Add implementation for leetcode problems 78, 131

leetcode_131.py

@@ -0,0 +1,63 @@
+#!/usr/bin/env python3
+# -*- coding: utf-8 -*-
+"""
+Created on Sun Mar 15 00:00:00 2026
+
+@author: rishigoswamy
+
+    LeetCode 131: Palindrome Partitioning
+    Link: https://leetcode.com/problems/palindrome-partitioning/
+
+    Problem:
+    Given a string s, partition s such that every substring of the partition is a
+    palindrome. Return all possible palindrome partitioning of s.
+
+    Approach:
+    Backtracking. At each step, try every substring s[pivot:i+1]. If it is a
+    palindrome, add it to path and recurse from i+1. When pivot reaches the end
+    of the string, a complete valid partition is found — append a deep copy to results.
+
+    1️⃣ Base case: pivot == len(s) → append deep copy of path to results.
+    2️⃣ Iterate i from pivot to end; extract s[pivot:i+1].
+    3️⃣ If substring is a palindrome, append to path, recurse with pivot = i+1, pop.
+    4️⃣ palindrome() uses two pointers to check in O(k) time.
+
+    // Time Complexity : O(n * 2^n)
+        At most 2^n partitions; palindrome check is O(n) per substring.
+    // Space Complexity : O(n)
+        Recursive call stack and path are at most O(n) deep.
+
+"""
+
+import copy
+from typing import List
+
+
+class Solution:
+    def partition(self, s: str) -> List[List[str]]:
+        self.result = []
+
+        def helper(s, pivot, path):
+            if pivot == len(s):
+                self.result.append(copy.deepcopy(path))
+                return
+
+            for i in range(pivot, len(s)):
+                subString = s[pivot:i + 1]
+                if palindrome(subString):
+                    path.append(subString)
+                    helper(s, i + 1, path)
+                    path.pop(-1)
+
+        def palindrome(s):
+            l = 0
+            r = len(s) - 1
+            while l < r:
+                if s[l] != s[r]:
+                    return False
+                l += 1
+                r -= 1
+            return True
+
+        helper(s, 0, [])
+        return self.result

leetcode_78.py

@@ -0,0 +1,47 @@
+#!/usr/bin/env python3
+# -*- coding: utf-8 -*-
+"""
+Created on Sun Mar 15 00:00:00 2026
+
+@author: rishigoswamy
+
+    LeetCode 78: Subsets
+    Link: https://leetcode.com/problems/subsets/
+
+    Problem:
+    Given an integer array nums of unique elements, return all possible subsets
+    (the power set). The solution set must not contain duplicate subsets.
+
+    Approach:
+    Backtracking. At each recursive call, the current path represents a valid subset
+    and is added to results immediately (including the empty set). Then iterate from
+    pivot onwards to avoid duplicate/reordered subsets, appending each element,
+    recursing, and backtracking.
+
+    1️⃣ Append a copy of path to results at every call (captures all subset sizes).
+    2️⃣ Iterate i from pivot to end; append nums[i], recurse with pivot = i+1, pop.
+
+    // Time Complexity : O(n * 2^n)
+        2^n subsets, each taking up to O(n) to copy into results.
+    // Space Complexity : O(n)
+        Recursive call stack and path are at most O(n) deep.
+
+"""
+
+from typing import List
+
+
+class Solution:
+    def subsets(self, nums: List[int]) -> List[List[int]]:
+        self.res = []
+
+        def helper(nums, pivot, path):
+            self.res.append(list(path))
+
+            for i in range(pivot, len(nums)):
+                path.append(nums[i])
+                helper(nums, i + 1, path)
+                path.pop()
+
+        helper(nums, 0, [])
+        return self.res