How do you solve Replace All Digits with Characters on LeetCode?

Replace All Digits with Characters (LeetCode #1844) 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: Each digit's replacement depends solely on the preceding character.

What is the brute force solution for Replace All Digits with Characters?

The brute force approach for Replace All Digits with Characters is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves iterating through the string and replacing each digit with the corresponding shifted character. This method is straightforward but inefficient, as it requires repeated calculations for each digit. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Replace All Digits with Characters?

Replace All Digits with Characters uses the following concepts: String. The recommended patterns to study are: Character Manipulation, String Processing.

What are the interview tips for Replace All Digits with Characters?

Always clarify the constraints and edge cases before starting to code. Think about the time complexity of your approach and how to optimize it. Practice explaining your thought process as you code, as this helps in interviews.

What are common mistakes when solving Replace All Digits with Characters?

Not handling the string indices correctly, especially when accessing previous characters. Forgetting to convert the character digit to an integer.

#1844

Replace All Digits with Characters

Easy

StringCharacter ManipulationString Processing

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n)Space O(n)

The optimal approach is similar to the brute-force method but avoids unnecessary calculations by directly calculating the new character based on the ASCII values. This allows us to achieve the desired result in linear time.

⚙️

Algorithm

4 steps

1Step 1: Initialize an empty result string.
2Step 2: Loop through each character in the string `s`.
3Step 3: If the index is even, append the character to the result.
4Step 4: If the index is odd, calculate the new character using ASCII values and append it to the result.

solution.py9 lines

1def replaceDigits(s):
2    result = ''
3    for i in range(len(s)):
4        if i % 2 == 0:
5            result += s[i]
6        else:
7            new_char = chr(ord(s[i - 1]) + int(s[i]))
8            result += new_char
9    return result

ℹ

Complexity note: The optimal solution runs in linear time because we only traverse the string once, performing constant-time operations for each character.

1Each digit's replacement depends solely on the preceding character.
2ASCII values can be used for efficient character shifting.

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