How do you solve Check Distances Between Same Letters on LeetCode?

Check Distances Between Same Letters (LeetCode #2399) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: Each letter appears exactly twice, which simplifies the problem.

What is the brute force solution for Check Distances Between Same Letters?

The brute force approach for Check Distances Between Same Letters is Brute Force, with O(n²) time complexity and O(1) space complexity. We can check the distance between the two occurrences of each letter by iterating through the string and comparing indices. This is straightforward but inefficient, as we will check each letter's distance separately. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Check Distances Between Same Letters?

Check Distances Between Same Letters uses the following concepts: Array, Hash Table, String. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Check Distances Between Same Letters?

Clearly explain your thought process and approach before coding. Consider edge cases and constraints while discussing your solution. Optimize your solution after the brute force approach to demonstrate problem-solving skills.

What are common mistakes when solving Check Distances Between Same Letters?

Not handling the case where a letter appears only once (though it won't happen here). Confusing the distance calculation by not subtracting 1 correctly.

#2399

Check Distances Between Same Letters

Easy

ArrayHash TableStringHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(1)

💡

Intuition

Time O(n)Space O(1)

We can optimize by storing the first occurrence of each letter in a single pass through the string, then check the distances in a second pass. This reduces unnecessary repeated searches.

⚙️

Algorithm

3 steps

1Step 1: Create an array to store the first occurrence index of each letter.
2Step 2: Iterate through the string and fill this array with the indices of the first occurrences.
3Step 3: For each letter, calculate the distance using the stored indices and compare it to the distance array.

solution.py10 lines

1def checkDistances(s, distance):
2    first_occurrence = [-1] * 26
3    for i, char in enumerate(s):
4        index = ord(char) - ord('a')
5        if first_occurrence[index] == -1:
6            first_occurrence[index] = i
7        else:
8            if i - first_occurrence[index] - 1 != distance[index]:
9                return False
10    return True

ℹ

Complexity note: The time complexity is O(n) because we only iterate through the string twice, and the space complexity is O(1) since we use a fixed-size array of size 26.

1Each letter appears exactly twice, which simplifies the problem.
2The distance between two occurrences can be calculated directly using their indices.

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