"Backtracking-2 done"

Problem-1.java

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+// Time Complexity: O(n * 2^n), n = length of input array, because each element can double the number of subsets
+// Space Complexity: O(2^n), to store all subsets
+// 1. Start with an empty subset and iteratively build all subsets by adding each number to existing subsets.
+// 2. For each number, copy all existing subsets, add the current number, and append to the result.
+// 3. At the end, the result contains all possible subsets of the input array.
+
+class Solution {
+    public List<List<Integer>> subsets(int[] nums) {
+        List<List<Integer>> result = new ArrayList<>();
+        result.add(new ArrayList<>()); // start with empty subset
+
+        for (int num : nums) {
+            int resultSize = result.size();
+            for (int j = 0; j < resultSize; j++) {
+                List<Integer> temp = new ArrayList<>(result.get(j)); // make a copy
+                temp.add(num); // add current number
+                result.add(temp);
+            }
+        }
+
+        return result;
+    }
+}

Problem-1.py

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+# Time Complexity: O(n * 2^n), n = length of input array, because each element can double the number of subsets
+# Space Complexity: O(2^n), to store all subsets
+# 1. Start with an empty subset and iteratively build all subsets by adding each number to existing subsets.
+# 2. For each number, copy all existing subsets, add the current number, and append to the result.
+# 3. At the end, the result contains all possible subsets of the input array.
+
+class Solution:
+    def subsets(self, nums: List[int]) -> List[List[int]]:
+        result = []
+        result.append([])
+        for i in range(len(nums)):
+            result_size = len(result)
+            for j in range(result_size):
+                temp = result[j].copy()
+                temp.append(nums[i])
+                result.append(temp)
+            print(result)
+        return result
+

Problem-2.java

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+// Time Complexity: O(n * 2^n) in the worst case, n = length of the string
+// Space Complexity: O(n) for recursion stack + O(total partitions) for result
+// 1. Use DFS/backtracking to explore all possible partitions of the string.
+// 2. For each substring, check if it is a palindrome before adding to the current path.
+// 3. Add the current path to the result when the end of the string is reached and backtrack.
+
+class Solution {
+    List<List<String>> result;
+
+    public List<List<String>> partition(String s) {
+        result = new ArrayList<>();
+        helper(s, new ArrayList<>(), 0);
+        return result;
+    }
+
+    private void helper(String s, List<String> path, int pivot) {
+        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 (isPalindrome(curr)) {
+                path.add(curr);
+                helper(s, path, i + 1);
+                path.remove(path.size() - 1); // backtrack
+            }
+        }
+    }
+
+    private boolean isPalindrome(String curr) {
+        int i = 0, j = curr.length() - 1;
+        while (i < j) {
+            if (curr.charAt(i) != curr.charAt(j)) return false;
+            i++;
+            j--;
+        }
+        return true;
+    }
+}

Problem-2.py

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+# Time Complexity: O(n * 2^n) in the worst case, n = length of the string
+# Space Complexity: O(n) for recursion stack + O(total partitions) for result
+# 1. Use DFS/backtracking to explore all possible partitions of the string.
+# 2. For each substring, check if it is a palindrome before adding to the current path.
+# 3. Add the current path to the result when the end of the string is reached and backtrack.
+
+class Solution:
+    def partition(self, s: str) -> List[List[str]]:
+        self.result = []
+        self.helper(s,[],0)
+        return self.result
+    def helper(self,s,path,pivot):
+        if pivot == len(s):
+            self.result.append(list(path))
+            return
+        for i in range(pivot, len(s)):
+            curr = s[pivot:i+1]
+            if self.isPalindrome(curr):
+                path.append(curr)
+                self.helper(s,path,i+1)
+                path.pop()
+    def isPalindrome(self,curr):
+        i = 0
+        j = len(curr)-1
+        while i < j:
+            if curr[i] != curr[j]:
+                return False
+            i += 1
+            j -= 1
+        return True
+