diff --git a/problem1.java b/problem1.java
new file mode 100644
index 00000000..b04c8ae0
--- /dev/null
+++ b/problem1.java
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+    /**
+    Time Complexity = O(2^N)
+    Explanation:
+    Each element has two choices: include or exclude.
+    This generates all possible subsets → total 2^N subsets.
+
+    Space Complexity = O(N)
+    Explanation:
+    Recursion depth is N and path list stores at most N elements.
+
+    Did this code successfully run on LeetCode : Yes
+
+    Any problem you faced while coding this :
+    Initially created a new list for every recursive call,
+    which increased space usage.
+    Optimized it using backtracking (add → recurse → remove),
+    which reuses the same list and improves efficiency.
+    Also ensured to copy the path when adding to result to avoid mutation issues.
+    */
+
+    class Solution {
+
+
+
+        List<List<Integer>> result;
+    
+        public List<List<Integer>> subsets(int[] nums) {
+            this.result = new ArrayList<>();
+            helper(nums, 0, new ArrayList<>());
+            return result;
+        }
+    
+        private void helper(int[] nums, int i, List<Integer> path) {
+    
+            if (i == nums.length) {
+                result.add(new ArrayList<>(path)); // important copy
+                return;
+            }
+    
+            // Not choose
+            helper(nums, i + 1, path);
+    
+            // Choose
+            path.add(nums[i]);
+            helper(nums, i + 1, path);
+    
+            // Backtrack
+            path.remove(path.size() - 1);
+        }
+    }
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diff --git a/problem2.java b/problem2.java
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--- /dev/null
+++ b/problem2.java
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+
+    /**
+    Time Complexity : O(N * 2^N)
+    Explanation:
+    - There are 2^N possible ways to partition the string.
+    - For each partition, we check substrings for palindrome (O(N)).
+    So overall complexity is O(N * 2^N).
+
+    Space Complexity : O(N)
+    Explanation:
+    Recursion depth can go up to N and path stores at most N substrings.
+
+    Did this code successfully run on LeetCode : Yes
+
+    Any problem you faced while coding this :
+    Initially had difficulty deciding how to split the string dynamically.
+    Fixed it by using backtracking:
+        - Try every possible substring starting from current index
+        - Check if it is a palindrome
+        - If yes, include it in path and recurse
+    Also needed to backtrack properly (remove last element)
+    and copy the path when adding to result.
+    */
+
+    class Solution {
+
+
+        public List<List<String>> partition(String s) {
+    
+            List<List<String>> result = new ArrayList<>();
+            helper(s, 0, new ArrayList<>(), result);
+            return result;
+        }
+    
+        private void helper(String s, int pivot, List<String> path, List<List<String>> result) {
+    
+            // Base case
+            if (pivot == s.length()) {
+                result.add(new ArrayList<>(path));
+                return;
+            }
+    
+            // Try all possible substrings
+            for (int i = pivot; i < s.length(); i++) {
+    
+                String curr = s.substring(pivot, i + 1);
+    
+                if (isPalindrome(curr)) {
+                    path.add(curr);
+                    helper(s, i + 1, path, result);
+                    path.remove(path.size() - 1); // backtrack
+                }
+            }
+        }
+    
+        // Check palindrome
+        private boolean isPalindrome(String s) {
+    
+            int i = 0;
+            int j = s.length() - 1;
+    
+            while (i < j) {
+                if (s.charAt(i) != s.charAt(j)) return false;
+                i++;
+                j--;
+            }
+    
+            return true;
+        }
+    }
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