# Shortest Way to Form String

From any string, we can form asubsequenceof that string by deleting some number of characters (possibly no deletions).

Given two strings`source`and`target`, return the minimum number of subsequences of`source`such that their concatenation equals`target`. If the task is impossible, return`-1`.

**Example 1:**

```
Input: 
source = 
"abc"
, target = 
"abcbc"
Output: 
2
Explanation: 
The target "abcbc" can be formed by "abc" and "bc", which are subsequences of source "abc".
```

**Example 2:**

```
Input: 
source = 
"abc"
, target = 
"acdbc"
Output: 
-1
Explanation: 
The target string cannot be constructed from the subsequences of source string due to the character "d" in target string.
```

**Example 3:**

```
Input: 
source = 
"xyz"
, target = 
"xzyxz"
Output: 
3
Explanation: 
The target string can be constructed as follows "xz" + "y" + "xz".
```

**Note:**

1. Both the

   `source`

   and

   `target`

   strings consist of only lowercase English letters from "a"-"z".
2. The lengths of

   `source`

   and

   `target`

   string are between

   `1`

   and

   `1000`

   .

分析

贪心算法

每次遍历完整个source，target有 == 的就移动。下次继续继续重新遍历source。 返回遍历完多少次source

```
class Solution:
    def shortestWay(self, s: str, t: str) -> int:
        n,m = len(s),len(t)
        res =i= 0
        while i < m:
            ti = i
            for c in s:
                if i < m and c == t[i]:
                    i += 1
            if ti == i:
                return -1
            res += 1
        return res
```

建inverted index:

这里i 其实可以当做source无限连接，i在source的Pos.在reverted Index list范围内就可以跳动，说明还在当前s内，范围外就重来再新一个s继续loop

```
Initialize i = -1 (i represents the smallest valid next offset) and loop_cnt = 1 (number of passes through source).
Iterate through the target
```

[https://leetcode.com/problems/shortest-way-to-form-string/discuss/304662/Python-O(M-%2B-N\*logM)-using-inverted-index-%2B-binary-search-(Similar-to-LC-792\\](https://leetcode.com/problems/shortest-way-to-form-string/discuss/304662/Python-O%28M-%2B-N*logM%29-using-inverted-index-%2B-binary-search-%28Similar-to-LC-792%29\\)

```
class Solution:
    def shortestWay(self, s: str, t: str) -> int:
        ii = collections.defaultdict(list) #invertedIndex

        for idx, v in enumerate(s):
            ii[v].append(idx)
        loopcnt = 1
        curidx = -1
        for tidx, tv in enumerate(t):
            if tv not in ii:
                return -1
            iidx = bisect.bisect_left(ii[tv],curidx)
            if iidx >= len(ii[tv]):
                loopcnt += 1
                curidx = ii[tv][0] + 1
            else:
                curidx = ii[tv][iidx] + 1
        return loopcnt
```


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