Decision tree, traversed with DFS.

Leet_Code/medium/39_Combination_Sum.py

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+"""
+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
+
+Takeaway:
+
+    Decision tree, traversed with DFS.
+"""
+
+class Solution:
+    def combinationSum(self, candidates: "list[int]", target: "int") -> "list[list[int]]":
+
+        # to solve the decision tree
+        # we approach it in a unique manner
+        # at every level, when we decide that we are not adding
+        # a value to a possible solution, we wont
+        # be adding that value anymore
+
+        # to track which elements we can choose
+        # we will have a pointer and after each decision we 
+        # will move the pointer
+
+        res = []
+
+        # which are the elements we are allowed to choose - i 
+        def dfs(i , current_combination, total):
+            # base case where we succeed
+            if total == target:
+                # make a copy because we will be modifying it
+                res.append(current_combination.copy())
+                return
+            # out of bounds OR we are over target:
+            if i >= len(candidates) or total > target:
+                return
+
+            # first decision
+            # choose to add the candidates[i]
+            current_combination.append(candidates[i])
+            # move on deeper - do not change i, choose to reuse it
+            dfs(i, current_combination, total + candidates[i])
+
+            # second decision
+            # pop the value before going to other decision
+            # total will not change, because we did not add the value
+            current_combination.pop()
+            dfs(i + 1, current_combination, total)
+
+        dfs(0, [], 0)
+        return res