How do you solve Reformat The String on LeetCode?

Reformat The String (LeetCode #1417) 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: The difference in counts of letters and digits must not exceed 1 for a valid reformat.

What is the brute force solution for Reformat The String?

The brute force approach for Reformat The String is Brute Force, with O(n²) time complexity and O(n!) space complexity. The brute-force approach involves generating all possible permutations of the string and checking each one to see if it meets the criteria of alternating characters. This is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Reformat The String?

Reformat The String uses the following concepts: String. The recommended patterns to study are: Counting and grouping, Two Pointers.

What are the interview tips for Reformat The String?

Always clarify the problem constraints and edge cases with the interviewer. Think about the character types and their counts before jumping into code. Explain your thought process as you code; it helps the interviewer understand your approach.

What are common mistakes when solving Reformat The String?

Failing to check the count difference between letters and digits before attempting to format the string. Not handling edge cases where the string contains only letters or only digits.

#1417

Reformat The String

Easy

StringCounting and groupingTwo Pointers

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n)Space O(n)

The optimal approach involves counting the number of letters and digits, and then constructing the result based on which type has more characters. This is efficient and avoids unnecessary permutations.

⚙️

Algorithm

3 steps

1Step 1: Count the number of letters and digits in the string.
2Step 2: Check if the absolute difference between counts is greater than 1; if so, return an empty string.
3Step 3: Create a result array of the appropriate size and fill it by alternating characters from letters and digits.

solution.py16 lines

1def reformat_string(s):
2    letters = [c for c in s if c.isalpha()]
3    digits = [c for c in s if c.isdigit()]
4    if abs(len(letters) - len(digits)) > 1:
5        return ''
6    result = []
7    if len(letters) > len(digits):
8        result.append(letters.pop())
9    for i in range(len(s)):
10        if i % 2 == 0:
11            if digits:
12                result.append(digits.pop())
13        else:
14            if letters:
15                result.append(letters.pop())
16    return ''.join(result)

ℹ

Complexity note: The time complexity is O(n) because we traverse the string a couple of times (once for counting and once for building the result). The space complexity is O(n) for storing letters and digits separately.

1The difference in counts of letters and digits must not exceed 1 for a valid reformat.
2Always start with the character type that has more characters to maintain the alternating pattern.

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