feat: Solve word-break problem

Author: hu6r1s

word-break/hu6r1s.java

@@ -0,0 +1,19 @@
+import java.util.*;
+
+class Solution {
+    public boolean wordBreak(String s, List<String> wordDict) {
+        Set<String> wordSet = new HashSet<>(wordDict);
+        boolean[] dp = new boolean[s.length() + 1];
+        dp[0] = true;
+
+        for (int i = 1; i < s.length() + 1; i++) {
+            for (int j = 0; j < i; j++) {
+                if (dp[j] && (wordSet.contains(s.substring(j, i)))) {
+                    dp[i] = true;
+                    break;
+                }
+            }
+        }
+        return dp[s.length()];
+    }
+}

word-break/hu6r1s.py

@@ -0,0 +1,14 @@
+class Solution:
+    def wordBreak(self, s: str, wordDict: List[str]) -> bool:
+        word_set = set(wordDict)   # O(1) 조회를 위해 set으로 변환
+        n = len(s)
+        dp = [False] * (n + 1)
+        dp[0] = True   # 공집합은 항상 가능
+
+        for i in range(1, n + 1):
+            for j in range(i):
+                if dp[j] and s[j:i] in word_set:
+                    dp[i] = True
+                    break   # i번째까지 나눌 수 있으면 더 확인할 필요 없음
+
+        return dp[-1]