How do you solve Decrypt String from Alphabet to Integer Mapping on LeetCode?

Decrypt String from Alphabet to Integer Mapping (LeetCode #1309) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Scanning from right to left allows for efficient decoding.

What is the time complexity of Decrypt String from Alphabet to Integer Mapping?

The optimal solution for Decrypt String from Alphabet to Integer Mapping runs in O(n) time and uses O(n) space. The time complexity is O(n) because we only make a single pass through the string. The space complexity is O(n) due to the space used for the result string.

What is the brute force solution for Decrypt String from Alphabet to Integer Mapping?

The brute force approach for Decrypt String from Alphabet to Integer Mapping is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves checking each character in the string and determining if it’s part of a single-digit or a two-digit mapping. This is straightforward but inefficient as it may require multiple passes over the string. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Decrypt String from Alphabet to Integer Mapping?

Decrypt String from Alphabet to Integer Mapping uses the following concepts: String. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Decrypt String from Alphabet to Integer Mapping?

Always clarify the input format and constraints before starting. Think about edge cases, such as the smallest or largest possible inputs. Practice explaining your thought process as you code.

What are common mistakes when solving Decrypt String from Alphabet to Integer Mapping?

Not checking for the '#' character correctly. Confusing single-digit and two-digit mappings.

#1309

Decrypt String from Alphabet to Integer Mapping

Easy

StringHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

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Intuition

Time O(n)Space O(n)

The optimal approach involves scanning the string from right to left. This allows us to easily check for the presence of a '#' character and decode the string in a single pass, making it more efficient.

⚙️

Algorithm

3 steps

1Step 1: Initialize an empty result string.
2Step 2: Start iterating from the end of the string.
3Step 3: If the current character is '#', decode the two preceding characters as a two-digit number; otherwise, decode the current character as a single-digit number.

solution.py11 lines

1def decrypt(s):
2    result = ''
3    i = len(s) - 1
4    while i >= 0:
5        if s[i] == '#':
6            result += chr(int(s[i - 2:i]) + 96)
7            i -= 3
8        else:
9            result += chr(int(s[i]) + 96)
10            i -= 1
11    return result[::-1]

ℹ

Complexity note: The time complexity is O(n) because we only make a single pass through the string. The space complexity is O(n) due to the space used for the result string.

1Scanning from right to left allows for efficient decoding.
2Recognizing patterns in the input string helps in determining the mapping.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.