Find minimum positive integer x such that a(x^2) + b(x) + c >= k

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Given four integers a, b, c, and k. The task is to find the minimum positive value of x such that ax² + bx + c ≥ k.

Examples:

Input: a = 3, b = 4, c = 5, k = 6
Output: 1
For x = 0, a * 0 + b * 0 + c = 5 < 6
For x = 1, a * 1 + b * 1 + c = 3 + 4 + 5 = 12 > 6

Input: a = 2, b = 7, c = 6, k = 3
Output: 0

Brute Force Approach:

The brute force approach to solve this problem would be to iterate over all possible values of x starting from 0 and check if ax^2 + bx + c is greater than or equal to k. If the condition is satisfied for any value of x, we return that x as the minimum positive integer satisfying the given equation.

Below is the implementation of the above approach:

C++

#include <bits/stdc++.h>
using namespace std;
 
// Function to return the minimum positive
// integer satisfying the given equation
int MinimumX(int a, int b, int c, int k)
{
    int x = 0;
    while(a*x*x + b*x + c < k) {
        x++;
    }
    return x;
}
 
// Driver code
int main()
{
    int a = 3, b = 2, c = 4, k = 15;
    cout << MinimumX(a, b, c, k);
 
    return 0;
}

Java

import java.util.*;
 
public class Main {
     
    // Function to return the minimum positive
    // integer satisfying the given equation
    public static int MinimumX(int a, int b, int c, int k) {
        int x = 0;
        while(a*x*x + b*x + c < k) {
            x++;
        }
        return x;
    }
 
    // Driver code
    public static void main(String[] args) {
        int a = 3, b = 2, c = 4, k = 15;
        System.out.println(MinimumX(a, b, c, k));
    }
}

Python3

# Function to return the minimum positive
# integer satisfying the given equation
def MinimumX(a, b, c, k):
    x = 0
    while a*x*x + b*x + c < k:
        x += 1
    return x
 
# Driver code
def main():
    a = 3
    b = 2
    c = 4
    k = 15
    print(MinimumX(a, b, c, k))
 
if __name__ == "__main__":
    main()

C#

using System;
 
public class Program
{
    // Function to return the minimum positive
    // integer satisfying the given equation
    static int MinimumX(int a, int b, int c, int k)
    {
        int x = 0;
        while (a * x * x + b * x + c < k)
        {
            x++;
        }
        return x;
    }
 
    // Driver code
    public static void Main()
    {
        int a = 3, b = 2, c = 4, k = 15;
        Console.WriteLine(MinimumX(a, b, c, k));
    }
}

Javascript

// Function to return the minimum positive
// integer satisfying the given equation
function MinimumX(a, b, c, k) {
    let x = 0;
    while (a * x * x + b * x + c < k) {
        x++;
    }
    return x;
}
 
// Driver code
    const a = 3, b = 2, c = 4, k = 15;
    console.log(MinimumX(a, b, c, k));

Output

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

Approach: The idea is to use binary search. The lower limit for our search will be 0 since x has to be minimum positive integer.

Below is the implementation of the above approach:

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to return the minimum positive
// integer satisfying the given equation
int MinimumX(int a, int b, int c, int k)
{
    int x = INT_MAX;
 
    if (k <= c)
        return 0;
 
    int h = k - c;
    int l = 0;
 
    // Binary search to find the value of x
    while (l <= h) {
        int m = (l + h) / 2;
        if ((a * m * m) + (b * m) > (k - c)) {
            x = min(x, m);
            h = m - 1;
        }
        else if ((a * m * m) + (b * m) < (k - c))
            l = m + 1;
        else
            return m;
    }
 
    // Return the answer
    return x;
}
 
// Driver code
int main()
{
    int a = 3, b = 2, c = 4, k = 15;
    cout << MinimumX(a, b, c, k);
 
    return 0;
}

Java

// Java implementation of the approach
class GFG
{
     
// Function to return the minimum positive
// integer satisfying the given equation
static int MinimumX(int a, int b, int c, int k)
{
    int x = Integer.MAX_VALUE;
 
    if (k <= c)
        return 0;
 
    int h = k - c;
    int l = 0;
 
    // Binary search to find the value of x
    while (l <= h) 
    {
        int m = (l + h) / 2;
        if ((a * m * m) + (b * m) > (k - c)) 
        {
            x = Math.min(x, m);
            h = m - 1;
        }
        else if ((a * m * m) + (b * m) < (k - c))
            l = m + 1;
        else
            return m;
    }
 
    // Return the answer
    return x;
}
 
// Driver code
public static void main(String[] args)
{
    int a = 3, b = 2, c = 4, k = 15;
    System.out.println(MinimumX(a, b, c, k));
}
}
 
// This code is contributed by Code_Mech. 

