Given four positive numbers **A, B, C, D**, such that A and B are the first term and the common difference of first Arithmetic sequence respectively, while C and D represent the same for the second Arithmetic sequence respectively as shown below:

First Arithmetic Sequence:

A, A + B, A + 2B, A + 3B, …….

Second Arithmetic Sequence:C, C + D, C + 2D, C + 3D, …….

The task is to find the least common element from the above AP sequences. If no such number exists, then print **-1**.

**Examples:**

Input:A = 2, B = 20, C = 19, D = 9Output:82Explanation:

Sequence 1: {2, 2 + 20, 2 + 2(20), 2 + 3(20), 2 + 4(20), …..} = {2, 22, 42, 62,82, …..}

Sequence 2: {19, 19 + 9, 19 + 2(9), 19 + 3(9), 19 + 4(9), 19 + 5(9), 19 + 6(9), 19 + 7(9) …..} = {19, 28, 37, 46, 55, 64, 73,82, …..}

Therefore, 82 is the smallest common element.

Input:A = 2, B = 18, C = 19, D = 9Output:-1

**Approach:**

Since any term of the given two sequences can be expressed as **A + x*B** and **C + y*D**, therefore, to solve the problem, we need to find the smallest values of x and y for which the two terms are equal.

Follow the steps below to solve the problem:

- In order to find the smallest value common in both the AP sequences, the idea is to find the smallest integer values of x and y satisfying the following equation:

A + x*B = C + y*D

=> The above equation can be rearranged as

x*B = C – A + y*D

=> The above equation can be further rearranged as

x = (C – A + y*D) / B

- Check if there exists any integer value
**y**such that**(C – A + y*D) % B**is**0**or not. If it exists, then the smallest number is**(C + y*D)**. - Check whether
**(C + y*D)**is the answer or not, where**y**will be in a range**(0, B)**because from**i = B, B+1,**…. the remainder values will be repeated. - If no such number is obtained from the above steps, then print
**-1**.

Below is the implementation of the above approach:

## C++

`// C++ program to implement` `// the above approach` `#include <bits/stdc++.h>` `using` `namespace` `std;` `// Function to find the smallest element` `// common in both the subsequences` `long` `smallestCommon(` `long` `a, ` `long` `b,` ` ` `long` `c, ` `long` `d)` `{` ` ` `// If a is equal to c` ` ` `if` `(a == c)` ` ` `return` `a;` ` ` `// If a exceeds c` ` ` `if` `(a > c) {` ` ` `swap(a, c);` ` ` `swap(b, d);` ` ` `}` ` ` `long` `first_term_diff = (c - a);` ` ` `long` `possible_y;` ` ` `// Check for the satisfying` ` ` `// equation` ` ` `for` `(possible_y = 0; possible_y < b; possible_y++) {` ` ` `// Least value of possible_y` ` ` `// satisfying the given equation` ` ` `// will yield true in the below if` ` ` `// and break the loop` ` ` `if` `((first_term_diff % b` ` ` `+ possible_y * d)` ` ` `% b` ` ` `== 0) {` ` ` `break` `;` ` ` `}` ` ` `}` ` ` `// If the value of possible_y` ` ` `// satisfying the given equation` ` ` `// lies in range [0, b]` ` ` `if` `(possible_y != b) {` ` ` `return` `c + possible_y * d;` ` ` `}` ` ` `// If no value of possible_y` ` ` `// satisfies the given equation` ` ` `return` `-1;` `}` `// Driver Code` `int` `main()` `{` ` ` `long` `A = 2, B = 20, C = 19, D = 9;` ` ` `cout << smallestCommon(A, B, C, D);` ` ` `return` `0;` `}` |

## Java

`// Java program to implement` `// the above approach` `import` `java.util.*;` `class` `GFG{` `// Function to find the smallest element` `// common in both the subsequences` `static` `int` `smallestCommon(` `int` `a, ` `int` `b,` ` ` `int` `c, ` `int` `d)` `{` ` ` `// If a is equal to c` ` ` `if` `(a == c)` ` ` `return` `a;` ` ` `// If a exceeds c` ` ` `if` `(a > c)` ` ` `{` ` ` `swap(a, c);` ` ` `swap(b, d);` ` ` `}` ` ` `int` `first_term_diff = (c - a);` ` ` `int` `possible_y;` ` ` `// Check for the satisfying` ` ` `// equation` ` ` `for` `(possible_y = ` `0` `;` ` ` `possible_y < b; possible_y++)` ` ` `{` ` ` `// Least value of possible_y` ` ` `// satisfying the given equation` ` ` `// will yield true in the below if` ` ` `// and break the loop` ` ` `if` `((first_term_diff % b +` ` ` `possible_y * d) % b == ` `0` `)` ` ` `{` ` ` `break` `;` ` ` `}` ` ` `}` ` ` `// If the value of possible_y` ` ` `// satisfying the given equation` ` ` `// lies in range [0, b]` ` ` `if` `(possible_y != b)` ` ` `{` ` ` `return` `c + possible_y * d;` ` ` `}` ` ` `// If no value of possible_y` ` ` `// satisfies the given equation` ` ` `return` `-` `1` `;` `}` ` ` `static` `void` `swap(` `int` `x, ` `int` `y)` `{` ` ` `int` `temp = x;` ` ` `x = y;` ` ` `y = temp;` `}` ` ` `// Driver Code` `public` `static` `void` `main(String[] args)` `{` ` ` `int` `A = ` `2` `, B = ` `20` `, C = ` `19` `, D = ` `9` `;` ` ` `System.out.print(smallestCommon(A, B, C, D));` `}` `}` `// This code is contributed by PrinciRaj1992` |

