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Find First Palindromic String in the Array - LeetCode Daily Challenge

— Prerna Sharma

Find First Palindromic String in the Array - LeetCode Daily Challenge

Problem Statement

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".

Try here before watching the video.

Video Explanation

Java Code

class Solution {
    private boolean isPalindrome(String word) {
        int left = 0, right = word.length() - 1;
        while(left < right) {
            if(word.charAt(left++) != word.charAt(right--)) {
                return false;
            }
        }
        return true;
    }

    public String firstPalindrome(String[] words) {
        for(String word : words) {
            if(isPalindrome(word)) {
                return word;
            }
        }
        return "";
    }
}

C++ Code

class Solution {
private:
    bool isPalindrome(string word) {
        int left = 0, right = word.length() - 1;
        while (left < right) {
            if (word[left++] != word[right--]) {
                return false;
            }
        }
        return true;
    }

public:
    string firstPalindrome(vector<string>& words) {
        for (const string& word : words) {
            if (isPalindrome(word)) {
                return word;
            }
        }
        return "";
    }
};

Python Code

class Solution:
    def isPalindrome(self, word: str) -> bool:
        left, right = 0, len(word) - 1
        while left < right:
            if word[left] != word[right]:
                return False
            left += 1
            right -= 1
        return True

    def firstPalindrome(self, words: List[str]) -> str:
        for word in words:
            if self.isPalindrome(word):
                return word
        return ""

Javascript Code

var firstPalindrome = function(words) {
    const isPalindrome = word => {
        let left = 0, right = word.length - 1;
        while (left < right) {
            if (word[left++] !== word[right--]) {
                return false;
            }
        }
        return true;
    };

    for (const word of words) {
        if (isPalindrome(word)) {
            return word;
        }
    }
    return "";
};

Go Code

func isPalindrome(word string) bool {
	left, right := 0, len(word)-1
	for left < right {
		if word[left] != word[right] {
			return false
		}
		left++
		right--
	}
	return true
}

func firstPalindrome(words []string) string {
	for _, word := range words {
		if isPalindrome(word) {
			return word
		}
	}
	return ""
}

Complexity Analysis

Time Complexity: O(N * K), where N is the length of the words array and K is the length of each word in words.
Space Complexity: O(1)