Backtracking 2 done

Problem1.java

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+// Time Complexity : O(2^n) *n where n is the size of the input array
+// Space Complexity : 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
+
+/**
+ * Using backtracking approach to create a subset at every level with fixed pivot and moving i based for loop.
+ */
+class Solution {
+    public List<List<Integer>> subsets(int[] nums) {
+        List<List<Integer>> result = new ArrayList<>();
+        helper(nums, 0, new ArrayList<>(), result);
+        return result;
+    }
+
+    private void helper(int[] nums, int pivot, List<Integer> path, List<List<Integer>> result) {
+
+        result.add(new ArrayList<>(path));
+
+        for (int i = pivot; i < nums.length; i++) {
+            path.add(nums[i]);
+            helper(nums, i + 1, path, result);
+            path.remove(path.size() - 1);
+        }
+    }
+}

Problem2.java

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+// Time Complexity : O(2^n) *n where n is the size of the input array
+// 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
+
+/**
+ * Backtracking approach with for loop to create a substring at each pivot and i combination. Check if the string is palindrom.
+ */
+class Solution {
+    List<List<String>> result;
+
+    public List<List<String>> partition(String s) {
+        this.result = new ArrayList<>();
+        backtrack(s, 0, new ArrayList<>(), result);
+        return result;
+    }
+
+    private void backtrack(String s, int pivot, List<String> path, List<List<String>> result) {
+        if (pivot == s.length()) {
+            result.add(new ArrayList<>(path));
+            return;
+        }
+
+        for (int i = pivot; i < s.length(); i++) {
+            String curr = s.substring(pivot, i + 1);
+            if (isPalindrom(curr)) {
+                path.add(curr);
+                backtrack(s, i + 1, path, result);
+                path.remove(path.size() - 1);
+            }
+        }
+    }
+
+    private boolean isPalindrom(String s) {
+        int i = 0;
+        int j = s.length() - 1;
+        while (i <= j) {
+            if (s.charAt(i++) != s.charAt(j--))
+                return false;
+        }
+
+        return true;
+    }
+}