What is the time complexity of Lexicographically Smallest String After Substring Operation?

The optimal solution for Lexicographically Smallest String After Substring Operation runs in O(n) time and uses O(n) space. The time complexity is O(n) because we only traverse the string a couple of times. The space complexity is O(n) due to the new string we create for the result.

What is the brute force solution for Lexicographically Smallest String After Substring Operation?

The brute force approach for Lexicographically Smallest String After Substring Operation is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves generating all possible substrings of the input string and replacing each substring with its preceding characters. We then compare all resulting strings to find the lexicographically smallest one. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Lexicographically Smallest String After Substring Operation?

Lexicographically Smallest String After Substring Operation uses the following concepts: String, Greedy. The recommended patterns to study are: Greedy, String Manipulation.

What are the interview tips for Lexicographically Smallest String After Substring Operation?

Always think about edge cases and how they affect your solution. Explain your thought process clearly, especially when transitioning from brute force to optimal solutions. Practice recognizing patterns in problems, such as when to use greedy approaches.

What are common mistakes when solving Lexicographically Smallest String After Substring Operation?

Failing to consider edge cases, such as strings that are entirely 'a's. Not optimizing the search for the first non-'a' character.

#2734

Lexicographically Smallest String After Substring Operation

Medium

StringGreedyGreedyString Manipulation

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)

Instead of checking all substrings, we can directly find the leftmost character that is not 'a' and replace all characters from that point onward. This is efficient because it guarantees the smallest lexicographical order by minimizing the first non-'a' character.

⚙️

Algorithm

3 steps

1Step 1: Find the index of the first character that is not 'a'.
2Step 2: Replace all characters from that index to the end of the string with their preceding characters.
3Step 3: If the string is all 'a's, replace the last character with 'z'.

solution.py6 lines

1def lexicographically_smallest_string(s):
2    n = len(s)
3    first_non_a = next((i for i, c in enumerate(s) if c != 'a'), n)
4    if first_non_a == n:
5        return s[:-1] + 'z'
6    return s[:first_non_a] + ''.join(chr(ord(c) - 1) if c != 'a' else 'z' for c in s[first_non_a:])

ℹ

Complexity note: The time complexity is O(n) because we only traverse the string a couple of times. The space complexity is O(n) due to the new string we create for the result.

1Finding the first non-'a' character is crucial for minimizing the string.
2Replacing characters efficiently can lead to significant performance improvements.

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