You are given a string s. Your task is to divide the string into as many parts as possible such that each letter appears in at most one part.
In other words, no character should occur in more than one partition. After concatenating all parts in order, the result should be the original string s.
For example, given s = "bcbcdd", a valid partition is ["bcbc", "dd"]. However, partitions like ["bcb", "cdd"] or ["bc", "bc", "dd"] are invalid because some letters appear in multiple parts.
Return a list of integers representing the sizes of these partitions.
Constraints:
s.length
s consists of lowercase English letters.
The solution employs a two pointer approach combined with a last-occurrence map to divide the string, s, into the maximum number of contiguous partitions, such that each character appears in at most one part. At first glance, the problem appears to be tricky because the challenge lies in determining where to split the string. If we try to make a partition at the first occurrence of a character, it may fail. For example, in s = "abacbc", if we cut right after 'a', we get "a" and "bacbc", which is invalid because 'a' appears again later. This observation indicates that a valid partition must extend until the last occurrence of every character it contains. To solve this, we first determine the last position of each ...
You are given a string s. Your task is to divide the string into as many parts as possible such that each letter appears in at most one part.
In other words, no character should occur in more than one partition. After concatenating all parts in order, the result should be the original string s.
For example, given s = "bcbcdd", a valid partition is ["bcbc", "dd"]. However, partitions like ["bcb", "cdd"] or ["bc", "bc", "dd"] are invalid because some letters appear in multiple parts.
Return a list of integers representing the sizes of these partitions.
Constraints:
s.length
s consists of lowercase English letters.
The solution employs a two pointer approach combined with a last-occurrence map to divide the string, s, into the maximum number of contiguous partitions, such that each character appears in at most one part. At first glance, the problem appears to be tricky because the challenge lies in determining where to split the string. If we try to make a partition at the first occurrence of a character, it may fail. For example, in s = "abacbc", if we cut right after 'a', we get "a" and "bacbc", which is invalid because 'a' appears again later. This observation indicates that a valid partition must extend until the last occurrence of every character it contains. To solve this, we first determine the last position of each ...