#1417

Easy

auto_awesomeString-driven solution strategy

LeetCode Problem Workspace

Reformat The String

Reformat The String involves rearranging alphanumeric characters in a string so that no adjacent characters have the same type (letters or digits).

string

Problem Statement

You are given an alphanumeric string s consisting of lowercase English letters and digits. Your task is to rearrange the string in such a way that no two adjacent characters are of the same type, meaning no two letters or two digits can be next to each other.

Return the reformatted string if it’s possible, or an empty string if no valid rearrangement exists. Consider edge cases where the string contains only letters or only digits, as these cannot be reformatted to satisfy the condition.

Examples

Example 1

Input: s = "a0b1c2"

Output: "0a1b2c"

No two adjacent characters have the same type in "0a1b2c". "a0b1c2", "0a1b2c", "0c2a1b" are also valid permutations.

Example 2

Input: s = "leetcode"

Output: ""

"leetcode" has only characters so we cannot separate them by digits.

Example 3

Input: s = "1229857369"

Output: ""

"1229857369" has only digits so we cannot separate them by characters.

Constraints

1 <= s.length <= 500
s consists of only lowercase English letters and/or digits.

Solution Approach

Count and Compare the Number of Letters and Digits

First, count the number of letters and digits in the string. If the absolute difference between the counts of letters and digits is greater than 1, return an empty string as it's impossible to alternate the characters.

Rearrange Based on the Larger Group

If the counts of letters and digits are balanced or differ by one, alternate between them starting with the larger group. This ensures that no two adjacent characters are of the same type.

Edge Case Handling

Handle edge cases such as strings with only letters or only digits early, returning an empty string immediately, as no valid reformatting is possible.

Complexity Analysis

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

The time complexity of this solution is O(n), where n is the length of the string, as we must iterate through it at least once to count the characters. The space complexity is O(n) in the worst case due to the storage required for the result string.

What Interviewers Usually Probe

Focus on whether the candidate can identify edge cases like strings containing only letters or digits.
Evaluate their understanding of string-driven solution strategies and balancing character types.
Check if they are able to explain the time and space complexities clearly and concisely.

Common Pitfalls or Variants

Common pitfalls

Forgetting to handle edge cases where there are no letters or no digits, leading to incorrect answers.
Incorrectly assuming that the order of characters doesn't matter in terms of type adjacency.
Not considering cases where the counts of letters and digits differ by more than 1, leading to wrong results.

Follow-up variants

Handling strings with a mix of uppercase and lowercase letters.
Optimizing the solution for very large strings close to the maximum input size.
Allowing spaces or other characters in the input string.

FAQ

How do I handle strings with only letters or only digits in Reformat The String?

In such cases, immediately return an empty string because no valid reformatting is possible without the presence of both letters and digits.

What is the time complexity of the solution for Reformat The String?

The time complexity is O(n), where n is the length of the string, due to the need to traverse the string and count characters.

Can I solve Reformat The String using recursion?

While recursion could be applied, an iterative approach is generally more efficient for this problem and avoids the potential overhead of recursive calls.

How do I know when the reformatting is impossible in Reformat The String?

Reformatting is impossible if the difference in counts between letters and digits exceeds 1, meaning you can't alternate them without two of the same type being adjacent.

What should be the approach when the counts of letters and digits are the same in Reformat The String?

When the counts are the same, you can start the reformatting with either letters or digits. The result will alternate successfully between the two types.

terminal

Solution

Solution 1: Simulation

We classify all characters in string $s$ into two categories: "digits" and "letters", and put them into arrays $a$ and $b$ respectively.

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class Solution:
    def reformat(self, s: str) -> str:
        a = [c for c in s if c.islower()]
        b = [c for c in s if c.isdigit()]
        if abs(len(a) - len(b)) > 1:
            return ''
        if len(a) < len(b):
            a, b = b, a
        ans = []
        for x, y in zip(a, b):
            ans.append(x + y)
        if len(a) > len(b):
            ans.append(a[-1])
        return ''.join(ans)

Continue Practicing

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