1055 Shortest Way to Form String

A subsequence of a string is a new string that is formed from the original string by deleting some (can be none) of the characters without disturbing the relative positions of the remaining characters. (i.e., "ace" is a subsequence of "abcde" while "aec" is not).

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".

Constraints:

• 1 <= source.length, target.length <= 1000
• source and target consist of lowercase English letters.

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class Solution:
    def shortestWay(self, source: str, target: str) -> int:
        char_to_indices = defaultdict(list)
        for i, c in enumerate(source):
            char_to_indices[c].append(i)

        i = 0  # source_iterator
        count = 1

        for c in target:
            if c not in char_to_indices:
                return -1

            # Binary Search to find the index of the character in source
            # next to the current i
            index = bisect.bisect_left(char_to_indices[c], i)

            # If we have reached the end of the list, we need to iterate
            # through source again, hence first index of character in source.
            if index == len(char_to_indices[c]):
                count += 1
                i = char_to_indices[c][0] + 1
            else:
                i = char_to_indices[c][index] + 1
        return count