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Here is my c++ O(n) solution using a hash to keep the position of the last time that letter appeared, and then keeping track after each letter the position of the last substr of that length, so that the longest L will be the position of the last character in my distinct substring.
It's a play off of an O(nlogn) algorithm used to solve an LIS except the hash check for next position to update makes the O(logn) from the LIS binary search to just O(1). Also since we don't have to keep track of each element of the sequence with this unique case, we can just go back (L-1) characters and take the substring after the loop.
- ming64 September 23, 2015