How to Solve Partition String Problem
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Partition String
Given a string s, partition it into unique segments according to the following procedure: Start building a segment beginning at index 0. Continue extending the current segment character by character u...
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Understanding the Partition String Problem
Let's break down this LeetCode problem and understand what makes it challenging in interview settings.
Problem Statement
Given a string s, partition it into unique segments according to the following procedure: Start building a segment beginning at index 0. Continue extending the current segment character by character until the current segment has not been seen before. Once the segment is unique, add it to your list of segments, mark it as seen, and begin a new segment from the next index. Repeat until you reach the end of s. Return an array of strings segments, where segments[i] is the ith segment created.
Partition String
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Examples
s = "abbccccd"["a","b","bc","c","cc","d"]Index Segment After Adding Seen Segments Current Segment Seen Before? New Segment Updated Seen Segments 0 "a" [] No "" ["a"] 1 "b" ["a"] No "" ["a", "b"] 2 "b" ["a", "b"] Yes "b" ["a", "b"] 3 "bc" ["a", "b"] No "" ["a", "b", "bc"] 4 "c" ["a", "b", "bc"] No "" ["a", "b", "bc", "c"] 5 "c" ["a", "b", "bc", "c"] Yes "c" ["a", "b", "bc", "c"] 6 "cc" ["a", "b", "bc", "c"] No "" ["a", "b", "bc", "c", "cc"] 7 "d" ["a", "b", "bc", "c", "cc"] No "" ["a", "b", "bc", "c", "cc", "d"] Hence, the final output is ["a", "b", "bc", "c", "cc", "d"].
s = "aaaa"["a","aa"]Index Segment After Adding Seen Segments Current Segment Seen Before? New Segment Updated Seen Segments 0 "a" [] No "" ["a"] 1 "a" ["a"] Yes "a" ["a"] 2 "aa" ["a"] No "" ["a", "aa"] 3 "a" ["a", "aa"] Yes "a" ["a", "aa"] Hence, the final output is ["a", "aa"].
Constraints
1 <= s.length <= 105s contains only lowercase English letters.How Interview Coder Helps with Leetcode Problems
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