## Algorithm

Problem Name: 691. Stickers to Spell Word

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.

## Code Examples

### #1 Code Example with Javascript Programming

```Code - Javascript Programming```

``````
const minStickers = function(stickers, target) {
const isEqual = (arr1, arr2) => {
for (let i = 0; i  <  arr1.length; ++i) if (arr1[i] !== arr2[i]) return false
return true
}

const minus = (arr1, arr2) => {
let res = []
for (let i = 0; i  <  arr1.length; ++i)
res[i] = arr1[i] <= 0 ? arr1[i] : arr1[i] - arr2[i]
return res
}

const isAllNonpositive = arr => {
return arr.every(item => item <= 0)
}

const getString = arr => {
return arr.reduce((acc, cur, idx) => {
if (cur > 0) return acc + String.fromCharCode(idx + 97).repeat(cur)
else return acc
}, '')
}

let ss = stickers.map(word => {
let tmp = new Array(26).fill(0)
for (let i = 0; i  <  word.length; ++i) tmp[word.charCodeAt(i) - 97]++
return tmp
})
let root = new Array(26).fill(0)
for (let i = 0; i  <  target.length; ++i) root[target.charCodeAt(i) - 97]++
let cache = new Set()
let queue = [root]
let size = 0,
level = 0,
front = null
while (queue.length !== 0) {
size = queue.length
while (size--) {
front = queue.shift()
for (let w of ss) {
let t = minus(front, w)
let str = getString(t)
if (isEqual(t, front) || cache.has(str)) continue
if (isAllNonpositive(t)) return level + 1
else {
queue.push(t)
}
}
}
level++
}
return -1
}
``````
Copy The Code &

Input

cmd
stickers = ["with","example","science"], target = "thehat"

Output

cmd
3

### #2 Code Example with Python Programming

```Code - Python Programming```

``````
class Solution:
def minStickers(self, stickers, target):
cnt, res, n = collections.Counter(target), [float("inf")], len(target)
def dfs(dic, used, i):
if i == n:
res[0] = min(res[0], used)
elif dic[target[i]] >= cnt[target[i]]:
dfs(dic, used, i + 1)
elif used < res[0] - 1:
for sticker in stickers:
if target[i] in sticker:
for s in sticker:
dic[s] += 1
dfs(dic, used + 1, i + 1)
for s in sticker:
dic[s] -= 1
dfs(collections.defaultdict(int), 0, 0)
return res[0] < float("inf") and res[0] or -1
``````
Copy The Code &

Input

cmd
stickers = ["with","example","science"], target = "thehat"

Output

cmd
3