# Finding a local peak in an array in O(log N)?

So I watched a couple of videos regarding this and the divide and conquer made sense to me. But I am still not convinced as to how recursing on the side with the larger number guarantees us the solution.

Please can someone explain it Mathematically.

• Do you mean a local peak (item greater than its neighbors) or a global peak (item greater than all others)? – Solomonoff's Secret Nov 5 '18 at 21:43
• Yes, I meant a local peek – Shivangi Singh Nov 6 '18 at 1:11
• – Solomonoff's Secret Nov 6 '18 at 18:31
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To reformulate the question, there is the following problem: given an array of numbers, find an index in the array that is a local maximum, meaning the value at that index $$\ge$$ the values at adjacent indices.

The algorithm suggested works as follows. Perform a binary search on the array. At each iteration choose a middle element of the array; examine its neighbors; if it is a local maximum or the binary search has narrowed down to just one item, output its index; otherwise recurse on a side with a greater neighbor. Clearly this algorithm runs in $$O(\log n)$$.

Your question is why this algorithm works.

The first observation is that every array has a local maximum. That is because an array of length 1 has a local maximum and if every array of length $$n$$ has a local maximum then given an array of length $$n+1$$, either its first element is a local maximum or each local maximum in the sub-array starting at the second element is a local maximum of the full array.

Now suppose there are two consecutive elements of the array $$a, b$$ and $$a < b$$. Every local maximum of the sub-array starting at $$b$$ is a local maximum of the full array and as shown in the previous paragraph, there must be at least one. Similarly if $$a > b$$, every local maximum of the sub-array ending at $$a$$ is a local maximum of the full array and there must be at least one. The algorithm always recurses to one of those mentioned sub-arrays and therefore when it gets to a single-element sub-array, whose only element is a local maximum of itself, that element is also a local maximum of the sub-array in the previous step, which must be a local maximum of the sub-array two steps prior, etc., all the way back to the full array.

No, it is not possible to find the max (peak) element in an unsorted array better than $$\mathcal{O}(n)$$.

When you executed your algorithm $$\mathcal{A}$$ that has $$c \log n$$ compare operations, that means; $$\mathcal{A}$$ only compared $$c \log n$$ elements. The remaining are not compared by $$\mathcal{A}$$.

We can easily construct an adversary argument for any $$\mathcal{O}(\log n)$$ algorithm by just placing the peak, within the elements for $$\mathcal{A}$$ didn't touch.

• The question asks for finding the local peak, not the global peak. – SOFe May 6 at 11:07
• @SOFe If you look at the original question, it was not local. The OP changed the question after the my answer, so I've kept it. – kelalaka May 6 at 11:48