Problem Description:
Given a string sticker that represents the set of characters available on a single sticker and a string word that represents the target word to spell out, return the minimum number of stickers that you need to spell out word. Each sticker can be used more than once, and you have an unlimited supply of stickers.
Meta version of https://leetcode.com/problems/stickers-to-spell-word/description/

If the word cannot be spelled out using the letters on the sticker, return -1.

Note:

The sticker and word consist of lowercase English letters only.
The lengths of the sticker and word strings are both in the range [1, 1000].
Function Signature:

public int minStickers(String sticker, String word) {
}

Example 1:
Input: sticker = "ban", word = "banana"
Output: 3
Explanation: We can use 3 stickers "bana" to spell out the word "banana". Each sticker provides one "b", one "a", and one "n". Three stickers provide all the letters needed to spell out "banana".

Example 2:
Input: sticker = "abc", word = "def"
Output: -1
Explanation: The sticker does not contain any of the letters needed to spell out "def". Therefore, it is impossible to spell out the word.

Constraints:
1 <= sticker.length, word.length <= 1000
sticker and word consist of lowercase English letters.

Solution Submission:
Below is the Java solution based on the earlier discussed logic. Ensure your code is well-commented, clean, and follows good coding practices before submission.

public class Solution {

    public int minStickers(String sticker, String word) {
        int[] freqWord = new int[26]; // Frequency array for the target word

        // Populate frequency array for the word
        for (char c : word.toCharArray()) {
            freqWord[c - 'a']++;
        }

        int[] freqSticker = new int[26]; // Frequency array for the sticker

        // Populate frequency array for the sticker
        for (char c : sticker.toCharArray()) {
            freqSticker[c - 'a']++;
        }

        int ans = 0; // Initialize answer

        // Loop through each letter in the alphabet
        for (int i = 0; i < 26; i++) {
            if (freqWord[i] > 0) { // If the letter is in the word
                if (freqSticker[i] == 0) return -1; // If the letter is not in the sticker, return -1

                // Calculate the number of stickers needed for this letter and update ans
                int required = (int) Math.ceil((double) freqWord[i] / freqSticker[i]);
                ans = Math.max(ans, required);
            }
        }

        return ans; // Return the minimum number of stickers required
    }
}