word-break solutions (ts, py)

Author: hi-rachel

word-break/hi-rachel.ts

@@ -0,0 +1,33 @@
+// wordDict 조합으로 s를 만들 수 있는지 반환해라
+// TC: O(n^2) SC: O(n + m * k) => s의 길이 + (단어 수 * 평균 단어 길이)
+// .has(word) => O(1)
+
+function wordBreak(s: string, wordDict: string[]): boolean {
+  const wordSet = new Set(wordDict);
+  const dp: boolean[] = Array(s.length + 1).fill(false);
+  dp[0] = true;
+
+  for (let i = 1; i <= s.length; i++) {
+    for (let j = 0; j < i; j++) {
+      if (dp[j] && wordSet.has(s.slice(j, i))) {
+        dp[i] = true;
+        break;
+      }
+    }
+  }
+  return dp[s.length];
+}
+
+// PY 풀이
+// class Solution:
+//     def wordBreak(self, s: str, wordDict: List[str]) -> bool:
+//         wordSet = set(wordDict)
+//         dp = [False] * (len(s)+1)
+//         dp[0] = True
+
+//         for i in range(1, len(s)+1):
+//             for j in range(0, i):
+//                 if dp[j] and s[j:i] in wordSet:
+//                     dp[i] = True
+
+//         return dp[len(s)]