How do you solve First Matching Character From Both Ends on LeetCode?

First Matching Character From Both Ends (LeetCode #3884) 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: Using two pointers reduces unnecessary comparisons.

What is the time complexity of First Matching Character From Both Ends?

The optimal solution for First Matching Character From Both Ends runs in O(n) time and uses O(1) space. We only traverse the string once, making it linear time complexity.

What is the brute force solution for First Matching Character From Both Ends?

The brute force approach for First Matching Character From Both Ends is Brute Force, with O(n²) time complexity and O(1) space complexity. We check every index from the start of the string to see if the character matches its counterpart from the end. This is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve First Matching Character From Both Ends?

First Matching Character From Both Ends uses the following concepts: Two Pointers, String. The recommended patterns to study are: Two Pointers, String Matching.

What are the interview tips for First Matching Character From Both Ends?

Clarify edge cases with the interviewer. Explain your thought process while coding. Consider both time and space complexity during your solution.

What are common mistakes when solving First Matching Character From Both Ends?

Not considering the case where the string is of length 1. Confusing indices when checking characters from both ends.

#3884

First Matching Character From Both Ends

Easy

Two PointersStringTwo PointersString Matching

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)

Using two pointers, we can efficiently check characters from both ends towards the center, reducing unnecessary comparisons.

⚙️

Algorithm

3 steps

1Step 1: Initialize two pointers, left at 0 and right at n-1.
2Step 2: While left < right, check if s[left] == s[right].
3Step 3: If they match, return left; if no match found, return -1.

solution.py8 lines

1def first_matching_char(s):
2    left, right = 0, len(s) - 1
3    while left < right:
4        if s[left] == s[right]:
5            return left
6        left += 1
7        right -= 1
8    return -1

ℹ

Complexity note: We only traverse the string once, making it linear time complexity.

1Using two pointers reduces unnecessary comparisons.
2Matching characters can occur at various positions, not just the middle.

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