You are given two strings order and s. All the characters of order are unique and were sorted in some custom order previously.

Permute the characters of s so that they match the order that order was sorted. More specifically, if a character x occurs before a character y in order, then x should occur before y in the permuted string.

Return any permutation of s that satisfies this property.

LeetCode Problem - 791

class Solution {
    // This method custom sorts the string `s` based on the order specified in `order`.
    public String customSortString(String order, String s) {
        // HashMap to store the frequency of each character in the string `s`
        HashMap<Character, Integer> freq = new HashMap<>();

        // Count the frequency of each character in the string `s`
        for (int i = 0; i < s.length(); i++) {
            char letter = s.charAt(i);
            freq.put(letter, freq.getOrDefault(letter, 0) + 1);
        }

        // StringBuilder to build the custom sorted string
        StringBuilder sb = new StringBuilder();

        // Append characters in the order specified in `order`
        for (char letter : order.toCharArray()) {
            if (freq.containsKey(letter)) {
                sb.append(String.valueOf(letter).repeat(freq.get(letter)));
                // Remove the character from the frequency map after appending
                freq.remove(letter);
            }
        }

        // Append remaining characters in `s` that are not in `order`
        for (char letter : freq.keySet()) {
            sb.append(String.valueOf(letter).repeat(freq.get(letter)));
        }

        // Return the custom sorted string
        return sb.toString();
    }
}