feat: add lc 46, 39, 78

work/1/Medium/Array/39.combination-sum.py

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+"""
+39. Combination Sum
+Difficulty: Medium
+https://leetcode.com/problems/combination-sum/
+
+──────────────────────────────────────────────────
+
+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.
+
+The same number may be chosen from candidates an unlimited number of
+times. Two combinations are unique if the frequency of at least one of
+the chosen numbers is different.
+
+The test cases are generated such that the number of unique
+combinations that sum up to target is less than 150 combinations for
+the given input.
+
+
+
+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
+"""
+
+
+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

work/1/Medium/Array/46.permutations.py

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+"""
+46. Permutations
+Difficulty: Medium
+https://leetcode.com/problems/permutations/
+
+──────────────────────────────────────────────────
+
+Given an array nums of distinct integers, return all the possible
+permutations. You can return the answer in any order.
+
+
+
+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.
+"""
+
+
+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

work/1/Medium/Array/78.subsets.py

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+"""
+78. Subsets
+Difficulty: Medium
+https://leetcode.com/problems/subsets/
+
+──────────────────────────────────────────────────
+
+Given an integer array nums of unique elements, return all possible
+subsets (the power set).
+
+The solution set must not contain duplicate subsets. Return the
+solution in any order.
+
+
+
+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.
+"""
+
+
+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