## Python3

`# Python3 program to implement` `# the above approach` `# Function to find the smallest element` `# common in both the subsequences` `def` `smallestCommon(a, b, c, d):` ` ` ` ` `# If a is equal to c` ` ` `if` `(a ` `=` `=` `c):` ` ` `return` `a;` ` ` `# If a exceeds c` ` ` `if` `(a > c):` ` ` `swap(a, c);` ` ` `swap(b, d);` ` ` `first_term_diff ` `=` `(c ` `-` `a);` ` ` `possible_y ` `=` `0` `;` ` ` `# Check for the satisfying` ` ` `# equation` ` ` `for` `possible_y ` `in` `range` `(b):` ` ` `# Least value of possible_y` ` ` `# satisfying the given equation` ` ` `# will yield True in the below if` ` ` `# and break the loop` ` ` `if` `((first_term_diff ` `%` `b ` `+` ` ` `possible_y ` `*` `d) ` `%` `b ` `=` `=` `0` `):` ` ` `break` `;` ` ` `# If the value of possible_y` ` ` `# satisfying the given equation` ` ` `# lies in range [0, b]` ` ` `if` `(possible_y !` `=` `b):` ` ` `return` `c ` `+` `possible_y ` `*` `d;` ` ` `# If no value of possible_y` ` ` `# satisfies the given equation` ` ` `return` `-` `1` `;` `def` `swap(x, y):` ` ` `temp ` `=` `x;` ` ` `x ` `=` `y;` ` ` `y ` `=` `temp;` `# Driver Code` `if` `__name__ ` `=` `=` `'__main__'` `:` ` ` `A ` `=` `2` `; B ` `=` `20` `; C ` `=` `19` `; D ` `=` `9` `;` ` ` `print` `(smallestCommon(A, B, C, D));` `# This code is contributed by Rajput-Ji` |

## C#

`// C# program to implement` `// the above approach` `using` `System;` `class` `GFG{` ` ` `// Function to find the smallest element` `// common in both the subsequences` `static` `int` `smallestCommon(` `int` `a, ` `int` `b,` ` ` `int` `c, ` `int` `d)` `{` ` ` `// If a is equal to c` ` ` `if` `(a == c)` ` ` `return` `a;` ` ` `// If a exceeds c` ` ` `if` `(a > c)` ` ` `{` ` ` `swap(a, c);` ` ` `swap(b, d);` ` ` `}` ` ` `int` `first_term_diff = (c - a);` ` ` `int` `possible_y;` ` ` `// Check for the satisfying` ` ` `// equation` ` ` `for` `(possible_y = 0;` ` ` `possible_y < b; possible_y++)` ` ` `{` ` ` `// Least value of possible_y` ` ` `// satisfying the given equation` ` ` `// will yield true in the below if` ` ` `// and break the loop` ` ` `if` `((first_term_diff % b +` ` ` `possible_y * d) % b == 0)` ` ` `{` ` ` `break` `;` ` ` `}` ` ` `}` ` ` `// If the value of possible_y` ` ` `// satisfying the given equation` ` ` `// lies in range [0, b]` ` ` `if` `(possible_y != b)` ` ` `{` ` ` `return` `c + possible_y * d;` ` ` `}` ` ` `// If no value of possible_y` ` ` `// satisfies the given equation` ` ` `return` `-1;` `}` ` ` `static` `void` `swap(` `int` `x, ` `int` `y)` `{` ` ` `int` `temp = x;` ` ` `x = y;` ` ` `y = temp;` `}` ` ` `// Driver Code` `public` `static` `void` `Main(String[] args)` `{` ` ` `int` `A = 2, B = 20, C = 19, D = 9;` ` ` `Console.Write(smallestCommon(A, B, C, D));` `}` `}` `// This code is contributed by Rajput-Ji` |

**Output:**

82

**Time Complexity:** O(B)**Auxiliary Space:** O(1)

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