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You are given a string s. We want to partition this string into as many parts as possible so that each letter appears in at most one part. Return a list of integers representing the size of these parts.

Example:

Input: s = "ababcbacadefegdehijhklij"
Output: [9, 7, 8]
Explanation:
The partitions are "ababcbaca", "defegde", "hijhklij".
This is because:
- "a", "b", and "c" appear in the first part.
- "d", "e", "f", and "g" appear in the second part.
- "h", "i", "j", "k", and "l" appear in the third part.

Clarifying Questions

1. Input Constraints:
   • Is the input string s guaranteed to be non-empty?
   • Can the input string contain any characters other than lowercase English letters?

Assumptions based on standard problem constraints:

• s contains only lowercase English letters.
• The length of s is at most 500.

1. Output Constraints:
   • Should the output list provide partition sizes in the same order as they appear in the input string?

Strategy

Steps:

1. Calculate Last Occurrences: Traverse the string to record the last position (index) of each character.
2. Initialize Pointers: Use two pointers to denote the current segment start and end. Initialize both at 0.
3. Expand and Determine Segments: Iterate through the string, expanding the current segment until it encapsulates the last occurrence of all characters in that segment.
4. Store Partition Size: Once the segment end is reached, record its size and move to the next segment.

Implementation Outline:

1. Traverse the string to create a dictionary of the last occurrence of each character.
2. Iterate through the string to identify partitions:
   • As each character is processed, update the segment end to the maximum of the current segment end or the last occurrence of the character.
   • When the current index matches the segment end, it signifies the end of a partition.

Time Complexity

• O(n) where n is the length of the string s. This is because:
  • Creating the dictionary of last occurrences requires O(n).
  • A single pass through the string to form partitions takes O(n).

def partition_labels(s):
    # Calculate the last occurrence of each character
    last_occurrence = {char: idx for idx, char in enumerate(s)}

    partitions = []
    start = 0
    end = 0

    # Traverse the string to find the partitions
    for idx, char in enumerate(s):
        end = max(end, last_occurrence[char])

        # When we reach the end of the current partition
        if idx == end:
            partitions.append(end - start + 1)
            start = idx + 1
    
    return partitions

# Example usage:
s = "ababcbacadefegdehijhklij"
print(partition_labels(s))  # Output: [9, 7, 8]

This code snippet achieves the desired partitioning of the string as described in the problem statement, ensuring that each character appears in at most one part of the partitioned string.

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