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

Given two integer arrays A[] and D[], where A_i and D_i 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: 7 
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 <bits/stdc++.h>
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

<script>
// Javascript implementation of the approach
var MAXN = 1000000;

// Function to return element common
// in maximum number of APs
function maxCommonElement(A, D, N)
{
    // Initialize the count variable
    var cnt = Array(MAXN).fill(0);

    for (var i = 0; i < N; i++) {

        // Increment count for every
        // element of an AP
        for (var j = A[i]; j < MAXN; j += D[i])
            cnt[j]++;
    }

    // Find the index with maximum count
    var ans = 0;
    var com = 0;

    for(var i = 0; i < MAXN; i++)
    {
        if (cnt[i] > ans)
        {
            ans = cnt[i] ;
            com = i ;
        }
    }

    // Return the maximum common element
    return com;
}

// Driver code
var A = [ 13, 1, 2, 5 ],
    D = [ 5, 10, 1, 12 ];
var N = A.length;
document.write( maxCommonElement(A, D, N));

// This code is contributed by rutvik_56.
</script>

Output:

41

Time Complexity: O(N * 10⁶)

Auxiliary Space: O(10⁶)