Add solution for Decode Ways problem

Author: KwonNayeon

decode-ways/KwonNayeon.py

@@ -0,0 +1,42 @@
+"""
+Conditions:
+- 1 <= s.length <= 100
+- s contains only digits and may contain leading zero(s).
+
+Time Complexity: O(n)
+- 각 문자를 한 번씩만 방문함 (문자열의 길이: n)
+
+Space Complexity: O(n)
+- dp 배열의 크기는 n+1
+- 이 배열 이외에 추가 공간을 사용하지 않음
+
+Base cases:
+- If string is empty or starts with '0', return 0 (impossible to decode)
+
+Dynamic Programming approach:
+- dp[i] represents the number of ways to decode first i characters
+- dp array has size n+1 to include the empty string case (dp[0])
+- For each position, consider both single-digit and two-digit decodings
+
+메모:
+- 다른 dp문제 풀어보기
+"""
+class Solution:
+    def numDecodings(self, s: str) -> int:
+        if not s or s[0] == '0':
+            return 0
+
+        n = len(s)
+        dp = [0] * (n + 1)
+        dp[0] = 1
+
+        for i in range(1, n + 1):
+            if s[i-1] != '0':
+                dp[i] += dp[i-1]
+
+            if i > 1 and s[i-2] != '0':
+                two_digit = int(s[i-2:i])
+                if 1 <= two_digit <= 26:
+                    dp[i] += dp[i-2]
+
+        return dp[n]