Palindrome Substring Queries in JavaScript
Last Updated :
17 Nov, 2024
Here are the various approaches to check if a substring is a palindrome (a string that reads the same forward and backwards).
Using for Loop - Basic Approach
for loop iterates through each substring character from beginning to end for comparing each pair. If all pairs match, the substring is a palindrome.
JavaScript
function isPalindrome(s, start, end) {
for (let i = 0; i < Math.floor((end - start + 1) / 2);
i++) {
if (s[start + i] !== s[end - i]) {
return false;
}
}
return true;
}
const s = "racecar";
console.log(
isPalindrome(s, 0, s.length - 1));
console.log(isPalindrome(s, 0, 3));
Output
true
false
Using reverse(), join(), and split() Methods
In this approach, we extract the substring, reverse it, and check if it matches the original substring.
JavaScript
function isPalindrome(s, start, end) {
const sub = s.substring(start, end + 1);
return sub === sub.split("").reverse().join("");
}
const s = "racecar";
console.log(isPalindrome(s, 0, s.length - 1));
console.log(isPalindrome(s, 0, 3));
Output
true
false
Using Two-Pointer Technique
The two-pointer technique is an optimized approach where two pointers move toward the center from both ends, comparing characters. This method is efficient and does not require additional memory.
JavaScript
function isPalindrome(s, start, end) {
while (start < end) {
if (s[start] !== s[end]) {
return false;
}
start++;
end--;
}
return true;
}
const s = "madam";
console.log(isPalindrome(s, 0, s.length - 1));
console.log(isPalindrome(s, 1, 3));
console.log(isPalindrome(s, 0, 3));
Output
true
true
false
Using substring(), every(), and charAt() Methods
This approach uses substring() to extract a substring and every() with charAt() to compare corresponding characters from the start and end.
JavaScript
function isPalindrome(s, start, end) {
const sub = s.substring(start, end + 1);
return [...sub].every((char, i) =>
char === sub.charAt(sub.length - 1 - i));
}
const s = "civic";
console.log(isPalindrome(s, 0, s.length - 1));
console.log(isPalindrome(s, 0, 3));
Output
true
false
Using Hashing and Rolling Hash - Advanced Approach
Hashing and rolling hash techniques are advanced ways to check for palindromes. They make checking efficient when there are many queries by storing hash values for quick comparisons.
JavaScript
// Helper function to calculate forward and backward hashes
function computeHashes(s) {
const n = s.length;
const base = 31;
const mod = 1e9 + 7;
// Arrays to store hash values and powers
let forwardHash = Array(n + 1).fill(0);
let backwardHash = Array(n + 1).fill(0);
let power = Array(n + 1).fill(1);
// Compute power of base for each position
for (let i = 1; i <= n; i++) {
power[i] = (power[i - 1] * base) % mod;
}
// Compute forward and backward hashes
for (let i = 0; i < n; i++) {
forwardHash[i + 1] = (forwardHash[i] * base + s.charCodeAt(i)) % mod;
backwardHash[i + 1] =
(backwardHash[i] * base + s.charCodeAt(n - 1 - i)) % mod;
}
return { forwardHash, backwardHash, power, mod };
}
// Function to check if a substring is palindrome using precomputed hashes
function isPalindromeWithHash(s, start, end, hashes) {
const { forwardHash, backwardHash, power, mod } = hashes;
// Compute the forward hash for the substring
let forwardSubHash =
(forwardHash[end + 1] -
((forwardHash[start] * power[end - start + 1]) % mod) +
mod) %
mod;
// Compute the backward hash for the substring
let backwardSubHash =
(backwardHash[s.length - start] -
((backwardHash[s.length - end - 1]
* power[end - start + 1]) % mod) + mod) %
mod;
// Compare forward and backward hashes
return forwardSubHash === backwardSubHash;
}
const s = "level";
const hashes = computeHashes(s);
console.log(isPalindromeWithHash(s, 0, 4, hashes));
console.log(isPalindromeWithHash(s, 1, 3, hashes));
console.log(isPalindromeWithHash(s, 0, 2, hashes));
Output
true
true
false
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