# Count common elements in two arrays which are in Arithmetic Progression

Given two arrays **arr1[]** and **arr2[]** of size **M** and **N** respectively. Both arrays are in Arithmetic Progression and the first element of both arrays is the same. The task is to find the number of common elements in arr1[] and arr2[].**Examples:**

Input:arr1[] = {2, 3, 4, 5, 6}, arr2[] = {2, 4, 6, 8, 10}Output:3Explanation:

Common elements are {2, 4, 6}Input:arr1[] = {1, 4, 7, 10, 13, 16}, arr2[] = {1, 3, 5, 7, 9}Output:2Explanation:

Common elements are {1, 7}

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

__{IDE}__**Approach:** The idea is to use Least Common Multiple of the common difference of the two Arithmetic Progression to solve this problem. Below is the illustration of the steps:

- Find the common difference of the two arithmetic progression with the help of the below formulae

diff1 = arr1[1] - arr1[0] diff2 = arr2[1] - arr2[0]

- Find the Least common multiple of the common difference of the two arithmetic progression.
- Common elements that are possible in the two arithmetic progression will be the difference of the last elements of the arithmetic progression to the first element divided by the LCM of the common difference.

elements1 = (arr1[m-1] - arr1[0]) / LCM(diff1, diff2) elements2 = (arr2[n-1] - arr2[0]) / LCM(diff1, diff2) // Common Elements ans = min(elements, elements2)

Below is the implementation of the above approach:

## C++

`// C++ implementation to count the` `// common elements of the two arithmetic` `// progression of the given sequence` `#include<bits/stdc++.h>` `using` `namespace` `std;` `// Function to find GCD` `int` `gcd(` `int` `a,` `int` `b)` `{` ` ` `if` `(b == 0)` ` ` `return` `a;` ` ` `return` `gcd(b, a % b);` `}` `// Function to find LCM` `int` `findlcm(` `int` `a, ` `int` `b)` `{` ` ` `int` `gc = gcd(a, b);` ` ` `return` `a * b / gc;` `}` `// Function to count common element` `// of arr1[] and arr2[]` `int` `CountCommon(` `int` `arr1[], ` `int` `arr2[],` ` ` `int` `m, ` `int` `n)` `{` ` ` ` ` `// Common Difference` ` ` `int` `diff1 = arr1[1] - arr1[0];` ` ` `int` `diff2 = arr2[1] - arr2[0];` ` ` `// Function calling` ` ` `int` `lcm = findlcm(diff1, diff2);` ` ` `int` `ans1 = (arr1[m - 1] - arr1[0]) / lcm;` ` ` `int` `ans2 = (arr2[n - 1] - arr2[0]) / lcm;` ` ` `int` `ans = min(ans1, ans2);` ` ` ` ` `return` `(ans + 1);` `}` `// Driver code` `int` `main()` `{` ` ` `int` `arr1[] = { 2, 5, 8, 11, 14, 17 };` ` ` `int` `arr2[] = { 2, 4, 6, 8, 10, 12 };` ` ` ` ` `int` `m = ` `sizeof` `(arr1) / ` `sizeof` `(arr1[0]);` ` ` `int` `n = ` `sizeof` `(arr2) / ` `sizeof` `(arr2[0]);` ` ` `// Function calling` ` ` `cout << CountCommon(arr1, arr2, m, n);` ` ` `return` `0;` `}` `// This code is contributed by amal kumar choubey` |

## Java

`// Java implementation to count the` `// common elements of the two arithmetic` `// progression of the given sequence` `import` `java.util.*;` `class` `GFG{` `// Function to find GCD` `static` `int` `gcd(` `int` `a,` `int` `b)` `{` ` ` `if` `(b == ` `0` `)` ` ` `return` `a;` ` ` `return` `gcd(b, a % b);` `}` `// Function to find LCM` `static` `int` `findlcm(` `int` `a, ` `int` `b)` `{` ` ` `int` `gc = gcd(a, b);` ` ` `return` `a * b / gc;` `}` `// Function to count common element` `// of arr1[] and arr2[]` `static` `int` `CountCommon(` `int` `[]arr1,` ` ` `int` `[]arr2,` ` ` `int` `m, ` `int` `n)` `{` ` ` ` ` `// Common Difference` ` ` `int` `diff1 = arr1[` `1` `] - arr1[` `0` `];` ` ` `int` `diff2 = arr2[` `1` `] - arr2[` `0` `];` ` ` `// Function calling` ` ` `int` `lcm = findlcm(diff1, diff2);` ` ` `int` `ans1 = (arr1[m - ` `1` `] - arr1[` `0` `]) / lcm;` ` ` `int` `ans2 = (arr2[n - ` `1` `] - arr2[` `0` `]) / lcm;` ` ` `int` `ans = Math.min(ans1, ans2);` ` ` ` ` `return` `(ans + ` `1` `);` `}` `// Driver code` `public` `static` `void` `main(String args[])` `{` ` ` `int` `[]arr1 = { ` `2` `, ` `5` `, ` `8` `, ` `11` `, ` `14` `, ` `17` `};` ` ` `int` `[]arr2 = { ` `2` `, ` `4` `, ` `6` `, ` `8` `, ` `10` `, ` `12` `};` ` ` ` ` `int` `m = arr1.length;` ` ` `int` `n = arr2.length;` ` ` `// Function calling` ` ` `System.out.print(CountCommon(arr1, arr2, m, n));` `}` `}` `// This code is contributed by Nidhi_biet` |

