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.

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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.

  f(0) 
  f(1)
  f(2)
  f(4)
  f(8)
  f(16)
  f(32)
  ....
  ....
  f(high)
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).

C++

// C++ code for binary search 
#include<bits/stdc++.h>  
using namespace std; 
  
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 code */
int main()  
{  
    cout<<"The value n where f() becomes" << 
        "positive first is "<< findFirstPositive();  
    return 0;  
} 
  
// This code is contributed by rathbhupendra 

C

#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", 
           findFirstPositive()); 
    return 0; 
}

Java

// Java program for Binary Search 
import java.util.*; 
  
class Binary 
{ 
    public static int f(int x)  
    { return (x*x - 10*x - 20); } 
  
    // Returns the value x where above  
    // function f() becomes positive 
    // first time. 
    public static 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 
    public static int binarySearch(int low, int high) 
    { 
        if (high >= low) 
        {    
            /* mid = (low + high)/2 */
            int mid = low + (high - low)/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 code 
    public static void main(String[] args) 
    { 
        System.out.print ("The value n where f() "+ 
                         "becomes positive first is "+ 
                         findFirstPositive()); 
    } 
} 
  
// This code is contributed by rishabh_jain 

Python3

# Python3 program for Unbound Binary search. 
  
# Let's take an example function as  
# f(x) = x^2 - 10*x - 20 
# Note that f(x) can be any monotonocally  
# increasing function 
def f(x):  
    return (x * x - 10 * x - 20) 
  
# Returns the value x where above function 
# f() becomes positive first time. 
def findFirstPositive() : 
      
    # When first value itself is positive 
    if (f(0) > 0): 
        return 0
  
    # Find 'high' for binary search 
    # by repeated doubling 
    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 
def binarySearch(low, high): 
    if (high >= low) : 
          
        # mid = (low + high)/2 
        mid = low + (high - low)/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 and (mid == low or 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 Code 
print ("The value n where f() becomes "+
      "positive first is ", findFirstPositive()); 
  
# This code is contributed by rishabh_jain 

C#

// C# program for Binary Search 
using System; 
  
class Binary 
{ 
    public static int f(int x)  
    {  
        return (x*x - 10*x - 20);  
    } 
  
    // Returns the value x where above  
    // function f() becomes positive 
    // first time. 
    public static 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 
    public static int binarySearch(int low, int high) 
    { 
        if (high >= low) 
        {  
            /* mid = (low + high)/2 */
            int mid = low + (high - low)/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 code 
    public static void Main() 
    { 
       Console.Write ("The value n where f() " + 
                      "becomes positive first is " + 
                       findFirstPositive()); 
    } 
} 
  
// This code is contributed by nitin mittal 

PHP

<?php 
// PHP program for Binary Search 
  
// Let's take an example function  
// as f(x) = x^2 - 10*x - 20 
// Note that f(x) can be any  
// monotonocally increasing function 
function f($x)  
{  
    return ($x * $x - 10 * $x - 20);  
} 
  
// Returns the value x where above  
// function f() becomes positive  
// first time. 
function findFirstPositive() 
{ 
    // When first value  
    // itself is positive 
    if (f(0) > 0) 
        return 0; 
  
    // Find 'high' for binary  
    // search by repeated doubling 
    $i = 1; 
    while (f($i) <= 0) 
        $i = $i * 2; 
  
    // Call binary search 
    return binarySearch(intval($i / 2), $i); 
} 
  
// Searches first positive value 
// of f(i) where low <= i <= high 
function binarySearch($low, $high) 
{ 
    if ($high >= $low) 
    { 
        /* mid = (low + high)/2 */
        $mid = $low + intval(($high -  
                              $low) / 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 Code 
echo "The value n where f() becomes ".  
                 "positive first is ".  
                 findFirstPositive() ; 
  
// This code is contributed by Sam007 
?> 

Output :

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

Related Article:
Exponential Search

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My Personal Notes

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

C++

C

Java

Python3

C#

PHP

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