# Find the integral roots of a given Cubic equation

Given 5 integers say A, B, C, D, and E which represents the cubic equation , the task is to find the integral solution for this equation. If there doesn’t exist any integral solution then print “NA”.

Examples:

Input: A = 1, B = 0, C = 0, D = 0, E = 27
Output: 3

Input: A = 1, B = 0, C = 0, D = 0, E = 16
Output: NA

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

Approach: The idea is to use binary search. Below are the steps:

1. Initialise the start and end variable as 0 & 105 respectively.
2. Find the middle(say mid) value of start and end check if it satisfy the given equation or not.
3. If current mid satisfy the given equation the print the mid value.
4. Else if the value of f(x) is less than E then update start as mid + 1.
5. Else Update end as mid – 1.
6. If we can’t find any integral solution for the above equation then print “-1”.

Below is the implementation of the above approach:

## C++

 `// C++ program for the above approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function to find the value at x of ` `// the given equation ` `long` `long` `int` `check(``int` `A, ``int` `B, ``int` `C, ` `                    ``int` `D, ``long` `long` `int` `x) ` `{ ` ` `  `    ``long` `long` `int` `ans; ` ` `  `    ``// Find the value equation at x ` `    ``ans = (A * x * x * x ` `           ``+ B * x * x ` `           ``+ C * x ` `           ``+ D); ` ` `  `    ``// Return the value of ans ` `    ``return` `ans; ` `} ` ` `  `// Function to find the integral ` `// solution of the given equation ` `void` `findSolution(``int` `A, ``int` `B, ``int` `C, ` `                  ``int` `D, ``int` `E) ` `{ ` ` `  `    ``// Initialise start and end ` `    ``int` `start = 0, end = 100000; ` ` `  `    ``long` `long` `int` `mid, ans; ` ` `  `    ``// Implement Binary Search ` `    ``while` `(start <= end) { ` ` `  `        ``// Find mid ` `        ``mid = start + (end - start) / 2; ` ` `  `        ``// Find the value of f(x) using ` `        ``// current mid ` `        ``ans = check(A, B, C, D, mid); ` ` `  `        ``// Check if current mid satisfy ` `        ``// the equation ` `        ``if` `(ans == E) { ` ` `  `            ``// Print mid and return ` `            ``cout << mid << endl; ` `            ``return``; ` `        ``} ` ` `  `        ``if` `(ans < E) ` `            ``start = mid + 1; ` `        ``else` `            ``end = mid - 1; ` `    ``} ` ` `  `    ``// Print "NA" if not found ` `    ``// any integral solution ` `    ``cout << ``"NA"``; ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` `    ``int` `A = 1, B = 0, C = 0; ` `    ``int` `D = 0, E = 27; ` ` `  `    ``// Function Call ` `    ``findSolution(A, B, C, D, E); ` `} `

## Java

 `// Java program for the above approach ` `import` `java.util.*; ` `class` `GFG{ ` ` `  `// Function to find the value at x of ` `// the given equation ` `static` `long` `check(``int` `A, ``int` `B, ``int` `C, ` `                         ``int` `D, ``long` `x) ` `{ ` `    ``long` `ans; ` ` `  `    ``// Find the value equation at x ` `    ``ans = (A * x * x * x +  ` `           ``B * x * x + C * x + D); ` ` `  `    ``// Return the value of ans ` `    ``return` `ans; ` `} ` ` `  `// Function to find the integral ` `// solution of the given equation ` `static` `void` `findSolution(``int` `A, ``int` `B, ``int` `C, ` `                         ``int` `D, ``int` `E) ` `{ ` ` `  `    ``// Initialise start and end ` `    ``long` `start = ``0``, end = ``100000``; ` ` `  `    ``long` `mid, ans; ` ` `  `    ``// Implement Binary Search ` `    ``while` `(start <= end)  ` `    ``{ ` ` `  `        ``// Find mid ` `        ``mid = start + (end - start) / ``2``; ` ` `  `        ``// Find the value of f(x) using ` `        ``// current mid ` `        ``ans = check(A, B, C, D, mid); ` ` `  `        ``// Check if current mid satisfy ` `        ``// the equation ` `        ``if` `(ans == E) ` `        ``{ ` ` `  `            ``// Print mid and return ` `            ``System.out.println(mid); ` `            ``return``; ` `        ``} ` ` `  `        ``if` `(ans < E) ` `            ``start = mid + ``1``; ` `        ``else` `            ``end = mid - ``1``; ` `    ``} ` ` `  `    ``// Print "NA" if not found ` `    ``// any integral solution ` `    ``System.out.println(``"NA"``); ` `} ` ` `  `// Driver Code ` `public` `static` `void` `main(String args[]) ` `{ ` `    ``int` `A = ``1``, B = ``0``, C = ``0``; ` `    ``int` `D = ``0``, E = ``27``; ` ` `  `    ``// Function Call ` `    ``findSolution(A, B, C, D, E); ` `} ` `} ` ` `  `// This code is contributed by Code_Mech `

