How to use Character Frequency Count In Javascript

In this approach, we’ll determine if the given string can be rearranged into a palindrome. If it can, then it must be a rotation of some palindrome. This is based on the fact that a string that can be rearranged into a palindrome must have at most one character with an odd frequency.

Example:

JavaScript
function canFormPalindrome(str) {
    const charCount = new Map();

    // Count the frequency of each character
    for (let char of str) {
        if (charCount.has(char)) {
            charCount.set(char, charCount.get(char) + 1);
        } else {
            charCount.set(char, 1);
        }
    }

    // Check the number of characters with odd frequency
    let oddCount = 0;
    for (let count of charCount.values()) {
        if (count % 2 !== 0) {
            oddCount++;
        }
    }

    // For a string to be rearranged into a palindrome,
    // there can be at most one character with an odd frequency
    return oddCount <= 1;
}

function isRotationOfPalindrome(str) {
    // If the string can form a palindrome, then
    // it must be a rotation of some palindrome
    return canFormPalindrome(str);
}

// Test cases
console.log(isRotationOfPalindrome("racecar")); 
console.log(isRotationOfPalindrome("abbab"));
console.log(isRotationOfPalindrome("baba"));
console.log(isRotationOfPalindrome("abcd"));
console.log(isRotationOfPalindrome("aab"));

Output
true
true
true
false
true




JavaScript Program to Check if a Given String is a Rotation of a Palindrome

In this article, we will see how to check a given string is a rotation of a palindrome. A palindrome is a string that remains the same when its characters are reversed (e.g., “racecar” is a palindrome).

Example:

Input:  str = "racecar"
Output: true
// "racecar" is a rotation of a palindrome "racecar"
Input: str = "abbab"
Output: true
// "abbab" is a rotation of "abbba".
Input: str = "baba"
Output: true
// "baba" is a rotation of a palindrome "baba".
Input: str = "abcd"
Output: false
// "abcd" is not a rotation of any palimdrome.

Table of Content

  • Naive Approach
  • Efficient Approach:
  • Using Hashing

Similar Reads

Naive Approach

In this approach, we generate all possible rotations of the string and check if any of them is a palindrome. It does this by generating all possible rotations of the string and checking if any of these rotations are palindromes using the isPalindrome function. If it finds a palindrome rotation, it returns true; otherwise, it returns false....

Efficient Approach:

In this approach, we will find a specific rotation that makes it easier to determine if the string is a rotation of a palindrome. The code checks if a given string is a rotation of a palindrome. It creates a doubled version of the input string to handle all possible rotations efficiently. Then, it iterates through the doubled string, extracting substrings of the same length as the original string and checking if any of these substrings are palindromes using the isPalindrome function. If it finds a palindrome rotation, it returns true, otherwise, it returns false....

Using Hashing

Using hashing, the approach compares the original string’s hash with the hashes of all its rotations. If any rotation’s hash matches, it verifies if the substring is a palindrome, determining if the string is a rotation of a palindrome....

Using Character Frequency Count

In this approach, we’ll determine if the given string can be rearranged into a palindrome. If it can, then it must be a rotation of some palindrome. This is based on the fact that a string that can be rearranged into a palindrome must have at most one character with an odd frequency....