[SunaDu] Week 5

best-time-to-buy-and-sell-stock/dusunax.py

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+'''
+# 121. Best Time to Buy and Sell Stock
+
+use **bottom-up dynamic programming** to solve the problem.
+
+## Time and Space Complexity
+```
+TC: O(n)
+SC: O(1)
+```
+
+## TC is O(n):
+- iterating through the list just once to find the maximum profit. = O(n)
+
+## SC is O(1):
+- using two variables to store the minimum price and maximum profit. = O(1)
+'''
+class Solution:
+    def maxProfit(self, prices: List[int]) -> int:
+        if len(prices) == 1:
+            return 0
+
+        lowest_price = prices[0] # SC: O(1)
+        max_profit = 0 # SC: O(1)
+
+        for price in prices: # TC: O(n)
+            lowest_price = min(price, lowest_price)
+            curr_profit = price - lowest_price
+            max_profit = max(curr_profit, max_profit)
+
+        return max_profit

encode-and-decode-strings/dusunax.py

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+'''
+# 271. Encode and Decode Strings
+
+## Time and Space Complexity
+
+Use ":" as a separator and also store the length of each string to decode the string correctly.
+
+
+### encode
+
+```
+TC: O(n * k)
+SC: O(m)
+```
+
+#### TC is O(n * k):
+
+- iterating through the list of strings and appending each string to the result. = O(n * k)
+  - f-string is O(k)
+
+#### SC is O(m):
+- storing the result in a string.
+
+### decode
+
+```
+TC: O(m)
+SC: O(m)
+```
+
+#### TC is O(m):
+- iterating over the string until the string length is 0. = O(m)
+- do list slicings for extract parts and removing the processed section = each operation takes O(k)
+
+#### SC is O(m):
+- storing the result in a list(total length of strings is m) = O(m)
+
+'''
+
+class Solution:
+    """
+    @param: strs: a list of strings
+    @return: encodes a list of strings to a single string.
+    """
+    def encode(self, strs):
+        result = ''
+        for str in strs: # TC: O(n)
+            result += f"{len(str)}:{str}" # TC: O(k)
+        return result
+
+    """
+    @param: str: A string
+    @return: decodes a single string to a list of strings
+    """
+    def decode(self, str):
+      result = []
+
+      while len(str) > 0: # TC: O(m)
+          length = int(str[:1]) # TC: O(k)
+          string = str[2:length+2] # TC: O(k)
+          str = str[length+2:] # TC: O(k)
+
+          result.append(string) # SC: O(m)
+
+      return result

group-anagrams/dusunax.py

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+'''
+# 49. Group Anagrams
+
+use **hash map** to solve the problem.
+
+## Time and Space Complexity
+```
+TC: O(N * K * Log(K))
+SC: O(N * K)
+```
+
+## TC is O(N * K * Log(K)):
+- iterating through the list of strings and sorting each string. = O(N * K * Log(K))
+
+## SC is O(N * K):
+- using a hash map to store the sorted strings. = O(N * K)
+'''
+class Solution:
+  def groupAnagrams(self, strs: List[str]) -> List[List[str]]:
+      result = defaultdict(list)
+
+      for str in strs: # TC: O(N)
+          sorted_key = ''.join(sorted(str)) # sorting 👉 TC: O(K * Log(K))
+          result[sorted_key].append(str) # TC: O(1) on average
+
+      return list(result.values())
+

implement-trie-prefix-tree/dusunax.py

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+'''
+# 208. Implement Trie (Prefix Tree)
+
+```
+- Node structure
+{children: {char: {children: { ... }, is_end: False}}, is_end: False}
+```
+
+## Time and Space Complexity
+
+```
+TC: O(n)
+SC: O(n)
+```
+
+#### TC is O(n):
+- insert, search, startsWith: iterating through the word just once. = O(n)
+
+#### SC is O(n):
+- insert, search, startsWith: using a hash map to store the children nodes. = O(n)
+'''
+
+class Trie:
+    def __init__(self):
+        self.root = {"children": {}, "is_end": False}
+
+    def insert(self, word: str) -> None:
+        node = self.root
+        for char in word:
+            if char not in node["children"]:
+                node["children"][char] = {"children": {}, "is_end": False}
+            node = node["children"][char]
+        node["is_end"] = True
+
+    def search(self, word: str) -> bool:
+        node = self.root
+        for char in word:
+            if char not in node["children"]:
+                return False
+            node = node["children"][char]
+        return node["is_end"]
+
+    def startsWith(self, prefix: str) -> bool:
+        node = self.root
+        for char in prefix:
+            if char not in node["children"]:
+                return False
+            node = node["children"][char]
+        return True

word-break/dusunax.py

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+'''
+# 139. Word Break
+
+use dynamic programming to check if the string is segmentable.
+
+> **use dp to avoid redundant computations:**
+> backtracking approach will check all possible combinations of words recursively, which has time complexity of O(2^n))
+
+## Time and Space Complexity
+
+```
+TC: O(n^2)
+SC: O(n)
+```
+
+#### TC is O(n^2):
+- nested loops for checking if the string is segmentable. = O(n^2)
+  - outer loop: iterate each char index from the start to the end. = O(n)
+  - inner loop: for each index in the outer loop, checks substrings within the range of valid words in wordDict. = worst case, O(n)
+
+#### SC is O(n):
+- using a dp array to store whether index i is segmentable. = O(n)
+'''
+class Solution:
+    def wordBreak(self, s: str, wordDict: List[str]) -> bool:
+        word_set = set(wordDict)
+        n = len(s)
+
+        segment_dp = [False] * (n + 1) # SC: O(n)
+        segment_dp[0] = True # Base case: an empty string is segmentable
+
+        for end in range(1, n + 1): # TC: O(n^2)
+            for start in range(end):
+                if segment_dp[start] and s[start:end] in word_set:
+                    segment_dp[end] = True
+                    break
+
+        return segment_dp[n]