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: 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

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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 

chevron_right


Python

filter_none

edit
close

play_arrow

link
brightness_4
code

# 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

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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

chevron_right


Output:

41


My Personal Notes arrow_drop_up

Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.



Improved By : mohit kumar 29, AnkitRai01

Article Tags :
Practice Tags :


Be the First to upvote.


Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.