String Palindrome Check - JavaScript Program

// Method 1: Compare with reversed string function isPalindrome1(str) { str = str.toLowerCase().replace(/[^a-z0-9]/g, ''); return str === str.split('').reverse().join(''); } console.log("madam:", isPalindrome1("madam")); console.log("hello:", isPalindrome1("hello")); console.log("A man a plan:", isPalindrome1("A man a plan a canal Panama")); // Method 2: Two-pointer approach function isPalindrome2(str) { str = str.toLowerCase().replace(/[^a-z0-9]/g, ''); let left = 0; let right = str.length - 1; while (left < right) { if (str[left] !== str[right]) { return false; } left++; right--; } return true; } console.log("\nTwo-pointer:"); console.log("racecar:", isPalindrome2("racecar")); console.log("world:", isPalindrome2("world")); // Method 3: Recursive approach function isPalindrome3(str) { str = str.toLowerCase().replace(/[^a-z0-9]/g, ''); if (str.length <= 1) return true; if (str[0] !== str[str.length - 1]) return false; return isPalindrome3(str.slice(1, -1)); } console.log("\nRecursive:"); console.log("level:", isPalindrome3("level")); console.log("hello:", isPalindrome3("hello")); // Method 4: Using every method function isPalindrome4(str) { str = str.toLowerCase().replace(/[^a-z0-9]/g, ''); return str.split('').every((char, i) => { return char === str[str.length - 1 - i]; }); } console.log("\nUsing every:"); console.log("deed:", isPalindrome4("deed"));

This program demonstrates different methods to check if a string is a palindrome.

Palindrome Definition

Reads same forwards and backwards:

"madam" → "madam" ✅
"hello" → "olleh" ❌

Method 1: Compare with Reversed

Simple approach:

javascript
str === str.split('').reverse().join('');

String Cleaning:

toLowerCase(): Case-insensitive
replace(/[^a-z0-9]/g, ''): Remove non-alphanumeric

Method 2: Two-Pointer

Most efficient:

javascript
let left = 0;
let right = str.length - 1;
while (left < right) {
    if (str[left] !== str[right]) return false;
    left++;
    right--;
}
return true;

Pros:

O(n) time, O(1) space
No string reversal needed
Most efficient

Method 3: Recursive

Elegant but less efficient:

javascript
if (str[0] !== str[str.length - 1]) return false;
return isPalindrome(str.slice(1, -1));

Method 4: Every Method

Functional approach:

javascript
str.split('').every((char, i) => {
    return char === str[str.length - 1 - i];
});

Every Method:

Tests if all elements pass test
Returns false on first failure

Time Complexity:

All methods: O(n)
Two-pointer: Best space efficiency

When to Use:

Reversed comparison: Simplest
Two-pointer: Most efficient
Recursive: Learning recursion
Every: Functional style

Output