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691. Stickers to Spell Word

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

BFS.

• Java
• C++
• Python
• Go
• RenderScript

• 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;
      }
  };
  

• 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
  
  

• 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
      }
  }
  
  

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