# Unbounded Binary Search

Given a monotonically increasing function f(x), find the value of x where f(x) becomes positive for the first time. In other words, find a positive integer x such that f(x-1), f(x-2),... are negative and f(x+1), f(x+2),... are positive.

A function is called monotonically increasing, if f(x) <= f(y) is true for all x and y such that
x <= y. For example, f(x) = 3x - 100 is a monotonically increasing function. It becomes positive for the first time when x = 34 as shown below:

A simple solution is to consider all positive numbers starting from 0, and find the first number for which f(x) is positive. The time complexity of this solution is O(x).

We can solve this problem in O(log(x)) time with the help of binary search algorithm. But we can't apply standard binary search on an unbounded search space since we don't know the upper limit of the search space.

The idea is to determine the range in which x resides using exponential search and perform a binary search within that range. The exponential search routine starts with i = 1, and keep on doubling i until f(i) becomes positive for the first time. When f(i) becomes positive, we perform a binary search within the search space [i/2, i] and find the target value x in O(log(x)) time.

Here's a C program that demonstrates it:

Output:

f(x) becomes positive for the first time when x = 34

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