[ayosecu] WEEK 05 Solutions

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

@@ -0,0 +1,25 @@
+from typing import List
+class Solution:
+    """
+        - Time Complexity: O(n), n = len(prices)
+        - Space Complexity: O(1)
+    """
+    def maxProfit(self, prices: List[int]) -> int:
+        min_price = float("inf")
+        max_profit = 0
+
+        for price in prices:
+            min_price = min(min_price, price)
+            max_profit = max(max_profit, price - min_price)
+
+        return max_profit
+
+tc = [
+        ([7,1,5,3,6,4], 5),
+        ([7,6,4,3,1], 0)
+]
+
+for i, (prices, e) in enumerate(tc, 1):
+    sol = Solution()
+    r = sol.maxProfit(prices)
+    print(f"TC {i} is Passed!" if r == e else f"TC {i} is Failed! - Expected: {e}, Result: {r}")

encode-and-decode-strings/ayosecu.py

@@ -0,0 +1,49 @@
+class Solution:
+    """
+    @param: strs: a list of strings
+    @return: encodes a list of strings to a single string.
+    - Time Complexity: O(N), N = All characters in strs
+    - Space Complexity: O(N)
+    """
+    def encode(self, strs):
+        # Encode Format: xx#str
+        # xx:len(s)
+        # strs = ["abc", "defg"]
+        # Encoded String = 3#abc4#defg
+        enc_list = []
+        for s in strs:
+            enc_list.append(f"{len(s)}#{s}")
+        return "".join(enc_list)
+
+    """
+    @param: str: A string
+    @return: decodes a single string to a list of strings
+    - Time Complexity: O(n), n = len(str)
+    - Space Complexity: O(1), if output is excluded.
+    """
+    def decode(self, str):
+        result = []
+
+        i = 0
+        while i < len(str):
+            j = i
+            while str[j] != "#":
+                j += 1
+            n = int(str[i:j])
+            i = j + 1
+            result.append(str[i:i + n])
+            i += n
+
+        return result
+
+tc = [
+        (["lint","code","love","you"], "4#lint4#code4#love3#you"),
+        (["we", "say", ":", "yes"], "2#we3#say1#:3#yes")
+]
+
+for i, (p1, p2) in enumerate(tc, 1):
+    sol = Solution()
+    r = sol.encode(p1)
+    print(f"TC {i} - encode() is Passed!" if r == p2 else f"TC {i} - encode() is Failed! - Expected:{p2}, Result: {r}")
+    r = sol.decode(p2)
+    print(f"TC {i} - decode() is Passed!" if r == p1 else f"TC {i} - decode() is Failed! - Expected:{p1}, Result: {r}")

group-anagrams/ayosecu.py

@@ -0,0 +1,32 @@
+from typing import List
+from collections import defaultdict
+
+class Solution:
+    """
+        - Time Complexity: O(nklogk), n = len(strs), k = Max of len(strs[i])
+            - for loop => O(n)
+            - sorted(s) => O(klogk)
+        - Space Complexity: O(nk), dictionary size
+    """
+    def groupAnagrams(self, strs: List[str]) -> List[List[str]]:
+        # dictionary
+        dic = defaultdict(list)
+
+        for s in strs:
+            ss = "".join(sorted(s))
+            dic[ss].append(s)
+
+        return list(dic.values())
+
+tc = [
+        (["eat","tea","tan","ate","nat","bat"], [["bat"],["nat","tan"],["ate","eat","tea"]]),
+        ([""], [[""]]),
+        (["a"], [["a"]])
+]
+
+for i, (strs, e) in enumerate(tc, 1):
+    sol = Solution()
+    r = sol.groupAnagrams(strs)
+    r_sorted = sorted([sorted(group) for group in r])
+    e_sorted = sorted([sorted(group) for group in e])    
+    print(f"TC {i} is Passed!" if r_sorted == e_sorted else f"TC {i} is Failed! - Expected: {e_sorted}, Result: {r_sorted}")

implement-trie-prefix-tree/ayosecu.py

@@ -0,0 +1,72 @@
+class TrieNode:
+    def __init__(self):
+        self.children = {}
+        self.isEnd = False
+
+class Trie:
+    # Init Structure
+    def __init__(self):
+        self.root = TrieNode()        
+
+    # Time Complexity: O(n), n = len(word)
+    # Space Complexity: O(n)
+    def insert(self, word: str) -> None:
+        current = self.root
+        for c in word:
+            if c not in current.children:
+                current.children[c] = TrieNode()
+            current = current.children[c]
+        current.isEnd = True        
+
+    # Time Complexity: O(n), n = len(word)
+    # Space Complexity: O(1)
+    def search(self, word: str) -> bool:
+        current = self.root
+        for c in word:
+            if c not in current.children:
+                return False
+            current = current.children[c]       
+        return current.isEnd    # Check whole characters were matched
+
+    # Time Complexity: O(n), n = len(prefix)
+    # Space Complexity: O(1)
+    def startsWith(self, prefix: str) -> bool:
+        current = self.root
+        for c in prefix:
+            if c not in current.children:
+                return False
+            current = current.children[c]       
+        return True
+
+
+tc = [
+    ("insert", "apple", None),
+    ("search", "apple", True),
+    ("search", "app", False),
+    ("startsWith", "app", True),
+    ("insert", "app", None),
+    ("search", "app", True),
+    ("insert", "banana", None),
+    ("search", "banana", True),
+    ("search", "bananas", False),
+    ("startsWith", "ban", True),
+    ("startsWith", "baz", False),
+    ("search", "", False),
+    ("startsWith", "", True)
+]
+
+trie = Trie()
+for i, (op, value, expected) in enumerate(tc, 1):
+    if op == "insert":
+        result = trie.insert(value)
+    elif op == "search":
+        result = trie.search(value)
+    elif op == "startsWith":
+        result = trie.startsWith(value)
+    else:
+        result = None  # unknown operation
+
+    if result == expected:
+        print(f"TC {i} is Passed!")
+    else:
+        print(f"TC {i} is Failed! - Op: {op}('{value}'), Expected: {expected}, Result: {result}")

word-break/ayosecu.py

@@ -0,0 +1,35 @@
+from typing import List
+
+class Solution:
+    """
+        - Time Complexity: O(n^2), n = len(s)
+        - Space Complexity: O(n + m)
+            - n = len(s) = dp size
+            - m = The number of characters in wordDict (wordDict size)
+    """
+    def wordBreak(self, s: str, wordDict: List[str]) -> bool:
+        n = len(s)
+        wordDict = set(wordDict)
+
+        # Use DP: dp[i] => s[0:i] is possible to be seperated by dictionary words
+        dp = [False] * (n + 1)
+        dp[0] = True
+
+        for i in range(1, n + 1):
+            for j in range(i):
+                if dp[j] and s[j:i] in wordDict:
+                    dp[i] = True
+                    break   # s[i-1] is possible => don't need to check forward string
+
+        return dp[-1]   # check s[0:n]
+
+tc = [
+        ("leetcode", ["leet","code"], True),
+        ("applepenapple", ["apple","pen"], True),
+        ("catsandog", ["cats","dog","sand","and","cat"], False)
+]
+
+sol = Solution()
+for i, (s, wordDict, e) in enumerate(tc, 1):
+    r = sol.wordBreak(s, wordDict)
+    print(f"TC {i} is Passed!" if r == e else f"TC {i} is Failed! - Expected: {e}, Result: {r}")