How do you solve Number of Different Integers in a String on LeetCode?

Number of Different Integers in a String (LeetCode #1805) 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: Leading zeros are ignored when converting strings to integers.

What is the time complexity of Number of Different Integers in a String?

The optimal solution for Number of Different Integers in a String runs in O(n) time and uses O(n) space. The time complexity is O(n) because we traverse the string once and use a set to store unique integers, which allows for efficient lookups.

What is the brute force solution for Number of Different Integers in a String?

The brute force approach for Number of Different Integers in a String is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves replacing non-digit characters with spaces and then splitting the string to extract integers. We will then check for uniqueness by comparing each integer with others, which can be inefficient. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Number of Different Integers in a String?

Number of Different Integers in a String uses the following concepts: Hash Table, String. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Number of Different Integers in a String?

Always clarify the problem requirements, especially around uniqueness. Think about edge cases, such as strings with only letters or only digits. Practice explaining your thought process while coding.

What are common mistakes when solving Number of Different Integers in a String?

Not trimming the string after replacement, leading to empty strings. Forgetting to handle leading zeros properly.

#1805

Number of Different Integers in a String

Easy

Hash TableStringHash MapArray

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 uses a set to store unique integers directly while processing the string. This avoids the need for nested comparisons and efficiently handles leading zeros.

⚙️

Algorithm

4 steps

1Step 1: Replace all non-digit characters with spaces.
2Step 2: Split the string into parts based on spaces.
3Step 3: Convert each part to an integer (this removes leading zeros) and add it to a set to ensure uniqueness.
4Step 4: Return the size of the set as the result.

solution.py5 lines

1# Full working Python code
2word = 'a123bc34d8ef34'
3modified = ''.join(c if c.isdigit() else ' ' for c in word)
4unique_integers = set(int(i) for i in modified.split() if i)
5result = len(unique_integers)

ℹ

Complexity note: The time complexity is O(n) because we traverse the string once and use a set to store unique integers, which allows for efficient lookups.

1Leading zeros are ignored when converting strings to integers.
2Using a set automatically handles uniqueness.

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