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Find initial integral solution of Linear Diophantine equation if finite solution exists
• Difficulty Level : Medium
• Last Updated : 23 Jul, 2020

Given three integers a, b and c representing a linear equation of the form: ax + by = c. The task is to find the initial integral solution of the given equation if finite solution exists.

A Linear Diophantine equation (LDE) is an equation with 2 or more integer unknowns and the integer unknowns are each to at most degree of 1. Linear Diophantine equation in two variables takes the form of ax+by=c, where x,y are integer variables and a, b, c are integer constants. x and y are unknown variables.

Examples:

Input: a = 4, b = 18, c = 10
Output: x = -20, y = 5

Explanation: (-20)*4 + (5)*18 = 10

Input: a = 9, b = 12, c = 5
Output: No Solutions exists

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

Approach:

1. First check if a and are non-zero.
2. If both of them are zero and c is non-zero then, no solution exists. If c is also zero then infinite solution exits.
3. For given a and b, calculate value of x1, y1 and gcd using Extended Euclidean Algorithm.
4. Now, for solution to exist gcd(a, b) should be multiple of c.
5. Calculate solution of the equation as follows:
```x = x1 * ( c / gcd )
y = y1 * ( c / gcd )
```

Below is the implementation of the above approach:

## C++

 `// C++ program for the above approach ` ` `  `#include   ` `using` `namespace` `std; ` ` `  `// Function to implement the extended ` `// euclid algorithm ` `int` `gcd_extend(``int` `a, ``int` `b, ` `               ``int``& x, ``int``& y) ` `{ ` `    ``// Base Case ` `    ``if` `(b == 0) { ` `        ``x = 1; ` `        ``y = 0; ` `        ``return` `a; ` `    ``} ` ` `  `    ``// Recursively find the gcd ` `    ``else` `{ ` `        ``int` `g = gcd_extend(b, ` `                           ``a % b, x, y); ` `        ``int` `x1 = x, y1 = y; ` `        ``x = y1; ` `        ``y = x1 - (a / b) * y1; ` `        ``return` `g; ` `    ``} ` `} ` ` `  `// Function to print the solutions of ` `// the given equations ax + by = c ` `void` `print_solution(``int` `a, ``int` `b, ``int` `c) ` `{ ` `    ``int` `x, y; ` `    ``if` `(a == 0 && b == 0) { ` ` `  `        ``// Condition for infinite solutions ` `        ``if` `(c == 0) { ` `            ``cout ` `                ``<< ``"Infinite Solutions Exist"` `                ``<< endl; ` `        ``} ` ` `  `        ``// Condition for no solutions exist ` `        ``else` `{ ` `            ``cout ` `                ``<< ``"No Solution exists"` `                ``<< endl; ` `        ``} ` `    ``} ` `    ``int` `gcd = gcd_extend(a, b, x, y); ` ` `  `    ``// Condition for no solutions exist ` `    ``if` `(c % gcd != 0) { ` `        ``cout ` `            ``<< ``"No Solution exists"` `            ``<< endl; ` `    ``} ` `    ``else` `{ ` ` `  `        ``// Print the solution ` `        ``cout << ``"x = "` `<< x * (c / gcd) ` `             ``<< ``", y = "` `<< y * (c / gcd) ` `             ``<< endl; ` `    ``} ` `} ` ` `  `// Driver Code ` `int` `main(``void``) ` `{ ` `    ``int` `a, b, c; ` ` `  `    ``// Given coefficients ` `    ``a = 4; ` `    ``b = 18; ` `    ``c = 10; ` ` `  `    ``// Function Call ` `    ``print_solution(a, b, c); ` `    ``return` `0; ` `} `

