Divide and conquer algorithm applied in finding a peak in an array.

Question

For an array a: a₁, a₂, … a_k, … a_n, a_k is a peak if and only if a_k-1 ≤ a_k ≥

Answer 1

The algorithm is correct, although it requires a bit of calculus to prove. First case is trivial → peak. Second case is a "half peak", meaning that it has the down slope, but not up.

We have 2 possibilities here:

• The function is monotonically decreasing till a_mid → a₁ ≥ a₂ is the peak.
• The function is not monotonically decreasing till a_mid → There is a 1≥k≥mid, for which a_k ≥ a_k-1. Let's choose the largest k for which this statement is true → a_k+1 < a_k ≥ a_k-1 → and that's the peak.

Similar argument can be applied for the third case with the opposite "half peak".