Create 2025-05-09-Leetcode-91.md

Author: hyeonukim

_posts/Leetcode/DP/2025-05-09-Leetcode-91.md

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+---
+title: Leetcode 91. Decode Ways
+description: Explanation for Leetcode 91 - Decode Ways, and its solution in Python.
+date: 2025-05-09
+categories: [Leetcode, Dynamic Programming, Medium]
+tags: [Leetcode, Python, Dynamic Programming, Medium]
+math: true
+---
+
+## Problem
+[Leetcode 91 - Decode Ways](https://leetcode.com/problems/decode-ways/description/)
+
+Example:
+```
+Input: s = "12"
+Output: 2
+Explanation: It can be decoded as "AB" (1 2) or "L" (12).
+
+Input: s = "226"
+Output: 3
+Explanation: It can be decoded as "BZ" (2 26), "VF" (22 6), or "BBF" (2 2 6).
+
+Input: s = "06"
+Output: 0
+Explanation: "06" cannot be mapped to any letter.
+```
+
+## Approach
+
+We can solve this problem using Dynamic Programming. The idea is to use a DP array where `dp[i]` represents the number of ways to decode the substring `s[:i]`. We iterate through the string and update the DP array based on valid single-digit and two-digit decodings.
+
+Here is the Python code for the solution:
+```python
+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  # Base case: empty string has one way to decode
+        dp[1] = 1  # Base case: single character (non-zero) has one way to decode
+        
+        for i in range(2, n + 1):
+            # Check single-digit decoding
+            if s[i - 1] != '0':
+                dp[i] += dp[i - 1]
+            
+            # Check two-digit decoding
+            two_digit = int(s[i - 2:i])
+            if 10 <= two_digit <= 26:
+                dp[i] += dp[i - 2]
+        
+        return dp[n]
+```
+
+## Time Complexity and Space Complexity
+
+Time Complexity: $O(n)$  
+
+Space Complexity: $O(n)$