What is the time complexity of Replace All ?'s to Avoid Consecutive Repeating Characters?

The optimal solution for Replace All ?'s to Avoid Consecutive Repeating Characters runs in O(n) time and uses O(n) space. The complexity is O(n) because we traverse the string once, and each replacement is done in constant time.

What is the brute force solution for Replace All ?'s to Avoid Consecutive Repeating Characters?

The brute force approach for Replace All ?'s to Avoid Consecutive Repeating Characters is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves replacing each '?' with every possible lowercase letter and checking if the resulting string has consecutive repeating 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 Replace All ?'s to Avoid Consecutive Repeating Characters?

Replace All ?'s to Avoid Consecutive Repeating Characters uses the following concepts: String. The recommended patterns to study are: Two Pointers, Greedy Algorithms.

What are the interview tips for Replace All ?'s to Avoid Consecutive Repeating Characters?

Think about how to minimize the number of passes through the string. Practice explaining your thought process as you code, as this helps clarify your approach. Consider edge cases and how they might affect your solution.

What are common mistakes when solving Replace All ?'s to Avoid Consecutive Repeating Characters?

Failing to consider both adjacent characters when replacing '?'. Not handling edge cases, such as '?' at the beginning or end of the string.

#1576

Replace All ?'s to Avoid Consecutive Repeating Characters

Easy

StringTwo PointersGreedy Algorithms

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 solution processes the string in a single pass, replacing '?' with a character that avoids consecutive duplicates by checking adjacent characters. This is efficient and straightforward.

⚙️

Algorithm

4 steps

1Step 1: Convert the string into a list for mutability.
2Step 2: Iterate through each character in the string.
3Step 3: When encountering a '?', choose a character that is different from the previous and next characters.
4Step 4: Replace '?' with the chosen character.

solution.py11 lines

1s = '?zs'
2result = list(s)
3for i in range(len(result)):
4    if result[i] == '?':
5        for c in 'abcdefghijklmnopqrstuvwxyz':
6            if (i > 0 and result[i-1] == c) or (i < len(result)-1 and result[i+1] == c):
7                continue
8            result[i] = c
9            break
10final_string = ''.join(result)
11print(final_string)

ℹ

Complexity note: The complexity is O(n) because we traverse the string once, and each replacement is done in constant time.

1Always check adjacent characters when replacing '?' to avoid duplicates.
2Iterating through the string in a single pass is more efficient than backtracking.

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