Backtracking-1

CombinationSum.py

@@ -0,0 +1,27 @@
+# Time Complexity : O(2 ^ (m+n)) where m is the candidates length and n is the target
+# Space Complexity : O(n)
+# Did this code successfully run on Leetcode : Yes
+# Any problem you faced while coding this : No
+# Approach : We try all combinations starting from a pivot index.
+# The loop represents both choices: picking i handles the choose case, and moving i forward handles the no-choose.
+# After choosing a number, we recurse with a reduced target and backtrack to explore other options.
+
+class Solution:
+    def combinationSum(self, candidates: List[int], target: int) -> List[List[int]]:
+        self.res = []
+        self.helper(candidates, target, 0, [])
+        return self.res
+
+    def helper(self, candidates, target, pivot, path):
+        if target < 0 or pivot == len(candidates):
+            return
+
+        if target == 0:
+            self.res.append(list(path))
+            return
+
+        for i in range(pivot, len(candidates)):
+            path.append(candidates[i])
+            self.helper(candidates, target - candidates[i], i, path)
+            path.pop()
+