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Hard |
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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.
Constraints:
n == stickers.length1 <= n <= 501 <= stickers[i].length <= 101 <= target.length <= 15stickers[i]andtargetconsist of lowercase English letters.
We notice that the length of the string target does not exceed 15. We can use a binary number of length 15 to represent whether each character of target has been spelled out. If the $i$th bit is 1, it means that the $i$th character of target has been spelled out; otherwise, it has not been spelled out.
We define an initial state 0, which means that all characters have not been spelled out. Then we use the Breadth-First Search (BFS) method, starting from the initial state. Each time we search, we enumerate all the stickers. For each sticker, we try to spell out each character of target. If we spell out a character, we set the $i$th bit of the corresponding binary number to 1, indicating that the character has been spelled out. Then we continue to search until we spell out all the characters of target.
The time complexity is target, and
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
}
}