Question

691. Stickers to Spell Word

中文文档

Description

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.length
• 1 <= n <= 50
• 1 <= stickers[i].length <= 10
• 1 <= target.length <= 15
• stickers[i] and target consist of lowercase English letters.

Solutions

Solution 1

class Solution:
  def minStickers(self, stickers: List[str], target: str) -> int:
    q = deque([0])
    ans = 0
    n = len(target)
    vis = [False] * (1 << n)
    vis[0] = True
    while q:
      for _ in range(len(q)):
        state = q.popleft()
        if state == (1 << n) - 1:
          return ans
        for s in stickers:
          nxt = state
          cnt = Counter(s)
          for i, c in enumerate(target):
            if not (nxt & (1 << i)) and cnt[c]:
              nxt |= 1 << i
              cnt[c] -= 1
          if not vis[nxt]:
            vis[nxt] = True
            q.append(nxt)
      ans += 1
    return -1

class Solution {
  public int minStickers(String[] stickers, String target) {
    Deque<Integer> q = new ArrayDeque<>();
    q.offer(0);
    int ans = 0;
    int n = target.length();
    boolean[] vis = new boolean[1 << n];
    vis[0] = true;
    while (!q.isEmpty()) {
      for (int t = q.size(); t > 0; --t) {
        int state = q.poll();
        if (state == (1 << n) - 1) {
          return ans;
        }
        for (String s : stickers) {
          int nxt = state;
          int[] cnt = new int[26];
          for (char c : s.toCharArray()) {
            ++cnt[c - 'a'];
          }
          for (int i = 0; i < n; ++i) {
            int idx = target.charAt(i) - 'a';
            if ((nxt & (1 << i)) == 0 && cnt[idx] > 0) {
              nxt |= 1 << i;
              --cnt[idx];
            }
          }
          if (!vis[nxt]) {
            vis[nxt] = true;
            q.offer(nxt);
          }
        }
      }
      ++ans;
    }
    return -1;
  }
}

class Solution {
public:
  int minStickers(vector<string>& stickers, string target) {
    queue<int> q{{0}};
    int ans = 0;
    int n = target.size();
    vector<bool> vis(1 << n);
    vis[0] = true;
    while (!q.empty()) {
      for (int t = q.size(); t; --t) {
        int state = q.front();
        if (state == (1 << n) - 1) return ans;
        q.pop();
        for (auto& s : stickers) {
          int nxt = state;
          vector<int> cnt(26);
          for (char& c : s) ++cnt[c - 'a'];
          for (int i = 0; i < n; ++i) {
            int idx = target[i] - 'a';
            if (!(nxt & (1 << i)) && cnt[idx]) {
              nxt |= 1 << i;
              --cnt[idx];
            }
          }
          if (!vis[nxt]) {
            vis[nxt] = true;
            q.push(nxt);
          }
        }
      }
      ++ans;
    }
    return -1;
  }
};

func minStickers(stickers []string, target string) int {
  q := []int{0}
  n := len(target)
  vis := make([]bool, 1<<n)
  vis[0] = true
  ans := 0
  for len(q) > 0 {
    for t := len(q); t > 0; t-- {
      state := q[0]
      if state == (1<<n)-1 {
        return ans
      }
      q = q[1:]
      for _, s := range stickers {
        nxt := state
        cnt := make([]int, 26)
        for _, c := range s {
          cnt[c-'a']++
        }
        for i, c := range target {
          idx := c - 'a'
          if (nxt&(1<<i)) == 0 && cnt[idx] > 0 {
            nxt |= 1 << i
            cnt[idx]--
          }
        }
        if !vis[nxt] {
          vis[nxt] = true
          q = append(q, nxt)
        }
      }
    }
    ans++
  }
  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
  }
}