#1309

Easy

auto_awesomeString-driven solution strategy

LeetCode Problem Workspace

Decrypt String from Alphabet to Integer Mapping

Decrypt a string using alphabet-to-integer mapping, considering the presence of '#' as a special delimiter.

Problem Statement

You are given a string s consisting of digits and the '#' character. You need to map s into a string of lowercase English letters where '1' maps to 'a', '2' maps to 'b', and so on. Additionally, any two-digit number followed by '#' should map to a corresponding letter, with '10#' mapping to 'j', '11#' to 'k', and so on.

Return the decoded string after mapping the digits and '#' symbols as specified. The input string will always be valid, and there will be exactly one valid mapping to form the output.

Examples

Example 1

Input: s = "10#11#12"

Output: "jkab"

"j" -> "10#" , "k" -> "11#" , "a" -> "1" , "b" -> "2".

Example 2

Input: s = "1326#"

Output: "acz"

Example details omitted.

Constraints

1 <= s.length <= 1000
s consists of digits and the '#' letter.
s will be a valid string such that mapping is always possible.

Solution Approach

Right-to-Left Scan

Start scanning the string from right to left. Check if the character is a single digit or part of a two-digit number followed by '#'. If it is, map it to the corresponding letter and skip two extra positions. If it's a single digit, map it to its letter and move one position.

Iterate Through String

As you scan, keep appending the decoded characters to a result string. Build the result by appending the characters directly to avoid reversing the string at the end.

Edge Case Handling

Handle edge cases such as the smallest input (a single digit) or large inputs with many '#'. Ensure the approach works efficiently even for the largest possible inputs, within the given constraints.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity is O(n), where n is the length of the input string, as we scan the string once. The space complexity is O(n) as well, since we store the decoded string during processing.

What Interviewers Usually Probe

Candidate scans the string from right to left.
Candidate correctly handles '#' for two-digit numbers.
Candidate avoids unnecessary string reversals, constructing the result efficiently.

Common Pitfalls or Variants

Common pitfalls

Forgetting to skip characters after decoding two-digit numbers with '#'.
Misinterpreting the '#' symbol, either as part of a digit or skipping it.
Inefficiently reversing the string after decoding, instead of building it from right to left.

Follow-up variants

Handling strings with only single-digit characters.
Adapting the solution to work with uppercase letters instead of lowercase.
Dealing with larger strings and optimizing for performance.

FAQ

How do I approach the Decrypt String from Alphabet to Integer Mapping problem?

Start by scanning from right to left. Check if the character is part of a two-digit number with '#', or a single digit, and map accordingly.

What should I do when encountering a '#' in the Decrypt String problem?

Treat it as a delimiter to identify a two-digit number that needs to be mapped to a letter. Skip the next character after handling the mapping.

What are the common pitfalls in solving the Decrypt String from Alphabet to Integer Mapping problem?

The main issues include misinterpreting the '#' symbol, failing to skip characters after decoding two-digit numbers, and inefficiently reversing the result string.

How can I optimize the solution for larger input sizes in the Decrypt String problem?

Optimize by scanning the string once from right to left and constructing the result string directly, avoiding unnecessary reversals or extra passes.

Is there any specific pattern I should focus on when solving the Decrypt String problem?

The pattern to focus on is scanning the string from right to left while handling two-digit numbers with '#'. This ensures that you map correctly and avoid skipping characters.

terminal

Solution

Solution 1: Simulation

We can directly simulate the process.

1

2

3

4

5

6

7

8

9

10

11

12

class Solution:
    def freqAlphabets(self, s: str) -> str:
        ans = []
        i, n = 0, len(s)
        while i < n:
            if i + 2 < n and s[i + 2] == "#":
                ans.append(chr(int(s[i : i + 2]) + ord("a") - 1))
                i += 3
            else:
                ans.append(chr(int(s[i]) + ord("a") - 1))
                i += 1
        return "".join(ans)

Continue Practicing

#1307 verbal arithmetic puzzle #1312 minimum insertion steps to make a string palindrome #1316 distinct echo substrings