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# Find an integer that is common in the maximum number of given arithmetic progressions

Given two integer arrays A[] and D[], where Ai and Di represent the first element and common difference of an arithmetic progression respectively, the task is to find an element that is common in the maximum number of given arithmetic progressions.
Examples:

Input: A[] = {3, 1, 2, 5}, D[] = {2, 3, 1, 2}
Output:
Explanation: The integer 7 is present in all the given APs.
Input: A[] = {13, 1, 2, 5}, D[] = {5, 10, 1, 12}
Output: 41

Approach: The problem can be easily solved using Hashing. Initialize a cnt[] array. For each element of every AP, increment the count of the element in the cnt[] array. Return the index of the cnt[] array with the maximum count.
Below is the implementation of the above approach:

## CPP

 `// C++ implementation of the approach``#include ``using` `namespace` `std;` `#define MAXN 1000000` `// Function to return element common``// in maximum number of APs``int` `maxCommonElement(``int` `A[], ``int` `D[], ``int` `N)``{``    ``// Initialize the count variable``    ``int` `cnt[MAXN] = { 0 };` `    ``for` `(``int` `i = 0; i < N; i++) {` `        ``// Increment count for every``        ``// element of an AP``        ``for` `(``int` `j = A[i]; j < MAXN; j += D[i])``            ``cnt[j]++;``    ``}` `    ``// Find the index with maximum count``    ``int` `com = max_element(cnt, cnt + MAXN) - cnt;` `    ``// Return the maximum common element``    ``return` `com;``}` `// Driver code``int` `main()``{``    ``int` `A[] = { 13, 1, 2, 5 },``        ``D[] = { 5, 10, 1, 12 };``    ``int` `N = ``sizeof``(A) / ``sizeof``(A[0]);` `    ``cout << maxCommonElement(A, D, N);` `    ``return` `0;``}`

## Java

 `// Java implementation of the approach``class` `GFG``{``    ` `    ``final` `static` `int` `MAXN = ``1000000` `;``    ` `    ``static` `int` `max_element(``int` `[]A, ``int` `n)``    ``{``        ``int` `max = A[``0``];``        ``for``(``int` `i = ``0``; i < n; i++)``            ``if` `(A[i] > max )``                ``max = A[i];``                ` `        ``return` `max;``    ``}` `    ``// Function to return element common``    ``// in maximum number of APs``    ``static` `int` `maxCommonElement(``int` `A[], ``int` `D[], ``int` `N)``    ``{``        ``// Initialize the count variable``        ``int` `cnt[] = ``new` `int``[MAXN] ;``    ` `        ``for` `(``int` `i = ``0``; i < N; i++)``        ``{``    ` `            ``// Increment count for every``            ``// element of an AP``            ``for` `(``int` `j = A[i]; j < MAXN; j += D[i])``                ``cnt[j]++;``        ``}``        ` `        ` `        ``// Find the index with maximum count``        ``int` `ans = ``0``;``        ``int` `com = ``0``;``        ` `        ``for``(``int` `i = ``0``; i < MAXN; i++)``        ``{``            ``if` `(cnt[i] > ans)``            ``{``                ``ans = cnt[i] ;``                ``com = i ;``            ``}``        ``}``    ` `        ``// Return the maximum common element``        ``return` `com;``    ``}``    ` `    ``// Driver code``    ``public` `static` `void` `main (String args[])``    ``{``        ``int` `A[] = { ``13``, ``1``, ``2``, ``5` `},``            ``D[] = { ``5``, ``10``, ``1``, ``12` `};``            ` `        ``int` `N = A.length;``    ` `        ``System.out.println(maxCommonElement(A, D, N));``    ``}``}` `// This code is contributed by AnkitRai01`

## Python

 `# Python implementation of the approach` `MAXN ``=` `1000000` `# Function to return element common``# in maximum number of APs``def` `maxCommonElement(A, D, N):``    ` `    ``# Initialize the count variable``    ``cnt ``=` `[``0``] ``*` `MAXN` `    ``for` `i ``in` `range``(N):` `        ``# Increment count for every``        ``# element of an AP``        ``for` `j ``in` `range``(A[i], MAXN, D[i]):``            ``cnt[j] ``+``=` `1` `    ``# Find the index with maximum count``    ``ans ``=` `0``    ``com ``=` `0``    ``for` `i ``in` `range``(MAXN):``        ``if` `cnt[i] > ans:``            ``ans ``=` `cnt[i]``            ``com ``=` `i` `    ``# Return the maximum common element``    ``return` `com` `# Driver code` `A ``=` `[``13``, ``1``, ``2``, ``5``]``D ``=` `[``5``, ``10``, ``1``, ``12``]``N ``=` `len``(A)` `print``(maxCommonElement(A, D, N))` `# This code is contributed by mohit kumar 29`

## C#

 `// C# implementation of the approach``using` `System;` `class` `GFG``{``    ` `    ``static` `int` `MAXN = 1000000 ;``    ` `    ``static` `int` `max_element(``int` `[]A, ``int` `n)``    ``{``        ``int` `max = A[0];``        ``for``(``int` `i = 0; i < n; i++)``            ``if` `(A[i] > max )``                ``max = A[i];``                ` `        ``return` `max;``    ``}` `    ``// Function to return element common``    ``// in maximum number of APs``    ``static` `int` `maxCommonElement(``int` `[]A, ``int` `[]D, ``int` `N)``    ``{``        ``// Initialize the count variable``        ``int` `[]cnt = ``new` `int``[MAXN] ;``    ` `        ``for` `(``int` `i = 0; i < N; i++)``        ``{``    ` `            ``// Increment count for every``            ``// element of an AP``            ``for` `(``int` `j = A[i]; j < MAXN; j += D[i])``                ``cnt[j]++;``        ``}``        ` `        ` `        ``// Find the index with maximum count``        ``int` `ans = 0;``        ``int` `com = 0;``        ` `        ``for``(``int` `i = 0; i < MAXN; i++)``        ``{``            ``if` `(cnt[i] > ans)``            ``{``                ``ans = cnt[i] ;``                ``com = i ;``            ``}``        ``}``    ` `        ``// Return the maximum common element``        ``return` `com;``    ``}``    ` `    ``// Driver code``    ``public` `static` `void` `Main ()``    ``{``        ``int` `[]A = { 13, 1, 2, 5 };``        ``int` `[]D = { 5, 10, 1, 12 };``            ` `        ``int` `N = A.Length;``    ` `        ``Console.WriteLine(maxCommonElement(A, D, N));``    ``}``}` `// This code is contributed by AnkitRai01`

## Javascript

 ``

Output:

`41`

Time Complexity: O(N * 106)

Auxiliary Space: O(106)

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