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困难 |
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我们有 n 种不同的贴纸。每个贴纸上都有一个小写的英文单词。
您想要拼写出给定的字符串 target ，方法是从收集的贴纸中切割单个字母并重新排列它们。如果你愿意，你可以多次使用每个贴纸，每个贴纸的数量是无限的。
返回你需要拼出 target 的最小贴纸数量。如果任务不可能，则返回 -1 。
注意：在所有的测试用例中，所有的单词都是从 1000 个最常见的美国英语单词中随机选择的，并且 target 被选择为两个随机单词的连接。
示例 1：
输入： stickers = ["with","example","science"], target = "thehat" 输出：3 解释： 我们可以使用 2 个 "with" 贴纸，和 1 个 "example" 贴纸。 把贴纸上的字母剪下来并重新排列后，就可以形成目标 “thehat“ 了。 此外，这是形成目标字符串所需的最小贴纸数量。
示例 2:
输入：stickers = ["notice","possible"], target = "basicbasic" 输出：-1 解释：我们不能通过剪切给定贴纸的字母来形成目标“basicbasic”。
提示:
n == stickers.length1 <= n <= 501 <= stickers[i].length <= 101 <= target.length <= 15stickers[i]和target由小写英文单词组成
我们注意到，字符串
我们定义一个初始状态
时间复杂度
class Solution:
def minStickers(self, stickers: List[str], target: str) -> int:
n = len(target)
q = deque([0])
vis = [False] * (1 << n)
vis[0] = True
ans = 0
while q:
for _ in range(len(q)):
cur = q.popleft()
if cur == (1 << n) - 1:
return ans
for s in stickers:
cnt = Counter(s)
nxt = cur
for i, c in enumerate(target):
if (cur >> i & 1) == 0 and cnt[c] > 0:
cnt[c] -= 1
nxt |= 1 << i
if not vis[nxt]:
vis[nxt] = True
q.append(nxt)
ans += 1
return -1class Solution {
public int minStickers(String[] stickers, String target) {
int n = target.length();
Deque<Integer> q = new ArrayDeque<>();
q.offer(0);
boolean[] vis = new boolean[1 << n];
vis[0] = true;
for (int ans = 0; !q.isEmpty(); ++ans) {
for (int m = q.size(); m > 0; --m) {
int cur = q.poll();
if (cur == (1 << n) - 1) {
return ans;
}
for (String s : stickers) {
int[] cnt = new int[26];
int nxt = cur;
for (char c : s.toCharArray()) {
++cnt[c - 'a'];
}
for (int i = 0; i < n; ++i) {
int j = target.charAt(i) - 'a';
if ((cur >> i & 1) == 0 && cnt[j] > 0) {
--cnt[j];
nxt |= 1 << i;
}
}
if (!vis[nxt]) {
vis[nxt] = true;
q.offer(nxt);
}
}
}
}
return -1;
}
}class Solution {
public:
int minStickers(vector<string>& stickers, string target) {
int n = target.size();
queue<int> q{{0}};
vector<bool> vis(1 << n);
vis[0] = true;
for (int ans = 0; q.size(); ++ans) {
for (int m = q.size(); m; --m) {
int cur = q.front();
q.pop();
if (cur == (1 << n) - 1) {
return ans;
}
for (auto& s : stickers) {
int cnt[26]{};
int nxt = cur;
for (char& c : s) {
++cnt[c - 'a'];
}
for (int i = 0; i < n; ++i) {
int j = target[i] - 'a';
if ((cur >> i & 1) == 0 && cnt[j] > 0) {
nxt |= 1 << i;
--cnt[j];
}
}
if (!vis[nxt]) {
vis[nxt] = true;
q.push(nxt);
}
}
}
}
return -1;
}
};func minStickers(stickers []string, target string) (ans int) {
n := len(target)
q := []int{0}
vis := make([]bool, 1<<n)
vis[0] = true
for ; len(q) > 0; ans++ {
for m := len(q); m > 0; m-- {
cur := q[0]
q = q[1:]
if cur == 1<<n-1 {
return
}
for _, s := range stickers {
cnt := [26]int{}
for _, c := range s {
cnt[c-'a']++
}
nxt := cur
for i, c := range target {
if cur>>i&1 == 0 && cnt[c-'a'] > 0 {
nxt |= 1 << i
cnt[c-'a']--
}
}
if !vis[nxt] {
vis[nxt] = true
q = append(q, nxt)
}
}
}
}
return -1
}function minStickers(stickers: string[], target: string): number {
const n = target.length;
const q: number[] = [0];
const vis: boolean[] = Array(1 << n).fill(false);
vis[0] = true;
for (let ans = 0; q.length; ++ans) {
const qq: number[] = [];
for (const cur of q) {
if (cur === (1 << n) - 1) {
return ans;
}
for (const s of stickers) {
const cnt: number[] = Array(26).fill(0);
for (const c of s) {
cnt[c.charCodeAt(0) - 97]++;
}
let nxt = cur;
for (let i = 0; i < n; ++i) {
const j = target.charCodeAt(i) - 97;
if (((cur >> i) & 1) === 0 && cnt[j]) {
nxt |= 1 << i;
cnt[j]--;
}
}
if (!vis[nxt]) {
vis[nxt] = true;
qq.push(nxt);
}
}
}
q.splice(0, q.length, ...qq);
}
return -1;
}use std::collections::{HashSet, VecDeque};
impl Solution {
pub fn min_stickers(stickers: Vec<String>, target: String) -> i32 {
let mut q = VecDeque::new();
q.push_back(0);
let mut ans = 0;
let n = target.len();
let mut vis = HashSet::new();
vis.insert(0);
while !q.is_empty() {
for _ in 0..q.len() {
let state = q.pop_front().unwrap();
if state == (1 << n) - 1 {
return ans;
}
for s in &stickers {
let mut nxt = state;
let mut cnt = [0; 26];
for &c in s.as_bytes() {
cnt[(c - b'a') as usize] += 1;
}
for (i, &c) in target.as_bytes().iter().enumerate() {
let idx = (c - b'a') as usize;
if (nxt & (1 << i)) == 0 && cnt[idx] > 0 {
nxt |= 1 << i;
cnt[idx] -= 1;
}
}
if !vis.contains(&nxt) {
q.push_back(nxt);
vis.insert(nxt);
}
}
}
ans += 1;
}
-1
}
}