[SunaDu] Week 6

container-with-most-water/dusunax.py

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+'''
+# 11. Container With Most Water
+
+use two pointers to find the maximum area.
+
+> **move the shorter line inward:**
+> - area is determined by the shorter line.
+> - move the shorter line inward => may find a taller line that can increase the area.
+
+## Time and Space Complexity
+
+```
+TC: O(n)
+SC: O(1)
+```
+
+### TC is O(n):
+- while loop iterates through the height array once. = O(n)
+
+### SC is O(1):
+- using two pointers and max_area variable. = O(1)
+'''
+
+class Solution:
+    def maxArea(self, height: List[int]) -> int:
+        left = 0
+        right = len(height) - 1
+        max_area = 0
+
+        while left < right: # TC: O(n)
+            distance = right - left
+            current_area = min(height[left], height[right]) * distance
+            max_area = max(current_area, max_area)
+
+            if height[left] < height[right]:
+                left += 1
+            else:
+                right -= 1
+
+        return max_area

design-add-and-search-words-data-structure/dusunax.py

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+'''
+# 211. Design Add and Search Words Data Structure
+
+use trie to perform add and search operations on words.
+use recursive dfs to search for words with "." wildcards.
+
+## Time and Space Complexity
+
+### addWord
+
+```
+TC: O(n)
+SC: O(n)
+```
+
+### TC is O(n):
+- iterating through each character of the word. = O(n)
+
+### SC is O(n):
+- storing the word in the trie. = O(n)
+
+### search
+
+```
+TC: O(n)
+SC: O(n)
+```
+
+### TC is O(n):
+- dfs iterates through each character of the word. = O(n)
+- if char is "."(wildcard) dfs iterates through all children of the current node. = O(n)
+
+> recursion for the wildcard could involve exploring several paths.
+> but time complexity is bounded by the length of the word, so it's still O(n).
+
+### SC is O(n):
+- length of the word as recursive call stack. = O(n)
+'''
+class WordDictionary:
+    def __init__(self):
+        self.trie = {}
+
+    def addWord(self, word: str) -> None:
+        node = self.trie
+        for char in word: # TC: O(n)
+            if char not in node:
+                node[char] = {}
+            node = node[char] 
+        node["$"] = True
+
+    def search(self, word: str) -> bool:
+        def dfs(node, i) -> bool:
+            if not isinstance(node, dict):
+                return False
+            if i == len(word):
+                return "$" in node if isinstance(node, dict) else False 
+            char = word[i]
+
+            if char == ".":
+                for child in node.values():
+                    if dfs(child, i + 1):
+                        return True
+                return False
+            else:
+                return dfs(node[char], i+1) if char in node else False
+
+        return dfs(self.trie, 0)

valid-parentheses/dusunax.py

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+'''
+# 20. Valid Parentheses
+
+use stack data structure to perform as a LIFO
+
+## Time and Space Complexity
+
+```
+TC: O(n)
+SC: O(n)
+```
+
+#### TC is O(n):
+- iterating through the string just once to check if the parentheses are valid. = O(n)
+
+#### SC is O(n):
+- using a stack to store the parentheses. = the worst case is O(n)
+- using a map to store the parentheses. = O(1)
+
+> for space complexity, fixed space is O(1).
+> 👉 parentheses_map is fixed and its size doesn't grow with the input size.
+> 👉 if the map has much larger size? the space complexity is still O(1).
+'''
+
+class Solution:
+    def isValid(self, s: str) -> bool:
+        stack = [] # SC: O(n)
+        parentheses_map = { # SC: O(1)
+            "(": ")",
+            "{": "}",
+            "[": "]"
+        }
+
+        for char in s: # TC: O(n) 
+            if char in parentheses_map:
+                stack.append(char)
+            else:
+                if len(stack) == 0 or parentheses_map[stack.pop()] != char:
+                    return False
+
+        return not stack