763. Partition Labels

Medium

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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.

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.

Example 1:

Input: s = "ababcbacadefegdehijhklij"
Output: [9,7,8]
Explanation:
The partition is "ababcbaca", "defegde", "hijhklij".
This is a partition so that each letter appears in at most one part.
A partition like "ababcbacadefegde", "hijhklij" is incorrect, because it splits s into less parts.

Example 2:

Input: s = "eccbbbbdec"
Output: [10]

Constraints:

• 1 <= s.length <= 500

• s consists of lowercase English letters.

class Solution:
    def partitionLabels(self, s: str) -> List[int]:
        # Looks like we have first create intervals and then merge them
        d = {}
        for index, char in enumerate(list(s)):
            if char not in d:
                d[char] = [index, index]
            else:
                start, end = d[char]
                d[char] = [start, index]
        intervals = list(d.values())
        merged_intervals = self.merge(intervals)
        result = []
        for start, end in merged_intervals:
            result.append(end-start+1)
        return result
        
    # This stub is taken from PB 56. Merge Intervals
    def merge(self, intervals: List[List[int]]) -> List[List[int]]:
        intervals.sort(key=lambda x : x[0])
        # intervals = sorted(intervals, key=cmp_to_key(self.comparator))
        # print(intervals)
        result = []
        for interval in intervals:
            if not result:
                result.append(interval)
                continue
            lastInterval = result[-1]
            thisTime = interval[0]
            lastTime = lastInterval[1]
            if thisTime <= lastTime:
                newInterval = [min(lastInterval[0], interval[0]), max(lastInterval[1], interval[1])]
                result.pop()
                result.append(newInterval)
            else:
                result.append(interval)
        return result