Last updated on October 9th, 2024 at 06:14 pm

Here, We see Stickers to Spell Word LeetCode Solution. This Leetcode problem is done in many programming languages like C++, Java, JavaScript, Python, etc. with different approaches.

List of all LeetCode Solution

Stickers to Spell Word LeetCode Solution

Problem Statement

We are given n different types of stickers. Each sticker has a lowercase English word on it.

You would like to spell out the given string target by cutting individual letters from your collection of stickers and rearranging them. You can use each sticker more than once if you want, and you have infinite quantities of each sticker.

Return the minimum number of stickers that you need to spell out target. If the task is impossible, return -1.

Note: In all test cases, all words were chosen randomly from the 1000 most common US English words, and target was chosen as a concatenation of two random words.

Example 1:
Input: stickers = [“with”,”example”,”science”], target = “thehat”
Output: 3
Explanation: We can use 2 “with” stickers, and 1 “example” sticker. After cutting and rearrange the letters of those stickers, we can form the target “thehat”. Also, this is the minimum number of stickers necessary to form the target string.

Example 2:
Input: stickers = [“notice”,”possible”], target = “basicbasic”
Output: -1
Explanation: We cannot form the target “basicbasic” from cutting letters from the given stickers.

1. Stickers to Spell Word LeetCode Solution C++

class Solution {
public:
    int minStickers(vector<string>& stickers, string& target) {
        int n = size(stickers);

        unordered_set<string> visited;

        vector<vector<int>> s_frequencies(n, vector<int>(26, 0));
        for (int i = 0; i < n; ++i)
            for (auto& c : stickers[i])
                ++s_frequencies[i][c - 'a'];
        
        vector<int> t_frequency(26, 0);
        for (auto& c : target)
            ++t_frequency[c - 'a'];

        queue<vector<int>> q;
        q.push(t_frequency);

        for (int res = 0; size(q); ++res) {
            for (int k = size(q); k > 0; --k) {
                auto t_freq = q.front(); q.pop();

                string t_str;
                for (int i = 0; i < 26; ++i)
                    if (t_freq[i] > 0)
                        t_str += string(t_freq[i], i);

                if (t_str == "") return res;

                if (visited.count(t_str)) continue;
                visited.insert(t_str);

                char seeking = t_str[0];
                for (auto& v : s_frequencies) {
                    if (v[seeking] > 0) {
                        q.push(t_freq); // Push first to copy t_freq
                        for (int i = 0; i < 26; ++i)
                            q.back()[i] -= v[i];
                    }
                }
            }
        }

        return -1;
    }
};

2. Stickers to Spell Word LeetCode Solution Java

class Solution {
    public int minStickers(String[] stickers, String target) {
        int n = stickers.length;

        target = sortChars(target);
        for (int i = 0; i < n; ++i) stickers[i] = sortChars(stickers[i]);
        Queue<String> q = new LinkedList();
        q.offer(target);
        int steps = 0;
        Set<String> visited = new HashSet<>();
        while (!q.isEmpty()) {
            steps++;
            int size = q.size();
            while(size-- > 0) {
                String x = q.poll();
                for (int i = 0; i < n; ++i) {
                    String now = filter(x, stickers[i]);
                    if (now.isEmpty()) return steps;
                    if (!now.equals(x) && !visited.contains(now)) {
                        visited.add(now);
                        q.offer(now);
                    }
                }
            }
        }
        return -1;
    }
    public String filter(String a, String b) {
        StringBuilder ret = new StringBuilder();
        int idx = 0;

        for (char c : a.toCharArray()) {
            boolean found = false;
            while (idx < b.length() && b.charAt(idx) <= c) {
                if (b.charAt(idx++) == c) {
                    found = true;
                    break;
                }
            }
            if (!found) ret.append(c);
        }
        return ret.toString();
    }
    private String sortChars(String s) {
        char[] chars = s.toCharArray();
        Arrays.sort(chars);
        return new String(chars);
    }
}

3. Stickers to Spell Word LeetCode Solution JavaScript

var minStickers = function(stickers, target) {
    let dp = new Map();
    const getStringDiff = (str1, str2) => {
        for(let c of str2) {
            if(str1.includes(c)) str1 = str1.replace(c, '');
        }
        return str1;
    }
    dp.set('', 0);

    function calcStickers(restStr) {
        if (dp.has(restStr)) return dp.get(restStr);
        let res = Infinity;
        for (let s of stickers.filter(s => s.includes(restStr[0]))) {
            let str = getStringDiff(restStr, s);
            res = Math.min(res, 1 + calcStickers(str));
        }
        dp.set(restStr, res === Infinity || res === 0 ? -1 : res);
        return dp.get(restStr);
    }
    return calcStickers(target)
}

4. Stickers to Spell Word LeetCode Solution Python

class Solution(object):
    def minStickers(self, stickers, target):
        n = len(target)
        maxMask = 1 << n
        dp = [float('inf')] * maxMask
        dp[0] = 0
        stickerCounts = []
        for sticker in stickers:
            stickerCounts.append(collections.Counter(sticker))
        for mask in range(maxMask):
            if dp[mask] == float('inf'):
                continue
            for i, stickerCount in enumerate(stickerCounts):
                if not any(c in stickerCount for c in target):
                    continue
                superMask = mask
                for c, freq in stickerCount.items():
                    for j, t in enumerate(target):
                        if c == t and not (superMask >> j & 1):
                            superMask |= 1 << j
                            freq -= 1
                            if freq == 0:
                                break
                dp[superMask] = min(dp[superMask], dp[mask] + 1)
        return -1 if dp[-1] == float('inf') else dp[-1]