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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]
andtarget
consist of lowercase English letters.
Solutions
BFS.
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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; } }
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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; } };
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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
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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 }
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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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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; }