Find local minima in an array

Question

Given an array of integers, find the local minima. An element A[i] is defined as a local minimum if A[i-1] > A[i] and A[i] < A[i+1] where i = 1...n-2. In case of boundar

Answer 1

Use a divide-and-conquer algorithm. Let m = n/2, and examine the value A[m] (that is, the element in the middle of the array).

Case 1: A[m−1] < A[m]. Then the left half of the array must contain a local minimum, so recurse on the left half. We can show this by contradiction: assume that A[i] is not a local minimum for each 0 ≤ i < m. Then A[m−1] is not a local minimum, which implies that A[m−2] < A[m−1]. Similarly, A[m −3] < A[m −2]. Continuing in this fashion, we obtain A[0] < A[1]. But then A[0] is a local minimum, contrary to our initial assumption.

Case 2: A[m + 1] > A[m]. Then the right half of the array must contain a local minimum, so recurse on the right half. This is symmetrical to Case 1.

Case 3: A[m − 1] > A[m] and A[m + 1] < A[m]. Then A[m] is a local minimum, so return it. The running time recurrence is T(n) = T(n/2) + Θ(1), which yields T(n) = Θ(log n).