8.8.2023

Author: ngocquanai

139. Word Break.py

@@ -0,0 +1,34 @@
+class Solution:
+    def wordBreak(self, s, wordDict):
+        dp = [False] * (len(s) + 1)
+        dp[len(s)] = True
+
+        # Change wordDict to dictionary to have 0(1) time for finding
+        word_dict = dict()
+        min_len = len(wordDict[0])
+        max_len = len(wordDict[0])
+        for element in wordDict:
+            part_len = len(element)
+            word_dict[element] = None
+            if min_len > part_len:
+                min_len = part_len
+            elif max_len < part_len:
+                max_len = part_len
+
+        for i in range(len(s) - 1, -1, -1):
+            for idx in range(min_len, max_len + 1):
+                if i + idx > len(s):
+                    break
+                if dp[i] == True:
+                    break
+
+                if s[i : i + idx] in word_dict:
+                    dp[i] = dp[i + idx]
+        return dp[0]
+        # Each element in word_dict range from min_len to max_len length
+
+
+wordDict = ["aaaa", "aaa"]
+s = "aaaaaaa"
+obj = Solution()
+print(obj.wordBreak(s, wordDict))

2124. Check if All A's Appears Before All B's.py

@@ -0,0 +1,29 @@
+class Solution:
+    def checkString(self, s):
+        first_b = -1
+        last_a = -1
+        status = False
+        for i in range(len(s)):
+            if s[i] == "a":
+                continue
+
+            first_b = i
+            break
+        for i in range(len(s) - 1, -1, -1):
+            if s[i] == "b":
+                continue
+            last_a = i
+            break
+        # Base case : there is not b in the array
+        if first_b == -1:
+            return False
+        # If last_a < first_b, then return True
+        if last_a < first_b:
+            return True
+        else:
+            return False
+
+
+obj = Solution()
+s = "bbb"
+print(obj.checkString(s))