add solution: word-break

Author: dusunax

word-break/dusunax.py

@@ -0,0 +1,38 @@
+'''
+# 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]