Write pseudocode for the binary search algorithm and state, with an explanation, it's worst case complexity in big-O notation

procedure binarySearch(A[], start, end, item): if start <= end: i = ceil((start + end) / 2)
if A[i] = item: return i if A[i] > item: return binarySearch(A[], i + 1, end, item) if A[i] < item: return binarySearch(A[], start, end -1, item) else: return -1 //item not found in A[]
Big-O worst case complexity is O(log(n)):Worst case occurs when the search reduces the array down to one element as this takes the most number of steps. In this case the number of steps taken is how many times the array, of length n, is halved until one element is left i.e. n/2^steps = 1. After rearranging this equation, the value of the steps variable is calculated by finding log2(n). In big-O notation the base of the log is irrelevant so it becomes O(log(n)).