Find First Palindromic String in the Array

Find First Palindromic String in the Array - Problem

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.

Input & Output

Example 1 — Basic Case

$ Input: words = ["abc","car","ada","racecar","cool"]

› Output: "ada"

💡 Note: The first palindromic string is "ada". Note that "racecar" is also palindromic but comes after "ada".

Example 2 — No Palindrome

$ Input: words = ["notlob","world"]

› Output: ""

💡 Note: None of the strings are palindromic, so return empty string.

Example 3 — Single Character

$ Input: words = ["def","ghi","a"]

› Output: "a"

💡 Note: Single character strings are palindromes. "a" is the first palindromic string.

Constraints

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

Visualization

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Asked in

a Amazon 35 M Microsoft 28

The key insight is to check each string for palindrome property using two pointers. Best approach uses two pointers from ends moving inward for efficient palindrome verification. Time: O(n×m), Space: O(1).

Common Approaches

✓ Brute Force - String Reversal

⏱️ Time: O(n×m) Space: O(m)

For each string in the array, create a reversed version and compare it with the original string. Return the first string that matches its reverse.

Two Pointers Palindrome Check

⏱️ Time: O(n×m) Space: O(1)

For each string, use two pointers starting from both ends and moving toward center. Compare characters at pointer positions. If all match, it's a palindrome.

Brute Force - String Reversal — Algorithm Steps

Iterate through each string in words
For each string, create reversed version
Compare original with reversed
Return first match or empty string

Visualization

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Step-by-Step Walkthrough

Check First Word

Compare 'abc' with 'cba' - not equal

Check Second Word

Compare 'car' with 'rac' - not equal

Check Third Word

Compare 'ada' with 'ada' - equal! Return 'ada'

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int isPalindrome(char* word) {
    int len = strlen(word);
    char reversed[len + 1];
    
    for (int i = 0; i < len; i++) {
        reversed[i] = word[len - 1 - i];
    }
    reversed[len] = '\0';
    
    return strcmp(word, reversed) == 0;
}

char* solution(char words[][100], int n) {
    static char result[100];
    
    for (int i = 0; i < n; i++) {
        if (isPalindrome(words[i])) {
            strcpy(result, words[i]);
            return result;
        }
    }
    
    strcpy(result, "");
    return result;
}

int main() {
    char line[1000];
    fgets(line, sizeof(line), stdin);
    
    char words[100][100];
    int count = 0;
    
    char* token = strtok(line, "[],\"\n");
    while (token != NULL) {
        if (strlen(token) > 0 && strcmp(token, " ") != 0) {
            strcpy(words[count], token);
            count++;
        }
        token = strtok(NULL, "[],\"\n");
    }
    
    char* result = solution(words, count);
    printf("%s\n", result);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n×m)

Visit each of n strings, reverse operation takes O(m) time where m is average string length

✓ Linear Growth

Space Complexity

O(m)

Create reversed string of length m for each comparison

✓ Linear Space

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Common Approaches

Brute Force - String Reversal — Algorithm Steps

Visualization

Code -

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