feat: add lc 46, 39, 78

apps/docs/content/backtracking/39-combination-sum.mdx

@@ -0,0 +1,53 @@
+---
+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
+sidebar:
+  label: 'Combination Sum'
+  badge: 'Medium'
+---
+
+<Badge variant="accent">Array</Badge>
+
+### 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
+```