O(nlogn) Algorithm - Find three evenly spaced ones within binary string

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刺人心
刺人心 2020-11-28 00:07

I had this question on an Algorithms test yesterday, and I can\'t figure out the answer. It is driving me absolutely crazy, because it was worth about 40 points. I figure

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  •  被撕碎了的回忆
    2020-11-28 00:32

    For the simple problem type (i.e. you search three "1" with only (i.e. zero or more) "0" between it), Its quite simple: You could just split the sequence at every "1" and look for two adjacent subsequences having the same length (the second subsequence not being the last one, of course). Obviously, this can be done in O(n) time.

    For the more complex version (i.e. you search an index i and an gap g>0 such that s[i]==s[i+g]==s[i+2*g]=="1"), I'm not sure, if there exists an O(n log n) solution, since there are possibly O(n²) triplets having this property (think of a string of all ones, there are approximately n²/2 such triplets). Of course, you are looking for only one of these, but I have currently no idea, how to find it ...

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