Least root of given quadratic equation for value greater than equal to K

Given the constants of quadratic equation F(x) = Ax2 + Bx + C as A, B, and C and an integer K, the task is to find the smallest value of root x such that F(x) ≥ K. If no such values exist then print “-1”. It is given that F(x) is a monotonically increasing function.

Examples: 

Input: A = 3, B = 4, C = 5, K = 6
Output: 1
Explanation:
For the given values F(x) = 3x2 + 4x + 5 the minimum value of x is 1, F(x) = 12, which is greater than the given value of K.

Input: A = 3, B = 4, C = 5, K = 150
Output: 7
Explanation: 
For the given values F(x) = 3x2 + 4x + 5 the minimum value of x is 7, F(x) = 180, which is greater than the given value of K.

Approach: The idea is to use Binary Search to find the minimum value of x. Below are the steps:



  • To get the value equal to or greater than K, the value of x must be in the range [1, sqrt(K)] as this is a quadratic equation.
  • Now, basically there is a need to search the appropriate element in the range, so for this binary search is implemented.
  • Computing F(mid), where mid is the middle value for the range [1, sqrt(K)]. Now the following three cases are possible:
    • If F(mid) ≥ K && F(mid) < K: This mean the current mid is the required answer.
    • If F(mid) < K: This means the current value of mid is less than the required value of x.  So, move towards the right, i.e., in the second half as F(x) is an increasing function.
    • If F(mid) > K: This means the current value of mid is greater than the required value of x. So, move towards the left, i.e., the first half as F(x) is an increasing function.

Below is the implementation of the above approach:

C++

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// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
  
// Function to calculate value of
// quadratic equation for some x
int func(int A, int B, int C, int x)
{
    return (A * x * x + B * x + C);
}
  
// Function to calculate the minimum
// value of x such that F(x) >= K using
// binary search
int findMinx(int A, int B, int C, int K)
{
  
    // Start and end value for
    // binary search
    int start = 1;
    int end = ceil(sqrt(K));
  
    // Binary Search
    while (start <= end) {
        int mid = start + (end - start) / 2;
  
        // Computing F(mid) and F(mid-1)
        int x = func(A, B, C, mid);
        int Y = func(A, B, C, mid - 1);
  
        // Checking the three cases
  
        // If F(mid) >= K  and
        // F(mid-1) < K return mid
        if (x >= K && Y < K) {
            return mid;
        }
  
        // If F(mid) < K go to mid+1 to end
        else if (x < K) {
            start = mid + 1;
        }
  
        // If F(mid) > K go to start to mid-1
        else {
            end = mid - 1;
        }
    }
  
    // If no such value exist
    return -1;
}
  
// Driver Code
int main()
{
    // Given coefficients of Equations
    int A = 3, B = 4, C = 5, K = 6;
  
    // Find minimum value of x
    cout << findMinx(A, B, C, K);
    return 0;
}

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Java

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// Java program for the above approach
import java.util.*;
  
class GFG{
  
// Function to calculate value of
// quadratic equation for some x
static int func(int A, int B, int C, int x)
{
    return (A * x * x + B * x + C);
}
  
// Function to calculate the minimum
// value of x such that F(x) >= K using
// binary search
static int findMinx(int A, int B, int C, int K)
{
      
    // Start and end value for
    // binary search
    int start = 1;
    int end = (int)Math.ceil(Math.sqrt(K));
  
    // Binary Search
    while (start <= end)
    {
        int mid = start + (end - start) / 2;
  
        // Computing F(mid) and F(mid-1)
        int x = func(A, B, C, mid);
        int Y = func(A, B, C, mid - 1);
  
        // Checking the three cases
        // If F(mid) >= K and
        // F(mid-1) < K return mid
        if (x >= K && Y < K) 
        {
            return mid;
        }
  
        // If F(mid) < K go to mid+1 to end
        else if (x < K) 
        {
            start = mid + 1;
        }
  
        // If F(mid) > K go to start to mid-1
        else 
        {
            end = mid - 1;
        }
    }
      
    // If no such value exist
    return -1;
}
  
// Driver code
public static void main(String[] args)
{
      
    // Given coefficients of Equations
    int A = 3, B = 4, C = 5, K = 6;
      
    // Find minimum value of x
    System.out.println(findMinx(A, B, C, K));
}
}
  
// This code is contributed by offbeat

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Python3

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# Python3 program for the above approach
import math
  
# Function to calculate value of
# quadratic equation for some x
def func(A, B, C, x):
      
    return (A * x * x + B * x + C)
  
# Function to calculate the minimum
# value of x such that F(x) >= K using
# binary search
def findMinx(A, B, C, K):
  
    # Start and end value for
    # binary search
    start = 1
    end = math.ceil(math.sqrt(K))
  
    # Binary Search
    while (start <= end):
        mid = start + (end - start) // 2
  
        # Computing F(mid) and F(mid-1)
        x = func(A, B, C, mid)
        Y = func(A, B, C, mid - 1)
  
        # Checking the three cases
  
        # If F(mid) >= K and
        # F(mid-1) < K return mid
        if (x >= K and Y < K):
            return mid
          
        # If F(mid) < K go to mid+1 to end
        elif (x < K):
            start = mid + 1
          
        # If F(mid) > K go to start to mid-1
        else:
            end = mid - 1
      
    # If no such value exist
    return -1
  
# Driver Code
  
# Given coefficients of Equations
A = 3
B = 4
C = 5
K = 6
  
# Find minimum value of x
print(findMinx(A, B, C, K))
  
# This code is contributed by code_hunt

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C#

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// C# program for the above approach
using System;
class GFG{
  
// Function to calculate value of
// quadratic equation for some x
static int func(int A, int B, int C, int x)
{
    return (A * x * x + B * x + C);
}
  
// Function to calculate the minimum
// value of x such that F(x) >= K using
// binary search
static int findMinx(int A, int B, int C, int K)
{
      
    // Start and end value for
    // binary search
    int start = 1;
    int end = (int)Math.Ceiling(Math.Sqrt(K));
  
    // Binary Search
    while (start <= end)
    {
        int mid = start + (end - start) / 2;
  
        // Computing F(mid) and F(mid-1)
        int x = func(A, B, C, mid);
        int Y = func(A, B, C, mid - 1);
  
        // Checking the three cases
        // If F(mid) >= K and
        // F(mid-1) < K return mid
        if (x >= K && Y < K) 
        {
            return mid;
        }
  
        // If F(mid) < K go to mid+1 to end
        else if (x < K) 
        {
            start = mid + 1;
        }
  
        // If F(mid) > K go to start to mid-1
        else 
        {
            end = mid - 1;
        }
    }
      
    // If no such value exist
    return -1;
}
  
// Driver code
public static void Main(String[] args)
{
      
    // Given coefficients of Equations
    int A = 3, B = 4, C = 5, K = 6;
      
    // Find minimum value of x
    Console.WriteLine(findMinx(A, B, C, K));
}
}
  
// This code is contributed by sapnasingh4991

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Output: 

1

Time Complexity: O(log(sqrt(K))
Auxiliary Space: O(1) 

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