
+                j++;
+            }
+            if (j == s.size()) break;
+            i++;
+            j++;
+        }
+        return i;
+    }
+};
+```
+
+```javascript
+class Solution {
+    /**
+     * @param {number[]} g
+     * @param {number[]} s
+     * @return {number}
+     */
+    findContentChildren(g, s) {
+        g.sort((a, b) => a - b);
+        s.sort((a, b) => a - b);
+
+        let i = 0, j = 0;
+        while (i < g.length) {
+            while (j < s.length && g[i] > s[j]) {
+                j++;
+            }
+            if (j === s.length) break;
+            i++;
+            j++;
+        }
+        return i;
+    }
+}
+```
+
+::tabs-end
+
+### Time & Space Complexity
+
+* Time complexity: $O(n \log n + m \log m)$
+* Space complexity: $O(1)$ or $O(n + m)$ depending on the sorting algorithm.
+
+> Where $n$ is the size of the array $g$ and $m$ is the size of the array $s$.
+
+---
+
+## 3. Two Pointers - II
+
+::tabs-start
+
+```python
+class Solution:
+    def findContentChildren(self, g: List[int], s: List[int]) -> int:
+        g.sort()
+        s.sort()
+
+        i = j = 0
+        while i < len(g) and j < len(s):
+            if g[i] <= s[j]:
+                i += 1
+            j += 1
+
+        return i
+```
+
+```java
+public class Solution {
+    public int findContentChildren(int[] g, int[] s) {
+        Arrays.sort(g);
+        Arrays.sort(s);
+
+        int i = 0;
+        for (int j = 0; i < g.length && j < s.length; j++) {
+            if (g[i] <= s[j]) i++;
+        }
+        return i;
+    }
+}
+```
+
+```cpp
+class Solution {
+public:
+    int findContentChildren(vector<int>& g, vector<int>& s) {
+        sort(g.begin(), g.end());
+        sort(s.begin(), s.end());
+
+        int i = 0;
+        for (int j = 0; i < g.size() && j < s.size(); j++) {
+            if (g[i] <= s[j]) i++;
+        }
+        return i;
+    }
+};
+```
+
+```javascript
+class Solution {
+    /**
+     * @param {number[]} g
+     * @param {number[]} s
+     * @return {number}
+     */
+    findContentChildren(g, s) {
+        g.sort((a, b) => a - b);
+        s.sort((a, b) => a - b);
+
+        let i = 0;
+        for (let j = 0; i < g.length && j < s.length; j++) {
+            if (g[i] <= s[j]) i++;
+        }
+        return i;
+    }
+}
+```
+
+::tabs-end
+
+### Time & Space Complexity
+
+* Time complexity: $O(n \log n + m \log m)$
+* Space complexity: $O(1)$ or $O(n + m)$ depending on the sorting algorithm.
+
+> Where $n$ is the size of the array $g$ and $m$ is the size of the array $s$.