## Python3

 `# Python3 program for the above approach ` ` `  `# Function to find the value at x of ` `# the given equation ` `def` `check(A, B, C, D, x) : ` ` `  `    ``ans ``=` `0``; ` ` `  `    ``# Find the value equation at x ` `    ``ans ``=` `(A ``*` `x ``*` `x ``*` `x ``+`  `           ``B ``*` `x ``*` `x ``+` `C ``*` `x ``+` `D); ` ` `  `    ``# Return the value of ans ` `    ``return` `ans; ` ` `  `# Function to find the integral ` `# solution of the given equation ` `def` `findSolution(A, B, C, D, E) : ` ` `  `    ``# Initialise start and end ` `    ``start ``=` `0``; end ``=` `100000``; ` ` `  `    ``mid ``=` `0``; ` `    ``ans ``=` `0``; ` ` `  `    ``# Implement Binary Search ` `    ``while` `(start <``=` `end) : ` ` `  `        ``# Find mid ` `        ``mid ``=` `start ``+` `(end ``-` `start) ``/``/` `2``; ` ` `  `        ``# Find the value of f(x) using ` `        ``# current mid ` `        ``ans ``=` `check(A, B, C, D, mid); ` ` `  `        ``# Check if current mid satisfy ` `        ``# the equation ` `        ``if` `(ans ``=``=` `E) : ` ` `  `            ``# Print mid and return ` `            ``print``(mid); ` `            ``return``; ` ` `  `        ``if` `(ans < E) : ` `            ``start ``=` `mid ``+` `1``; ` `        ``else` `: ` `            ``end ``=` `mid ``-` `1``; ` ` `  `    ``# Print "NA" if not found ` `    ``# any integral solution ` `    ``print``(``"NA"``); ` ` `  `# Driver Code ` `if` `__name__ ``=``=` `"__main__"` `: ` ` `  `    ``A ``=` `1``; B ``=` `0``; C ``=` `0``; ` `    ``D ``=` `0``; E ``=` `27``; ` ` `  `    ``# Function Call ` `    ``findSolution(A, B, C, D, E); ` `     `  `# This code is contributed by AnkitRai01 `

## C#

 `// C# program for the above approach ` `using` `System; ` `class` `GFG{ ` ` `  `// Function to find the value at x of ` `// the given equation ` `static` `long` `check(``int` `A, ``int` `B, ``int` `C, ` `                         ``int` `D, ``long` `x) ` `{ ` `    ``long` `ans; ` ` `  `    ``// Find the value equation at x ` `    ``ans = (A * x * x * x +  ` `           ``B * x * x + C * x + D); ` ` `  `    ``// Return the value of ans ` `    ``return` `ans; ` `} ` ` `  `// Function to find the integral ` `// solution of the given equation ` `static` `void` `findSolution(``int` `A, ``int` `B, ``int` `C, ` `                         ``int` `D, ``int` `E) ` `{ ` ` `  `    ``// Initialise start and end ` `    ``long` `start = 0, end = 100000; ` ` `  `    ``long` `mid, ans; ` ` `  `    ``// Implement Binary Search ` `    ``while` `(start <= end)  ` `    ``{ ` ` `  `        ``// Find mid ` `        ``mid = start + (end - start) / 2; ` ` `  `        ``// Find the value of f(x) using ` `        ``// current mid ` `        ``ans = check(A, B, C, D, mid); ` ` `  `        ``// Check if current mid satisfy ` `        ``// the equation ` `        ``if` `(ans == E) ` `        ``{ ` ` `  `            ``// Print mid and return ` `            ``Console.WriteLine(mid); ` `            ``return``; ` `        ``} ` ` `  `        ``if` `(ans < E) ` `            ``start = mid + 1; ` `        ``else` `            ``end = mid - 1; ` `    ``} ` ` `  `    ``// Print "NA" if not found ` `    ``// any integral solution ` `    ``Console.Write(``"NA"``); ` `} ` ` `  `// Driver Code ` `public` `static` `void` `Main() ` `{ ` `    ``int` `A = 1, B = 0, C = 0; ` `    ``int` `D = 0, E = 27; ` ` `  `    ``// Function Call ` `    ``findSolution(A, B, C, D, E); ` `} ` `} ` ` `  `// This code is contributed by Code_Mech `

Output:

```3
```

Time Complexity: O(log N) My Personal Notes arrow_drop_up Check out this Author's contributed articles.

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Improved By : AnkitRai01, Code_Mech