5-line C++ solution no KMP with detailed time complexity analysis O(n^(1+1/loglog(n))) or o(n^(1+e))

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    As you may know using KMP to pre-calculate the longest length L of proper prefix which is also a suffix costs O(n) time and O(n) space, and the answer for this problem will simply be L && n%(n-L)==0.

    Without using KMP, we can implement a more straightforward algorithm with a little more time spending. As many has posted the solution to check whether s has a proper prefix which is also a suffix with length L such that n%(n-L)==0 simply using straightforward string comparison.

        bool repeatedSubstringPattern(string s) {
          for (int i = 1, n = s.size(), sqt = sqrt(n); i <= sqt; ++i)
            if (n%i == 0)
              for (int L : {n-i, n-n/i}) 
                // O(n) space or char-wise comparison for O(1) space
                if (L && s.substr(n-L) == s.substr(0, L)) return true;
          return false;        

    The interesting question is what is the "a little more" time spending? The loop index i goes up to sqrt(n) to check substrings of length n-n%i only for n%i==0, so the time complexity will be

    • O(√n + Σd|n (n-d)) ≤ O(√n + Σd|n n) = O(nd(n))

    where d(n) is the divisor function of n (number of divisors). So we got the upper asymptotic bound.

    For lower asymptotic bound, note that any proper divisor d ≤ n/2, so

    • O(√n + Σd|n (n-d)) ≥ O(Σd|n n/2) = O(nd(n)).

    Function d(n) has upper asymptotic bound O(n1/loglog(n)), so the overall time complexity of the problem is

    • O(n1+1/loglog(n)) or o(n1+e) for any small e > 0,

    which means even though the KMP-less method is slower than KMP, but it is just infinitely tiny slower for worst case :-)

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    You used O(n) as the cost of the substring comparisons, right? But their cost isn't n, it's L. If you consider that, does it lead to an even better complexity?

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    @StefanPochmann Yes, actually, I tried to use a tighter bound L for cost of s.substr(n-L) == s.substr(0, L), which will lead to overall time complexity of

    • O(nd(n) - σ(n)), where σ(n) = Σd|n d is the order 1 divisor function (sum of divisors).

    This is indeed tighter than O(nd(n)). But I don't have a compact growth rate estimate for nd(n) - σ(n).

    My intuitive guess is that nd(n) - σ(n) will be pretty close to nd(n) since the largest proper divisor of n is at most n/2. I tried n = 2k which leads to nd(n) - σ(n) = O(nlog(n)).

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    Ah, never mind, I hadn't thought it through. L is at least n/2, so it doesn't matter.

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    @zzg_zzm Unfortunately this is not O(1) space, since every substr() call creates a copy of the original string, which is O(n) space at least.
    To achieve O(1) space, you have to use two pointers to do the comparison.

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    @ayuanx You are right. I have updated the post with comments. For O(1) space, it needs to do directly the char-wise comparison

    for (int j = 0; j < L && s[j] == s[j+n-L]; ) if (++j == L) return true;

    to save the substr() call in s.substr(n-L) == s.substr(0, L) since the algorithm does not have to make copies of chars.

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