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). For example, "11106" can be mapped into:
"AAJF" with the grouping (1 1 10 6)"KJF" with the grouping (11 10 6)Note that the grouping (1 11 06) is invalid because "06" cannot be mapped into 'F' since "6" is different from "06".
In addition to the mapping above, an encoded message may contain the '*' character, which can represent any digit from '1' to '9' ('0' is excluded). For example, the encoded message "1*" may represent any of the encoded messages "11", "12", "13", "14", "15", "16", "17", "18", or "19". Decoding "1*" is equivalent to decoding any of the encoded messages it can represent.
Given a string s consisting of digits and '*' characters, return the number of ways to decode it.
Since the answer may be very large, return it modulo 109 + 7.
Example 1:
Input: s = "*" Output: 9 Explanation: 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 2:
Input: s = "1*" Output: 18 Explanation: 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*".
Example 3:
Input: s = "2*" Output: 15 Explanation: The encoded message can represent any of the encoded messages "21", "22", "23", "24", "25", "26", "27", "28", or "29". "21", "22", "23", "24", "25", and "26" have 2 ways of being decoded, but "27", "28", and "29" only have 1 way. Hence, there are a total of (6 * 2) + (3 * 1) = 12 + 3 = 15 ways to decode "2*".
Constraints:
1 <= s.length <= 105s[i] is a digit or '*'.Problem Overview: You receive a string of digits representing encoded letters where '1' -> 'A' through '26' -> 'Z'. The twist is the character *, which can represent any digit from 1 to 9. Your task is to count how many valid decoding ways exist for the entire string. The answer can grow large, so return it modulo 1e9 + 7. This problem combines tricky case handling with classic dynamic programming over a string.
Approach 1: Dynamic Programming (O(n) time, O(n) space)
Use a DP array where dp[i] represents the number of ways to decode the prefix ending at index i. At each position, evaluate two possibilities: decoding the current character alone, and decoding the current character together with the previous one. Handling * is the key challenge because it expands into multiple valid digits. For example, a single * contributes 9 possibilities, while combinations like 1* produce 9 valid pairs and 2* produce 6. Similarly, ** creates 15 valid two-digit combinations (11–19 and 21–26). Iterate through the string, update dp[i] using these rules, and apply modulo arithmetic to prevent overflow. This approach is straightforward to reason about because every state explicitly stores the number of ways up to that index.
Approach 2: Space Optimized Dynamic Programming (O(n) time, O(1) space)
The recurrence only depends on the previous two states, so you can eliminate the full DP array. Maintain two variables: one for the number of ways to decode up to the previous position and one for the position before that. As you scan the string from left to right, compute the current number of ways using the same transition logic as the DP solution: single-character decoding and two-character decoding. Carefully handle cases involving *, such as *0, *7, or **, by counting the valid digit combinations that map to numbers between 10 and 26. After computing the new value, shift the variables forward. This keeps memory constant while preserving the same logic and correctness.
Both approaches rely on recognizing valid one-digit and two-digit decoding patterns while expanding wildcard possibilities. The core observation is that every index contributes ways derived from the previous one or two indices. The DP transition encodes all legal combinations defined by the mapping rules.
Recommended for interviews: Interviewers typically expect the space-optimized dynamic programming approach. It demonstrates that you understand the recurrence and can reduce memory usage from O(n) to O(1). Starting with the full DP table is a good way to explain your thinking. Then compress the state to show deeper mastery of dynamic programming patterns on string decoding problems.
This approach utilizes a dynamic programming (DP) technique to solve the problem. We define a DP array where dp[i] represents the number of ways to decode the substring of length i. We iterate through the string, updating the DP array based on the current character and its possible pairings with the previous character.
This solution introduces a dynamic programming approach. The dp[i] array stores the number of decoding ways for the substring up to the ith character. The initial state is set based on the first character, considering '.' as all digits 1-9. Each subsequent character adjusts dp[i] depending on its value and the preceding character(s), also accounting for the '*' possibilities. All computations are done modulo 10^9 + 7 to limit overflow.
Time Complexity: O(n), where n is the length of the input string.
Space Complexity: O(n), due to the dynamic programming array used.
This is a space-optimized version of the dynamic programming solution. Instead of using a full dp array, we keep track of only the last two states needed, thereby reducing space complexity.
This C solution optimizes space by maintaining only two integer variables, prev and curr, to store results of the previous and current positions respectively. This replaces the need for an entire dp array, reducing space usage while maintaining the dynamic programming logic.
C
JavaScript
Time Complexity: O(n), where n is the length of the input string.
Space Complexity: O(1), as we use only constant extra space.
| Approach | Complexity |
|---|---|
| Dynamic Programming | Time Complexity: O(n), where |
| Space Optimized Dynamic Programming | Time Complexity: O(n), where |
| Default Approach | — |
| Approach | Time | Space | When to Use |
|---|---|---|---|
| Dynamic Programming | O(n) | O(n) | Best for understanding transitions clearly and debugging wildcard decoding rules |
| Space Optimized Dynamic Programming | O(n) | O(1) | Preferred in interviews or memory-constrained environments since only two previous states are required |
花花酱 LeetCode 639. Decode Ways II - 刷题找工作 EP110 • Hua Hua • 4,004 views views
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