Suppose the t's length is n, there K smaller strings and the smaller string is of length m.
The binary search solution would be of time complexity O(n + KMlogn).
But if we use a brute force method to build the dictionary for each char with the indices of the all the next chars, the time complexity is O(n^2). But we only need O(m) to decide if a string is a subsequence. The overall time complexity is O(n^2 + KM).
It is easy to test that the second is more efficient when K is really large.
But the code would not pass the test cases here. Down vote this question.