The Art Gallery Guardian

Find the minimum of an array with a non-increasing and a non-decreasing part

1 Problem

Consider an array a of n entries from a totally ordered set. There exist a index j, such that a[i]\geq a[i+1] for all i < j, and a[i]\leq a[i+1] for all i\geq j.

How fast can we find such j? The worst time is O(n). When all the elements are equal, you must transverse the entire array to figure that out.

If m is the maximum time an element occurs in the array, then we can find an algorithm that solves this problem in O(m+\log n) time.

2 Algorithm

The idea is to use ternary search, which works well when all values are distinct, and go for linear search once we figure there are a lot of repeated elements.

First, we present a lemma.


If there exist a index i < j < k, such that a[i]=a[j]=a[k], then at least \min(j-i,k-j) positions in the array has the same value as a[i].

The specified positions
Image Credit: Vanessa Li.

Consider we want to find the minima from index l to r (not including r), and r-l>3. Let m_1=l+\lfloor \frac{r-l}{3} \rfloor and m_2=r-\lfloor \frac{r-l}{3} \rfloor such that l < m_1 < m_2 < r.

  • If a[m_1] < a[m_2], then we know the minima is in a[l..m_2]. Recurse.
  • If a[m_1] > a[m_2], then we know the minima is in a[m_1..r]. Recurse.
  • Otherwise a[m_1]=a[m_2]. If a[l]=a[m_1] or a[r]=a[m_1], then by the lemma, at least 1/3 of the values between position l and r is a[m_1]. We can also test if a[\lfloor \frac{m_1+m_2}{2} \rfloor]=a[m_1], if it is, then 1/6 of the values between position l and r is a[m_1]. Since there are so many repeated values, we just do a linear search for the minima.
  • Finally, if all above fails, we must have some value between position m_1 and m_2 take a different value from them. It must be a smaller value, and no value smaller than a[m_1] can exist outside a[m_1..m_2]. Recurse on a[m_1..m_2].

Here is the code in Go:

3 Complexity

T(m,n) is the time to run the algorithm on an array of length n with m repeats.

\displaystyle T(m,n) = \begin{cases} \frac{2}{3}T(m,n) + O(1) &\text{if strict inequality} \\ O(n) & \text{if } n \leq \frac{m}{6} \\ \frac{1}{3}T(m,n) + O(1) & \text{otherwise} \end{cases} For n larger than \frac{m}{6}, the algorithm will have O(\log \frac{n}{m}) recursive calls, each one cost O(1) time. Once it reaches small n, it will spend O(n)=O(m) time on a linear search. The algorithm spends a total of O(m+ \log \frac{n}{m}) = O(m+\log n) time.

4 Notes

Brosef Stalin offered an alternative logic that offers cleaner code.

Can we generalize this to first increase then decrease then increase arrays? One can show O(n) is best possible by considering an array with a[i]=i for all i\neq j, and a[j]=-1. There is no way to find j with out looking over every position.

Posted by Chao Xu on 2013-07-27.
Tags: Algorithm.