Leetcode #639: Decode Ways II
In this guide, we solve Leetcode #639 Decode Ways II in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
A message containing letters from A-Z can be encoded into numbers using the following mapping: 'A' -> "1" 'B' -> "2" ... 'Z' -> "26" To decode an encoded message, all the digits must be grouped then mapped back into letters using the reverse of the mapping above (there may be multiple ways).
Quick Facts
- Difficulty: Hard
- Premium: No
- Tags: String, Dynamic Programming
Intuition
The problem breaks into overlapping subproblems, so caching results prevents exponential repetition.
A carefully chosen DP state captures exactly what we need to build the final answer.
Approach
Define the DP state and recurrence, then compute states in the correct order.
Optionally compress space once the recurrence is clear.
Steps:
- Choose a DP state definition.
- Write the recurrence and base cases.
- Compute states in the correct order.
Example
'A' -> "1"
'B' -> "2"
...
'Z' -> "26"
Python Solution
class Solution:
def numDecodings(self, s: str) -> int:
mod = int(1e9 + 7)
n = len(s)
# dp[i - 2], dp[i - 1], dp[i]
a, b, c = 0, 1, 0
for i in range(1, n + 1):
# 1 digit
if s[i - 1] == "*":
c = 9 * b % mod
elif s[i - 1] != "0":
c = b
else:
c = 0
# 2 digits
if i > 1:
if s[i - 2] == "*" and s[i - 1] == "*":
c = (c + 15 * a) % mod
elif s[i - 2] == "*":
if s[i - 1] > "6":
c = (c + a) % mod
else:
c = (c + 2 * a) % mod
elif s[i - 1] == "*":
if s[i - 2] == "1":
c = (c + 9 * a) % mod
elif s[i - 2] == "2":
c = (c + 6 * a) % mod
elif (
s[i - 2] != "0"
and (ord(s[i - 2]) - ord("0")) * 10 + ord(s[i - 1]) - ord("0") <= 26
):
c = (c + a) % mod
a, b = b, c
return c
Complexity
The time complexity is O(n·m) (typical). The space complexity is O(n·m) or optimized.
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.