## Python3

`# Python3 implementation to count the` `# common elements of the two arithmetic` `# progression of the given sequence` `# Function to find GCD` `def` `gcd(a, b):` ` ` `if` `b ` `=` `=` `0` `:` ` ` `return` `a` ` ` `return` `gcd(b, a ` `%` `b)` `# Function to find LCM` `def` `findlcm(a, b):` ` ` `return` `a ` `*` `b ` `/` `/` `gcd(a, b)` `# Function to count Common Element` `# of arr1[] and arr2[]` `def` `CountCommon(arr1, arr2, m, n):` ` ` ` ` `# Common Difference` ` ` `diff1 ` `=` `arr1[` `1` `] ` `-` `arr1[` `0` `]` ` ` `diff2 ` `=` `arr2[` `1` `] ` `-` `arr2[` `0` `]` ` ` `# Function calling` ` ` `lcm ` `=` `findlcm(diff1, diff2)` ` ` `ans1 ` `=` `(arr1[m ` `-` `1` `] ` `-` `arr1[` `0` `]) ` `/` `/` `lcm` ` ` `ans2 ` `=` `(arr2[n ` `-` `1` `] ` `-` `arr2[` `0` `]) ` `/` `/` `lcm` ` ` `ans ` `=` `min` `(ans1, ans2)` ` ` `# Print the total Common Element` ` ` `print` `(ans ` `+` `1` `)` `# Driver Code` `if` `__name__ ` `=` `=` `"__main__"` `:` ` ` `arr1 ` `=` `[ ` `2` `, ` `5` `, ` `8` `, ` `11` `, ` `14` `, ` `17` `]` ` ` `arr2 ` `=` `[ ` `2` `, ` `4` `, ` `6` `, ` `8` `, ` `10` `, ` `12` `]` ` ` `m ` `=` `len` `(arr1)` ` ` `n ` `=` `len` `(arr2)` ` ` `# Function calling` ` ` `CountCommon(arr1, arr2, m, n)` |

## C#

`// C# implementation to count the` `// common elements of the two arithmetic` `// progression of the given sequence` `using` `System;` `class` `GFG{` `// Function to find GCD` `static` `int` `gcd(` `int` `a,` `int` `b)` `{` ` ` `if` `(b == 0)` ` ` `return` `a;` ` ` `return` `gcd(b, a % b);` `}` `// Function to find LCM` `static` `int` `findlcm(` `int` `a, ` `int` `b)` `{` ` ` `int` `gc = gcd(a, b);` ` ` `return` `a * b / gc;` `}` `// Function to count common element` `// of arr1[] and arr2[]` `int` `CountCommon(` `int` `[]arr1,` ` ` `int` `[]arr2,` ` ` `int` `m, ` `int` `n)` `{` ` ` ` ` `// Common Difference` ` ` `int` `diff1 = arr1[1] - arr1[0];` ` ` `int` `diff2 = arr2[1] - arr2[0];` ` ` `// Function calling` ` ` `int` `lcm = findlcm(diff1, diff2);` ` ` `int` `ans1 = (arr1[m - 1] - arr1[0]) / lcm;` ` ` `int` `ans2 = (arr2[n - 1] - arr2[0]) / lcm;` ` ` `int` `ans = min(ans1, ans2);` ` ` ` ` `return` `(ans + 1);` `}` `// Driver code` `public` `static` `void` `Main()` `{` ` ` `int` `[]arr1 = { 2, 5, 8, 11, 14, 17 };` ` ` `int` `[]arr2 = { 2, 4, 6, 8, 10, 12 };` ` ` ` ` `int` `m = arr1.Length;` ` ` `int` `n = arr2.Length;` ` ` `// Function calling` ` ` `Console.Write(CountCommon(arr1, arr2, m, n));` `}` `}` `// This code is contributed by Code_Mech` |

## Javascript

`<script>` `// Javascript implementation to count the` `// common elements of the two arithmetic` `// progression of the given sequence` `// Function to find GCD` `function` `gcd(a, b)` `{` ` ` `if` `(b == 0)` ` ` `return` `a;` ` ` `return` `gcd(b, a % b);` `}` `// Function to find LCM` `function` `findlcm(a, b)` `{` ` ` `let gc = gcd(a, b);` ` ` `return` `a * b / gc;` `}` `// Function to count common element` `// of arr1[] and arr2[]` `function` `CountCommon(arr1, arr2,` ` ` `m, n)` `{` ` ` ` ` `// Common Difference` ` ` `let diff1 = arr1[1] - arr1[0];` ` ` `let diff2 = arr2[1] - arr2[0];` ` ` `// Function calling` ` ` `let lcm = findlcm(diff1, diff2);` ` ` `let ans1 = Math.floor((arr1[m - 1] - arr1[0]) / lcm);` ` ` `let ans2 = Math.floor((arr2[n - 1] - arr2[0]) / lcm);` ` ` `let ans = Math.min(ans1, ans2);` ` ` ` ` `return` `(ans + 1);` `}` `// Driver code` ` ` `let arr1 = [ 2, 5, 8, 11, 14, 17 ];` ` ` `let arr2 = [ 2, 4, 6, 8, 10, 12 ];` ` ` ` ` `let m = arr1.length;` ` ` `let n = arr2.length;` ` ` `// Function calling` ` ` `document.write(CountCommon(arr1, arr2, m, n));` `// This code is contributed by Mayank Tyagi` `</script>` |

**Output:**

2

**Time Complexity:**O(log(max(diff1, diff2)))

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