completed problem 1 & 2

problem1.cpp

@@ -0,0 +1,29 @@
+// Time Complexity : O(n*2^n)
+// Space Complexity : O(h) = O(n) 
+// Did this code successfully run on Leetcode : yes
+// Any problem you faced while coding this : no
+
+
+// Your code here along with comments explaining your approach
+// Start at idx 0 and at every recursive call, add the path to the result
+// Start the for loop from the pivot and add the element at idx to path and recurse
+// After the recursive call, backtrack the last element added 
+
+class Solution {
+private:
+    void helper(vector<int>& nums, int idx, vector<int>& path, vector<vector<int>>& res) {
+        res.push_back(path);
+        for(int i=idx; i<nums.size(); i++) {
+            path.push_back(nums[i]);
+            helper(nums, i+1, path, res);
+            path.pop_back();
+        }
+    }
+public:
+    vector<vector<int>> subsets(vector<int>& nums) {
+        vector<int> path;
+        vector<vector<int>> res;
+        helper(nums, 0, path, res);
+        return res;
+    }
+};

problem2.cpp

@@ -0,0 +1,48 @@
+// Time Complexity : O(n*n*2^n)
+// Space Complexity : O(n^2)
+// Did this code successfully run on Leetcode : yes
+// Any problem you faced while coding this : no
+
+
+// Your code here along with comments explaining your approach
+// start at index 0, break string into substrings at each index
+// only if the substring is a palindrome, make a recursive call to process the rest of the string
+// backtrack after the recursive call to explore different length substrings
+// push the partitions into the result when we reach the end of the string
+
+class Solution {
+private:
+    bool isPal(string str) {
+        int left = 0;
+        int right = str.length()-1;
+        while(left < right) {
+            if(str[left] != str[right]) {
+                return false;
+            }
+            left++;
+            right--;
+        }
+        return true;
+    }
+    void helper(string s, int idx, vector<string>& part, vector<vector<string>>& res) {
+        if(idx == s.length()) {
+            res.push_back(part);
+            return;
+        }
+        for(int i=idx; i<s.length(); i++) {
+            string sub = s.substr(idx, i-idx+1);
+            if(isPal(sub)) {
+                part.push_back(sub);
+                helper(s, i+1, part, res);
+                part.pop_back();
+            }
+        }
+    }
+public:
+    vector<vector<string>> partition(string s) {
+        vector<string> part;
+        vector<vector<string>> res;
+        helper(s, 0, part, res);
+        return res;    
+    }
+};