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Decode Ways II
Decode Ways II is a challenging dynamic programming problem that involves decoding messages with digits and wildcard characters.
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Practice Focus
Hard · State transition dynamic programming
Answer-first summary
Decode Ways II is a challenging dynamic programming problem that involves decoding messages with digits and wildcard characters.
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The problem requires determining the number of ways to decode a string with digits and wildcard characters ('*'). Using dynamic programming, we maintain a state that tracks the number of ways to decode the string up to each position, considering both valid digit-to-letter mappings and the wildcard's multiple possibilities. The core challenge involves handling wildcard transitions and calculating all valid decoding combinations at each step.
Problem Statement
You are given a string s consisting of digits ('0' to '9') and the wildcard character '*' representing any digit between 1 and 9. The goal is to determine the number of ways to decode the string based on a mapping of digits to letters (1 maps to 'A', 2 to 'B', ..., 26 to 'Z'). For example, "11106" can be mapped into multiple ways, but certain combinations like '06' are invalid since a leading zero cannot form a valid letter.
For each position in the string, consider both the current digit or wildcard and the previous digit for possible valid decodings. Your task is to compute the total number of ways to decode the entire string. The result should be returned modulo 10^9 + 7. Note that the problem involves managing state transitions using dynamic programming, considering wildcard behavior and transitions between adjacent digits.
Examples
Example 1
Input: See original problem statement.
Output: See original problem statement.
'A' -> "1" 'B' -> "2" ... 'Z' -> "26"
Example 2
Input: s = "*"
Output: 9
The encoded message can represent any of the encoded messages "1", "2", "3", "4", "5", "6", "7", "8", or "9". Each of these can be decoded to the strings "A", "B", "C", "D", "E", "F", "G", "H", and "I" respectively. Hence, there are a total of 9 ways to decode "*".
Example 3
Input: s = "1*"
Output: 18
The encoded message can represent any of the encoded messages "11", "12", "13", "14", "15", "16", "17", "18", or "19". Each of these encoded messages have 2 ways to be decoded (e.g. "11" can be decoded to "AA" or "K"). Hence, there are a total of 9 * 2 = 18 ways to decode "1*".
Constraints
- 1 <= s.length <= 105
- s[i] is a digit or '*'.
Solution Approach
Dynamic Programming State Transition
We can solve the problem efficiently using dynamic programming (DP). Define a DP array where dp[i] stores the number of ways to decode the string up to index i. At each position, consider both single-character and two-character decodings, accounting for possible wildcards. For wildcard characters, multiply the possibilities for each valid digit it can represent (1-9).
Handling Wildcards
Wildcards ('*') are treated differently since they can represent any digit from '1' to '9'. For each wildcard, we need to count all valid possibilities for both single and two-digit mappings, ensuring the decoding respects the range and constraints. Proper handling of these cases is key to solving the problem.
Modulo Operation
Since the number of ways can grow very large, the result is required modulo 10^9 + 7. This ensures that intermediate calculations do not overflow, and the final result remains within the bounds of typical integer operations.
Complexity Analysis
| Metric | Value |
|---|---|
| Time | Depends on the final approach |
| Space | Depends on the final approach |
The time complexity of this approach is O(n), where n is the length of the input string s, because we process each character in the string once. The space complexity is O(n) due to the DP array used to store intermediate results for each position in the string.
What Interviewers Usually Probe
- Candidate should show understanding of dynamic programming and state transition techniques.
- Look for clarity in handling wildcard transitions and edge cases, especially leading zeros or invalid combinations.
- Check if the candidate understands modular arithmetic to manage large numbers during the computation.
Common Pitfalls or Variants
Common pitfalls
- Misinterpreting the behavior of the '*' character and failing to count all valid digit possibilities.
- Forgetting to handle the modulo operation, potentially causing overflow errors or incorrect results.
- Confusing valid two-character decodings, especially when handling transitions between digits.
Follow-up variants
- What if the string only contains digits (no '*')? This simplifies the problem to a straightforward dynamic programming approach without the need for wildcard handling.
- What if the input string is extremely large (up to 10^5)? The solution should be optimized to handle such large inputs efficiently using dynamic programming.
- What if the '*' wildcard can also represent '0'? This would change the valid range and require additional checks for validity during decoding.
FAQ
How do I handle the wildcard '*' in Decode Ways II?
The '' character can represent any digit from 1 to 9, so for each occurrence of ' ', you need to account for all valid mappings for both single and two-digit combinations.
What is the time complexity of solving Decode Ways II?
The time complexity is O(n), where n is the length of the string. This is due to the need to process each character in the input string once while building up the DP array.
Can I solve Decode Ways II without dynamic programming?
While dynamic programming provides an optimal solution, solving the problem without it would be inefficient, potentially leading to exponential time complexity due to redundant calculations.
What does the modulo operation do in Decode Ways II?
The modulo operation ensures that the number of ways to decode the string does not exceed the limits of typical integer operations, keeping the result manageable and avoiding overflow.
How does GhostInterview help with Decode Ways II?
GhostInterview offers structured guidance to approach the problem, focusing on dynamic programming techniques and helping you understand state transitions and edge cases.
Solution
Solution 1
#### Python3
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 cContinue Practicing
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