2108. Find First Palindromic String in the Array 

Description

Given an array of strings words, return the first palindromic string in the array. If there is no such string, return an empty string "".

A string is palindromic if it reads the same forward and backward.

Example 1:

Input: words = ["abc","car","ada","racecar","cool"]
Output: "ada"
Explanation: The first string that is palindromic is "ada".
Note that "racecar" is also palindromic, but it is not the first.

Example 2:

Input: words = ["notapalindrome","racecar"]
Output: "racecar"
Explanation: The first and only string that is palindromic is "racecar".

Example 3:

Input: words = ["def","ghi"]
Output: ""
Explanation: There are no palindromic strings, so the empty string is returned.

Constraints:

1 <= words.length <= 100
1 <= words[i].length <= 100
words[i] consists only of lowercase English letters.

Solutions

Solution 1: Simulation

We iterate through the array words, for each string w, we determine if it is a palindrome. If it is, then we return w; otherwise, we continue to iterate.

To determine if a string is a palindrome, we can use two pointers, one pointing to the start and the other to the end of the string, moving towards the center, and checking if the corresponding characters are equal. If, after traversing the entire string, no unequal characters are found, then the string is a palindrome.

The time complexity is $O(L)$, where $L$ is the sum of the lengths of all strings in the array words. The space complexity is $O(1)$.

Python3

class Solution:
    def firstPalindrome(self, words: List[str]) -> str:
        return next((w for w in words if w == w[::-1]), "")

Java

class Solution {
    public String firstPalindrome(String[] words) {
        for (var w : words) {
            boolean ok = true;
            for (int i = 0, j = w.length() - 1; i < j && ok; ++i, --j) {
                if (w.charAt(i) != w.charAt(j)) {
                    ok = false;
                }
            }
            if (ok) {
                return w;
            }
        }
        return "";
    }
}

C++

class Solution {
public:
    string firstPalindrome(vector<string>& words) {
        for (auto& w : words) {
            bool ok = true;
            for (int i = 0, j = w.size() - 1; i < j; ++i, --j) {
                if (w[i] != w[j]) {
                    ok = false;
                }
            }
            if (ok) {
                return w;
            }
        }
        return "";
    }
};

Go

func firstPalindrome(words []string) string {
	for _, w := range words {
		ok := true
		for i, j := 0, len(w)-1; i < j && ok; i, j = i+1, j-1 {
			if w[i] != w[j] {
				ok = false
			}
		}
		if ok {
			return w
		}
	}
	return ""
}

TypeScript

function firstPalindrome(words: string[]): string {
    return words.find(w => w === w.split('').reverse().join('')) || '';
}

Rust

impl Solution {
    pub fn first_palindrome(words: Vec<String>) -> String {
        for w in words {
            if w == w.chars().rev().collect::<String>() {
                return w;
            }
        }
        String::new()
    }
}

C

char* firstPalindrome(char** words, int wordsSize) {
    for (int i = 0; i < wordsSize; ++i) {
        char* w = words[i];
        int len = strlen(w);
        bool ok = true;
        for (int j = 0, k = len - 1; j < k && ok; ++j, --k) {
            if (w[j] != w[k]) {
                ok = false;
            }
        }
        if (ok) {
            return w;
        }
    }
    return "";
}