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Simpler solution with two passes over the string. Time and space complexity: O(n) where n = len(s).
1. First, mark the start and end of each character.
2. Calculate the locations at which the number of starts <= location equals number of ends <= location.
3. Output their differences.
There's a mistake in the question data. The last string's partition is [8, 7, 1, 5] because 'k' is a loner.
Output:
- leo January 31, 2018