feat(content): add new problem pages, reorganize docs into (problems) subfolders, and update meta display to page

apps/docs/content/(05-decision-space)/backtracking/(problems)/39_combination-sum.mdx

@@ -0,0 +1,56 @@
+---
+title: '39. Combination Sum'
+description: Given an array of distinct integers candidates and a target integer target, return a list of all unique combinations of candidates where the chosen numbers sum to target. You may return the combinations in any order
+icon: dot
+topics:
+  - { name: "Array", slug: "array" }
+  - { name: "Backtracking", slug: "backtracking" }
+issue: "https://github.com/prdlk/leetcode/issues/150"
+sidebar:
+  label: 'Combination Sum'
+  badge: 'Medium'
+---
+
+### Example 1:
+- Input: `candidates = [2,3,6,7], target = 7`
+- Output: `[[2,2,3],[7]]`
+- Explanation: `2` and `3` are `candidates`, and `2 + 2 + 3 = 7`. Note that `2` can be used multiple times. `7` is a candidate, and `7 = 7`. These are the only two combinations.
+
+### Example 2:
+- Input: `candidates = [2,3,5], target = 8`
+- Output: `[[2,2,2,2],[2,3,3],[3,5]]`
+
+### Example 3:
+- Input: `candidates = [2], target = 1`
+- Output: `[]`
+
+### Constraints:
+
+- `1 <= candidates.length <= 30`
+- `2 <= candidates[i] <= 40`
+- All elements of `candidates` are distinct.
+- `1 <= target <= 40`
+
+## Solution
+
+```py
+class Solution:
+    def combinationSum(self, candidates: list[int], target: int) -> list[list[int]]:
+        candidates.sort()
+        result, path = [], []
+
+        def backtrack(start, remaining):
+            if remaining == 0:
+                result.append(path[:])
+                return
+            for i in range(start, len(candidates)):
+                if candidates[i] > remaining:
+                    break
+
+                path.append(candidates[i])
+                backtrack(i, remaining - candidates[i])
+                path.pop()
+
+        backtrack(0, target)
+        return result
+```

apps/docs/content/(05-decision-space)/backtracking/(problems)/46_permutations.mdx

@@ -0,0 +1,55 @@
+---
+title: '46. Permutations'
+description: Given an array nums of distinct integers, return all the possible permutations. You can return the answer in any order
+icon: dot
+topics:
+  - { name: "Array", slug: "array" }
+  - { name: "Backtracking", slug: "backtracking" }
+issue: "https://github.com/prdlk/leetcode/issues/151"
+sidebar:
+  label: 'Permutations'
+  badge: 'Medium'
+---
+
+### Example 1:
+- Input: `nums = [1,2,3]`
+- Output: `[[1,2,3],[1,3,2],[2,1,3],[2,3,1],[3,1,2],[3,2,1]]`
+
+### Example 2:
+- Input: `nums = [0,1]`
+- Output: `[[0,1],[1,0]]`
+
+### Example 3:
+- Input: `nums = [1]`
+- Output: `[[1]]`
+
+### Constraints:
+
+- `1 <= nums.length <= 6`
+- `-10 <= nums[i] <= 10`
+- All the integers of `nums` are unique.
+
+## Solution
+
+```py
+class Solution:
+    def permute(self, nums: list[int]) -> list[list[int]]:
+        result, path = [], []
+        used = [False] * len(nums)
+
+        def backtrack():
+            if len(path) == len(nums):
+                result.append(path[:])
+                return
+            for i in range(len(nums)):
+                if used[i]:
+                    continue
+                used[i] = True
+                path.append(nums[i])
+                backtrack()
+                path.pop()
+                used[i] = False
+
+        backtrack()
+        return result
+```

apps/docs/content/(05-decision-space)/backtracking/(problems)/78_subsets.mdx

@@ -0,0 +1,45 @@
+---
+title: '78. Subsets'
+description: Given an integer array nums of unique elements, return all possible subsets (the power set)
+icon: dot
+topics:
+  - { name: "Array", slug: "array" }
+  - { name: "Backtracking", slug: "backtracking" }
+  - { name: "Bit Manipulation", slug: "bit-manipulation" }
+issue: "https://github.com/prdlk/leetcode/issues/149"
+sidebar:
+  label: 'Subsets'
+  badge: 'Medium'
+---
+
+### Example 1:
+- Input: `nums = [1,2,3]`
+- Output: `[[],[1],[2],[1,2],[3],[1,3],[2,3],[1,2,3]]`
+
+### Example 2:
+- Input: `nums = [0]`
+- Output: `[[],[0]]`
+
+### Constraints:
+
+- `1 <= nums.length <= 10`
+- `-10 <= nums[i] <= 10`
+- All the numbers of `nums` are unique.
+
+## Solution
+
+```py
+class Solution:
+    def subsets(self, nums: list[int]) -> list[list[int]]:
+        res = []
+
+        def backtrack(i, path):
+            res.append(path[:])
+            for num_idx in range(i, len(nums)):
+                path.append(nums[num_idx])
+                backtrack(num_idx + 1, path)
+                path.pop()
+
+        backtrack(0, [])
+        return res
+```