docs(content): add 435 non‑overlapping intervals doc, condense meta, bump guide badges

apps/docs/content/(01-linear-structures)/binary-search/meta.ts

@@ -1,8 +1,3 @@
 import { defineMeta } from "blume";
 
-export default defineMeta({
+export default defineMeta({ title: "Binary Search", order: 5, icon: "search", display: "group" });
-  title: "Binary Search",
-  order: 5,
-  icon: "search",
-  display: "group",
-});

apps/docs/content/(01-linear-structures)/hash-table/meta.ts

@@ -1,8 +1,3 @@
 import { defineMeta } from "blume";
 
-export default defineMeta({
+export default defineMeta({ title: "Hash Table", order: 1, icon: "hash", display: "group" });
-  title: "Hash Table",
-  order: 1,
-  icon: "hash",
-  display: "group",
-});

apps/docs/content/(05-decision-space)/intervals/435-non-overlapping-intervals.mdx

@@ -0,0 +1,53 @@
+---
+title: '435. Non-overlapping Intervals'
+description: Given an array of intervals intervals where intervals[i] = [starti, endi], return the minimum number of intervals you need to remove to make the rest of the intervals non-overlapping
+icon: dot
+topics:
+  - { name: "Array", slug: "array" }
+  - { name: "Dynamic Programming", slug: "dynamic-programming" }
+  - { name: "Greedy", slug: "greedy" }
+  - { name: "Sorting", slug: "sorting" }
+sidebar:
+  label: 'Non-overlapping Intervals'
+  badge: 'Medium'
+---
+
+### Example 1:
+- Input: `intervals = [[1,2],[2,3],[3,4],[1,3]]`
+- Output: `1`
+- Explanation: `[1,3]` can be removed and the rest of the `intervals` are non-overlapping.
+
+### Example 2:
+- Input: `intervals = [[1,2],[1,2],[1,2]]`
+- Output: `2`
+- Explanation: You need to remove two `[1,2]` to make the rest of the `intervals` non-overlapping.
+
+### Example 3:
+- Input: `intervals = [[1,2],[2,3]]`
+- Output: `0`
+- Explanation: You don't need to remove any of the `intervals` since they're already non-overlapping.
+
+### Constraints:
+
+- `1 <= intervals.length <= 10^5`
+- `intervals[i].length == 2`
+- `-5 * 10^4 <= starti < endi <= 5 * 10^4`
+
+## Solution
+
+```py
+class Solution:
+    def eraseOverlapIntervals(self, intervals: list[list[int]]) -> int:
+        intervals.sort(key=lambda x: x[0])
+        result = []
+
+        for start, end in intervals:
+            # Use '<' to avoid intervals which touch
+            if result and start < result[-1][1]:
+                result[-1][1] = min(result[-1][1], end)
+            else:
+                result.append([start, end])
+
+        # Difference between original and result is to remove
+        return len(intervals) - len(result)
+```

apps/docs/content/guides/(1-data-structures)/array.mdx

@@ -3,7 +3,7 @@ title: "Array"
 list: "https://leetcode.com/problem-list/array/"
 icon: list-ordered
 sidebar:
-  badge: '40'
+  badge: '41'
 ---
 
 :::note

apps/docs/content/guides/(2-algorithms)/sorting.mdx

@@ -3,7 +3,7 @@ title: "Sorting"
 list: "https://leetcode.com/problem-list/sorting/"
 icon: arrow-down-wide-narrow
 sidebar:
-  badge: '11'
+  badge: '12'
 ---
 
 :::note