# Least common element in given two Arithmetic sequences

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 = 9
Output: 82
Explanation:
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 = 9
Output: -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 ` `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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Improved By : princiraj1992, Rajput-Ji