Is First Matching Character From Both Ends easy or hard?

First Matching Character From Both Ends is considered an easy problem. It mainly tests basic string traversal and understanding of the two-pointer pattern with O(n) time and O(1) space.

First Matching Character From Both Ends Python/Java solution

The solution is typically implemented with two indices scanning inward. In Python, Java, C++, Go, or TypeScript, the logic is identical: initialize left and right pointers, compare characters, and move inward until a matching pair is found.

How to solve First Matching Character From Both Ends in O(n)?

Use two pointers: one at the start and one at the end of the string. Compare the characters at both pointers. If they match, return that character; otherwise move both pointers inward and repeat until a match is found or the pointers cross.

What is the best approach for First Matching Character From Both Ends?

The two-pointer simulation approach is the best solution. Start one pointer at the beginning of the string and another at the end, compare characters, and move inward until a match is found. This method runs in O(n) time and uses O(1) extra space.

Is First Matching Character From Both Ends asked at Google/Amazon/Meta?

This specific problem is categorized as an easy string simulation question. Variants using two pointers and symmetric comparisons frequently appear in interviews at companies like Amazon, Google, and Meta because they test basic string manipulation and pointer logic.

What data structure is used in First Matching Character From Both Ends?

The problem primarily uses a string with a two-pointer technique. No additional data structures such as hash maps or stacks are required, keeping the space complexity constant.

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

The optimal solution runs in O(n) time because each mirrored pair of characters is checked at most once while scanning from both ends. The space complexity is O(1) since only a few pointer variables are used.

Description Hints 1 Solution Video 1:49

3884. First Matching Character From Both Ends

Easy81.5% acceptance

You are given a string s of length n consisting of lowercase English letters.

Return the smallest index i such that s[i] == s[n - i - 1].

If no such index exists, return -1.

Example 1:

Input: s = "abcacbd"

Output: 1

Explanation:

At index i = 1, s[1] and s[5] are both 'b'.

No smaller index satisfies the condition, so the answer is 1.

Example 2:

Input: s = "abc"

Output: 1

Explanation:

At index i = 1, the two compared positions coincide, so both characters are 'b'.

No smaller index satisfies the condition, so the answer is 1.

Example 3:

Input: s = "abcdab"

Output: -1

Explanation:

For every index i, the characters at positions i and n - i - 1 are different.

Therefore, no valid index exists, so the answer is -1.

Constraints:

1 <= n == s.length <= 100
s consists of lowercase English letters.

Topics

String Two Pointers

Hints open one at a time

Hint 1

Use two pointers

1:49

Still stuck?Read the written solution, or watch a walkthrough.

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1class Solution:
2    def firstMatchingIndex(self, s: str) -> int:
3        

Starter codePython 3Spaces: 4

Examples

Case 1Case 2Case 3

Input

"abcacbd"

Input

"abc"

Input

"abcdab"

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