Python3

# Python3 implementation of the approach
 
# Function to return the minimum positive
# integer satisfying the given equation
def MinimumX(a, b, c, k):
 
    x = 10**9
 
    if (k <= c):
        return 0
 
    h = k - c
    l = 0
 
    # Binary search to find the value of x
    while (l <= h):
        m = (l + h) // 2
        if ((a * m * m) + (b * m) > (k - c)):
            x = min(x, m)
            h = m - 1
 
        elif ((a * m * m) + (b * m) < (k - c)):
            l = m + 1
        else:
            return m
 
    # Return the answer
    return x
 
# Driver code
a, b, c, k = 3, 2, 4, 15
print(MinimumX(a, b, c, k))
 
# This code is contributed by mohit kumar

C#

// C# implementation of the approach
using System;
 
class GFG
{
     
// Function to return the minimum positive
// integer satisfying the given equation
static int MinimumX(int a, int b, int c, int k)
{
    int x = int.MaxValue;
 
    if (k <= c)
        return 0;
 
    int h = k - c;
    int l = 0;
 
    // Binary search to find the value of x
    while (l <= h) 
    {
        int m = (l + h) / 2;
        if ((a * m * m) + (b * m) > (k - c)) 
        {
            x = Math.Min(x, m);
            h = m - 1;
        }
        else if ((a * m * m) + (b * m) < (k - c))
            l = m + 1;
        else
            return m;
    }
 
    // Return the answer
    return x;
}
 
// Driver code
public static void Main()
{
    int a = 3, b = 2, c = 4, k = 15;
    Console.Write(MinimumX(a, b, c, k));
}
}
 
// This code is contributed by Akanksha Rai

Javascript

<script>
// Javascript implementation of the approach
 
// Function to return the minimum positive
// integer satisfying the given equation
function MinimumX(a,b,c,k)
{
    let x = Number.MAX_VALUE;
   
    if (k <= c)
        return 0;
   
    let h = k - c;
    let l = 0;
   
    // Binary search to find the value of x
    while (l <= h) 
    {
        let m = Math.floor((l + h) / 2);
        if ((a * m * m) + (b * m) > (k - c)) 
        {
            x = Math.min(x, m);
            h = m - 1;
        }
        else if ((a * m * m) + (b * m) < (k - c))
            l = m + 1;
        else
            return m;
    }
   
    // Return the answer
    return x;
}
 
// Driver code
let a = 3, b = 2, c = 4, k = 15;
document.write(MinimumX(a, b, c, k));
 
// This code is contributed by patel2127
</script>

PHP

<?php
// PHP implementation of the approach 
 
// Function to return the minimum positive 
// integer satisfying the given equation 
function MinimumX($a, $b, $c, $k) 
{ 
    $x = PHP_INT_MAX; 
 
    if ($k <= $c) 
        return 0; 
 
    $h = $k - $c; 
    $l = 0; 
 
    // Binary search to find the value of x 
    while ($l <= $h) 
    { 
        $m = floor(($l + $h) / 2); 
        if (($a * $m * $m) + 
            ($b * $m) > ($k - $c))
        { 
            $x = min($x, $m); 
            $h = $m - 1; 
        } 
        else if (($a * $m * $m) + 
                 ($b * $m) < ($k - $c)) 
            $l = $m + 1; 
        else
            return $m; 
    } 
 
    // Return the answer 
    return $x; 
} 
 
// Driver code 
$a = 3; $b = 2; $c = 4; $k = 15; 
 
echo MinimumX($a, $b, $c, $k);
 
// This code is contributed by Ryuga
?>

Output

Time Complexity : O(log(k-c))
Auxiliary Space : O(1)

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