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Partition Labels

Difficulty: Medium

Problem Description

You are given a string s. We want to partition the string into as many parts as possible so that each letter appears in at most one part. For example, the string "ababcc" can be partitioned into ["abab", "cc"], but partitions such as ["aba", "bcc"] or ["ab", "ab", "cc"] are invalid. Note that the partition is done so that after concatenating all the parts in order, the resultant string should be s. Return a list of integers representing the size of these parts.

Key Insights

  • Each character in the string can only appear in one partition.
  • The last occurrence of each character determines the boundary of the partition.
  • A greedy approach can be used to extend the partition until all characters have been accounted for.

Space and Time Complexity

Time Complexity: O(n), where n is the length of the string s. Space Complexity: O(1), since we are using a fixed-size array to track character positions.

Solution

To solve the problem, we can use a greedy algorithm with the following steps:

  1. First, we will create a hash table (or array) to store the last occurrence of each character in the string.
  2. We will iterate through the string while maintaining a partition starting from the last seen index of the current character.
  3. Every time we reach the end of a partition (when the current index equals the maximum last occurrence of any character in the partition), we will record the size of this partition.
  4. Finally, we will return the list of partition sizes.

Code Solutions

def partition_labels(s: str) -> list:
    # Step 1: Create a dictionary to store the last occurrence of each character
    last_occurrence = {char: i for i, char in enumerate(s)}
    
    partitions = []
    start, end = 0, 0
    
    # Step 2: Iterate through the string
    for i, char in enumerate(s):
        end = max(end, last_occurrence[char])  # Update the end index of the partition
        
        # Step 3: If we've reached the end of a partition
        if i == end:
            partitions.append(end - start + 1)  # Append the size of the partition
            start = i + 1  # Move to the next partition start
    
    return partitions
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