In this Kata, we are going to see how a Hash (or Map or dict) can be used to keep track of characters in a string. 
Consider two strings `"aabcdefg"` and `"fbd"`. How many characters do we have to remove from the first string to get the second string?  Although not the only way to solve this, we could create a Hash of counts for each string and see which character counts are different. That should get us close to the answer. I will leave the rest to you. 
For this example, `solve("aabcdefg","fbd") = 5`. Also, `solve("xyz","yxxz") = 0`, because we cannot get second string from the first since the second string is longer.
More examples in the test cases.
Good luck!
solve=lambda a,b,c=__import__('collections').Counter:not c(b)-c(a)and len(a)-len(b)
solve=lambda a,b,c=__import__('collections').Counter:not c(b)-c(a)and len(a)-len(b)