Shortest Way to Form a Target String

You may choose any subsequence of source, preserving character order, and concatenate any number of chosen subsequences.
Return the minimum number of source subsequences whose concatenation equals target. Return -1 when formation is impossible.

Function
shortestWay(source: String, target: String) → int

Examples
Example 1
source = "abc"
target = "abcbc"
return = 2
Use abc and then bc.
Example 2
source = "abc"
target = "acdbc"
return = -1
The character d never appears in the source.
Example 3
source = "xyz"
target = "xzyxz"
return = 3
One optimal decomposition is xz, y, and xz.

Constraints
1 <= source.length, target.length <= 100000.
Both strings contain lowercase English letters.

The trick is greedy. Walk through target one character at a time and scan source forward with a pointer, matching as many target characters as you can in a single pass. When the source pointer runs off the end, count one more subsequence and restart the source pointer at zero. Greedy works because taking the earliest match never hurts later matches. Before any of that, check that every target character exists in source, otherwise return -1 immediately. That check is also how you avoid an infinite loop. The naive two-pointer rescan is O(source * target) in the worst case, which at 100000 each is too slow. Fix it by precomputing, for each position and letter, the next index of that letter in source, then binary search or table lookup per target character. The common pitfall is forgetting the impossible case or resetting the pointer wrongly. StealthCoder is your hedge on the live OA if the next-index table slips your mind.

Drill it cold or hedge it with StealthCoder. Either way, don't walk into the OA hoping you remember the trick.