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Program to find number of squareful arrays in Python
Suppose we want to make a target string of lowercase letters. At first, we have the sequence as n '?' marks (n is the length of target string). We also have a stamp of lowercase letters. On each turn, we can place the stamp over the sequence, and replace every letter in the with the corresponding letter from that stamp. You can make up to 10 * n turns.
As an example consider the initial sequence is "?????", and the stamp is "abc", then we may make strings like "abc??", "?abc?", "??abc" in the first turn. If the sequence is possible to stamp, then return an array of the index with the left-most letter being stamped at each turn. If that is not possible then return an empty array. So when the sequence is "ababc", and the stamp is "abc", then the answer can be like [0, 2], Because we can form like "?????" -> "abc??" - > "ababc".
So, if the input is like s = "abcd" t = "abcdbcd", then the output will be [3,0]
To solve this, we will follow these steps −
if size of s is same as 1, then
return a list from 0 to t when all characters in t are same and they are s[0], otherwise a new blank list
ans := a new list
while t is not same as size of t number of "?" marks, do
tmp := t
for i in range 0 to size of s, do
for j in size of s down to i+1:
search := i number of "?" concatenate substring of s[from index i to j-1] concatenate (size of s - j)number of "?"
while search is in t, do
insert where search is present in t at the end of ans
t := replace search with size of s number of "?" only once
if t is same as size of t number of "?", then
come out from loop
if t is same as size of t number of "?", then
come out from loop
if tmp is same as t, then
come out from loop
return reverse of ans.
Example
Let us see the following implementation to get better understanding
def solve(s, t): if len(s) == 1: return [i for i in range(len(t))] if all(t==s[0] for t in t)else [] ans = [] while t != "?" * len(t): tmp = t for i in range(len(s)): for j in reversed(range(i+1, len(s)+1)): search = "?" * i + s[i:j] + "?" * (len(s)-j) while t.find(search) != -1: ans.append(t.find(search)) t = t.replace(search, "?"*len(s), 1) if t == "?" * len(t): break if t == "?" * len(t): break if tmp == t: return [] return ans[::-1] s = "abcd" t = "abcdbcd" print(solve(s, t))
Input
"abcd", "abcdbcd"
Output
[3,0]
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