Unbounded Binary Search Example (Find the point where a monotonically increasing function becomes positive first time)


Given a function ‘int f(unsigned int x)’ which takes a non-negative integer ‘x’ as input and returns an integer as output. The function is monotonically increasing with respect to value of x, i.e., the value of f(x+1) is greater than f(x) for every input x. Find the value ‘n’ where f() becomes positive for the first time. Since f() is monotonically increasing, values of f(n+1), f(n+2),… must be positive and values of f(n-2), f(n-3), .. must be negative.
Find n in O(logn) time, you may assume that f(x) can be evaluated in O(1) time for any input x.

A simple solution is to start from i equals to 0 and one by one calculate value of f(i) for 1, 2, 3, 4 .. etc until we find a positive f(i). This works, but takes O(n) time.

Can we apply Binary Search to find n in O(Logn) time? We can’t directly apply Binary Search as we don’t have an upper limit or high index. The idea is to do repeated doubling until we find a positive value, i.e., check values of f() for following values until f(i) becomes positive.

Let 'high' be the value of i when f() becomes positive for first time.

Can we apply Binary Search to find n after finding ‘high’? We can apply Binary Search now, we can use ‘high/2’ as low and ‘high’ as high indexes in binary search. The result n must lie between ‘high/2’ and ‘high’.

Number of steps for finding ‘high’ is O(Logn). So we can find ‘high’ in O(Logn) time. What about time taken by Binary Search between high/2 and high? The value of ‘high’ must be less than 2*n. The number of elements between high/2 and high must be O(n). Therefore, time complexity of Binary Search is O(Logn) and overall time complexity is 2*O(Logn) which is O(Logn).

#include <stdio.h>
int binarySearch(int low, int high); // prototype

// Let's take an example function as f(x) = x^2 - 10*x - 20
// Note that f(x) can be any monotonocally increasing function
int f(int x) { return (x*x - 10*x - 20); }

// Returns the value x where above function f() becomes positive
// first time.
int findFirstPositive()
    // When first value itself is positive
    if (f(0) > 0)
        return 0;

    // Find 'high' for binary search by repeated doubling
    int i = 1;
    while (f(i) <= 0)
        i = i*2;

    //  Call binary search
    return binarySearch(i/2, i);

// Searches first positive value of f(i) where low <= i <= high
int binarySearch(int low, int high)
    if (high >= low)
        int mid = low + (high - low)/2; /* mid = (low + high)/2 */

        // If f(mid) is greater than 0 and one of the following two
        // conditions is true:
        // a) mid is equal to low
        // b) f(mid-1) is negative
        if (f(mid) > 0 && (mid == low || f(mid-1) <= 0))
            return mid;

        // If f(mid) is smaller than or equal to 0
        if (f(mid) <= 0)
            return binarySearch((mid + 1), high);
        else // f(mid) > 0
            return binarySearch(low, (mid -1));

    /* Return -1 if there is no positive value in given range */
    return -1;

/* Driver program to check above functions */
int main()
    printf("The value n where f() becomes positive first is %d",
    return 0;


The value n where f() becomes positive first is 12

Related Article:
Exponential Search

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above

GATE CS Corner    Company Wise Coding Practice

Recommended Posts:

2.4 Average Difficulty : 2.4/5.0
Based on 49 vote(s)

Writing code in comment? Please use ide.geeksforgeeks.org, generate link and share the link here.