Completed Backtracking-2

Palindrome Partitioning

@@ -0,0 +1,48 @@
+// We’re doing a 0/1 recursion, where at every index we either pick the current substring or skip it.
+// If we pick, we check if it's a palindrome and recurse from the next pivot.
+// When the full size adds up to the original string, we save that partition.
+
+// time o(2^n * n)
+//space o(n^2)
+
+class Solution {
+    List<List<String>> result;
+    public List<List<String>> partition(String s) {
+        this.result = new ArrayList<>();
+        helper(s, 0, 0, 0, new ArrayList<>());
+        return result;
+    }
+
+    private void helper(String s, int pivot, int i, int size, List<String> path){
+        if(i == s.length()){
+            if(size == s.length()){
+                result.add(new ArrayList<>(path));
+            }
+            return;
+        }
+
+        //no choose
+        helper(s, pivot, i+1, size, path);
+
+        //choose
+        String subStr = s.substring(pivot, i+1);
+        if(isPalindrome(subStr)){
+            //action
+            path.add(subStr);
+            //recurse
+            helper(s, i+1, i+1, size + subStr.length(), path);
+            //backtrack
+            path.remove(path.size()-1);
+        }
+    }
+
+    private boolean isPalindrome(String s){
+        int left = 0, right = s.length()-1;
+        while(left < right){
+            if(s.charAt(left) != s.charAt(right))
+                return false;
+            left++; right--;
+        }
+        return true;
+    }
+}

Subsets

@@ -0,0 +1,24 @@
+// We start with the empty subset in the result list.
+// For each number, we copy all current subsets and append the number to create new subsets.
+// This way, every element doubles the size of the list by being added to existing subsets.
+
+// time o(n * 2^n)
+// space o(n)
+
+class Solution {
+    public List<List<Integer>> subsets(int[] nums) {
+        List<List<Integer>> result = new ArrayList<>();
+        result.add(new ArrayList<>());
+
+        for (int i = 0; i < nums.length; i++) {
+            int size = result.size();
+            for (int j = 0; j < size; j++) {
+                List<Integer> temp = new ArrayList<>(result.get(j));
+                temp.add(nums[i]);
+                result.add(temp);
+            }
+        }
+
+        return result;
+    }
+}