How do you solve Find First Palindromic String in the Array on LeetCode?

Find First Palindromic String in the Array (LeetCode #2108) 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: Palindromes can be checked efficiently using two pointers.

What is the brute force solution for Find First Palindromic String in the Array?

The brute force approach for Find First Palindromic String in the Array is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach checks each string in the array to see if it is a palindrome by reversing it and comparing it to the original. This is straightforward but can be inefficient for larger inputs. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Find First Palindromic String in the Array?

Find First Palindromic String in the Array uses the following concepts: Array, Two Pointers, String. The recommended patterns to study are: Two Pointers, String Manipulation.

What are the interview tips for Find First Palindromic String in the Array?

Always clarify the problem statement and constraints before jumping into coding. Think about edge cases, such as empty strings or single-character strings. Explain your thought process while coding; it helps the interviewer understand your approach.

What are common mistakes when solving Find First Palindromic String in the Array?

Not checking for the first palindromic string and returning the last one found. Using unnecessary data structures or methods that increase time complexity.

#2108

Find First Palindromic String in the Array

Easy

ArrayTwo PointersStringTwo PointersString Manipulation

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)

The optimal solution leverages a two-pointer technique to check for palindromes in a single pass, which is more efficient than reversing the string.

⚙️

Algorithm

3 steps

1Step 1: Iterate through each string in the array.
2Step 2: For each string, use two pointers to check if the string is a palindrome by comparing characters from both ends towards the center.
3Step 3: If a match is found, return that string. If no palindromic string is found by the end of the iteration, return an empty string.

solution.py16 lines

1# Full working Python code
2
3def first_palindromic_string(words):
4    for word in words:
5        if is_palindrome(word):
6            return word
7    return ""
8
9def is_palindrome(s):
10    left, right = 0, len(s) - 1
11    while left < right:
12        if s[left] != s[right]:
13            return False
14        left += 1
15        right -= 1
16    return True

ℹ

Complexity note: The time complexity is O(n) because we only traverse each word once with two pointers, making it more efficient than the brute force method. The space complexity is O(1) since we are not using any additional data structures that grow with input size.

1Palindromes can be checked efficiently using two pointers.
2The first occurrence is what matters, so we can stop as soon as we find one.

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