## Java

 `// Java program for the above approach ` `class` `GFG{ ` ` `  `static` `int` `x, y;  ` ` `  `// Function to implement the extended ` `// euclid algorithm ` `static` `int` `gcd_extend(``int` `a, ``int` `b) ` `{ ` `     `  `    ``// Base Case ` `    ``if` `(b == ``0``) ` `    ``{ ` `        ``x = ``1``; ` `        ``y = ``0``; ` `        ``return` `a; ` `    ``} ` ` `  `    ``// Recursively find the gcd ` `    ``else` `    ``{ ` `        ``int` `g = gcd_extend(b, a % b); ` `        ``int` `x1 = x, y1 = y; ` `        ``x = y1; ` `        ``y = x1 - (a / b) * y1; ` `        ``return` `g; ` `    ``} ` `} ` ` `  `// Function to print the solutions of ` `// the given equations ax + by = c ` `static` `void` `print_solution(``int` `a, ``int` `b, ``int` `c) ` `{ ` `    ``if` `(a == ``0` `&& b == ``0``)  ` `    ``{ ` `         `  `        ``// Condition for infinite solutions ` `        ``if` `(c == ``0``) ` `        ``{ ` `            ``System.out.print(``"Infinite Solutions "` `+ ` `                             ``"Exist"` `+ ``"\n"``); ` `        ``} ` ` `  `        ``// Condition for no solutions exist ` `        ``else`  `        ``{ ` `            ``System.out.print(``"No Solution exists"` `+  ` `                             ``"\n"``); ` `        ``} ` `    ``} ` `    ``int` `gcd = gcd_extend(a, b); ` ` `  `    ``// Condition for no solutions exist ` `    ``if` `(c % gcd != ``0``) ` `    ``{ ` `        ``System.out.print(``"No Solution exists"` `+ ``"\n"``); ` `    ``} ` `    ``else` `    ``{ ` `         `  `        ``// Print the solution ` `        ``System.out.print(``"x = "` `+ x * (c / gcd) + ` `                       ``", y = "` `+ y * (c / gcd) + ``"\n"``); ` `    ``} ` `} ` ` `  `// Driver Code ` `public` `static` `void` `main(String[] args)  ` `{ ` `    ``int` `a, b, c; ` ` `  `    ``// Given coefficients ` `    ``a = ``4``; ` `    ``b = ``18``; ` `    ``c = ``10``; ` ` `  `    ``// Function Call ` `    ``print_solution(a, b, c); ` `} ` `} ` ` `  `// This code is contributed by Rajput-Ji `

## C#

 `// C# program for the above approach ` `using` `System; ` ` `  `class` `GFG{ ` ` `  `static` `int` `x, y;  ` ` `  `// Function to implement the extended ` `// euclid algorithm ` `static` `int` `gcd_extend(``int` `a, ``int` `b) ` `{ ` `     `  `    ``// Base Case ` `    ``if` `(b == 0) ` `    ``{ ` `        ``x = 1; ` `        ``y = 0; ` `        ``return` `a; ` `    ``} ` ` `  `    ``// Recursively find the gcd ` `    ``else` `    ``{ ` `        ``int` `g = gcd_extend(b, a % b); ` `        ``int` `x1 = x, y1 = y; ` `        ``x = y1; ` `        ``y = x1 - (a / b) * y1; ` `        ``return` `g; ` `    ``} ` `} ` ` `  `// Function to print the solutions of ` `// the given equations ax + by = c ` `static` `void` `print_solution(``int` `a, ``int` `b, ``int` `c) ` `{ ` `    ``if` `(a == 0 && b == 0)  ` `    ``{ ` `         `  `        ``// Condition for infinite solutions ` `        ``if` `(c == 0) ` `        ``{ ` `            ``Console.Write(``"Infinite Solutions "` `+ ` `                          ``"Exist"` `+ ``"\n"``); ` `        ``} ` ` `  `        ``// Condition for no solutions exist ` `        ``else` `        ``{ ` `            ``Console.Write(``"No Solution exists"` `+  ` `                          ``"\n"``); ` `        ``} ` `    ``} ` `    ``int` `gcd = gcd_extend(a, b); ` ` `  `    ``// Condition for no solutions exist ` `    ``if` `(c % gcd != 0) ` `    ``{ ` `        ``Console.Write(``"No Solution exists"` `+ ``"\n"``); ` `    ``} ` `    ``else` `    ``{ ` `         `  `        ``// Print the solution ` `        ``Console.Write(``"x = "` `+ x * (c / gcd) + ` `                    ``", y = "` `+ y * (c / gcd) + ``"\n"``); ` `    ``} ` `} ` ` `  `// Driver Code ` `public` `static` `void` `Main(String[] args)  ` `{ ` `    ``int` `a, b, c; ` ` `  `    ``// Given coefficients ` `    ``a = 4; ` `    ``b = 18; ` `    ``c = 10; ` ` `  `    ``// Function call ` `    ``print_solution(a, b, c); ` `} ` `} ` ` `  `// This code contributed by amal kumar choubey `

Output:

```x = -20, y = 5
```

Time Complexity: O(log(max(A, B))), where A and B are the coefficient of x and y in the given linear equation.
Auxiliary Space: O(1)

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