Count of pairs of Array elements which are divisible by K when concatenated

Given an array arr[] and an integer K, the task is to count the pair of indices (i, j) such that i !=j and concatenation of a[i] and a[j] is divisible by K.

Example: 

Input: arr[] = [4, 5, 2], K = 2 
Output:
Explanation: 
All possible concatenations are {45, 42, 54, 52, 24, 25}. 
Out of these, the numbers divisible by 2 are {42, 52, 24, 54}. 
Therefore, the count is 4.

Input: arr[] =[45, 1, 10, 12, 11, 7], K = 11 
Output:

Naive Approach: 
The simplest approach to solve the problem is as follows:  



  • Iterate over the array using nested loops with variables i and j.
  • Concatenate arr[i] and arr[j], for every i != j, by the equation:

Concatenation of arr[i] and arr[j] = (arr[i] * 10lenj) + arr[j], where lenj is the number of digits in arr[j] 

  • Check the divisibility of the concatenated number by K.

Concatenated Number is divisible by k if and only if the sum of (arr[j] % k) and ((arr[i] * 10lenj) % k) is 0 mod k

  • For all such pairs of (arr[i], arr[j]), increase count.
  • Print the final value of count.

Time complexity: O(N2*len(maxm), where maxm denotes the maximum element in the array and len(maxm) denotes the count of digits of maxm. 

Auxilairy Space: O(1)
Efficient Approach: 

To optimize the above approach follow the steps below: 

  • To apply the above formula, maintain a Map for each length from 1 to 10.
  • Store {len[a[i]], a[i] % k } in the map.
  • To count the pairs, for each j in [1, 10], increase the count by the frequency of (k – ((arr[i] * 10^j) % k)) stored in the Map as {j, k – ((arr[i] * 10^j) % k)} mapping.
  • If the pair (arr[i], arr[i]) is counted, decrease the count by 1.
  • After complete traversal of the array, print the final count

Below is the implementation of the above approach:

C++

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// C++ Program to count pairs
// of array elements which are
// divisible by K when concatenated
#include <bits/stdc++.h>
using namespace std;
 
map<int, int> rem[11];
 
// Function to calculate and return the
// count of pairs
int countPairs(vector<int> a, int n, int k)
{
 
    vector<int> len(n);
 
    // Compute power of 10 modulo k
    vector<int> p(11);
    p[0] = 1;
    for (int i = 1; i <= 10; i++) {
        p[i] = (p[i - 1] * 10) % k;
    }
 
    for (int i = 0; i < n; i++) {
        int x = a[i];
 
        // Calculate length of a[i]
        while (x > 0) {
            len[i]++;
            x /= 10;
        }
 
        // Increase count of remainder
        rem[len[i]][a[i] % k]++;
    }
 
    int ans = 0;
 
    for (int i = 0; i < n; i++) {
 
        for (int j = 1; j <= 10; j++) {
 
            // Calculate (a[i]* 10^lenj) % k
            int r = (a[i] * p[j]) % k;
 
            // Calculate (k - (a[i]* 10^lenj)% k) % k
            int xr = (k - r) % k;
 
            // Increase answer by count
            ans += rem[j][xr];
 
            // If a pair (a[i], a[i]) is counted
            if (len[i] == j
                && (r + a[i] % k) % k == 0)
                ans--;
        }
    }
 
    // Return the count of pairs
    return ans;
}
 
// Driver Code
int main()
{
    vector<int> a = { 4, 5, 2 };
    int n = a.size(), k = 2;
    cout << countPairs(a, n, k);
}

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Java

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// Java program to count pairs
// of array elements which are
// divisible by K when concatenated
import java.util.*;
import java.lang.*;
 
class GFG{
     
static int[][] rem = new int[11][11];
 
// Function to calculate and return the
// count of pairs
static int countPairs(int[] a, int n, int k)
{
    int[] len = new int[n];
     
    // Compute power of 10 modulo k
    int[] p = new int[11];
    p[0] = 1;
     
    for(int i = 1; i <= 10; i++)
    {
        p[i] = (p[i - 1] * 10) % k;
    }
     
    for(int i = 0; i < n; i++)
    {
        int x = a[i];
 
        // Calculate length of a[i]
        while (x > 0)
        {
            len[i]++;
            x /= 10;
        }
         
        // Increase count of remainder
        rem[len[i]][a[i] % k]++;
    }
 
    int ans = 0;
 
    for(int i = 0; i < n; i++)
    {
        for(int j = 1; j <= 10; j++)
        {
             
            // Calculate (a[i]* 10^lenj) % k
            int r = (a[i] * p[j]) % k;
 
            // Calculate (k - (a[i]* 10^lenj)% k) % k
            int xr = (k - r) % k;
 
            // Increase answer by count
            ans += rem[j][xr];
 
            // If a pair (a[i], a[i]) is counted
            if (len[i] == j &&
             (r + a[i] % k) % k == 0)
                ans--;
        }
    }
 
    // Return the count of pairs
    return ans;
}  
 
// Driver code
public static void main (String[] args)
{
    int[] a = { 4, 5, 2 };
    int n = a.length, k = 2;
     
    System.out.println(countPairs(a, n, k));
}
}
 
// This code is contributed by offbeat

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Python3

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# Python3 program to count pairs
# of array elements which are
# divisible by K when concatenated
rem = [ [ 0 for x in range(11) ]
            for y in range(11) ]
 
# Function to calculate and return the
# count of pairs
def countPairs(a, n, k):
 
    l = [0] * n
 
    # Compute power of 10 modulo k
    p = [0] * (11)
    p[0] = 1
     
    for i in range(1, 11):
        p[i] = (p[i - 1] * 10) % k
 
    for i in range(n):
        x = a[i]
 
        # Calculate length of a[i]
        while (x > 0):
            l[i] += 1
            x //= 10
         
        # Increase count of remainder
        rem[l[i]][a[i] % k] += 1
     
    ans = 0
 
    for i in range(n):
        for j in range(1, 11):
 
            # Calculate (a[i]* 10^lenj) % k
            r = (a[i] * p[j]) % k
 
            # Calculate (k - (a[i]* 10^lenj)% k) % k
            xr = (k - r) % k
 
            # Increase answer by count
            ans += rem[j][xr]
 
            # If a pair (a[i], a[i]) is counted
            if (l[i] == j and
               (r + a[i] % k) % k == 0):
                ans -= 1
 
    # Return the count of pairs
    return ans
 
# Driver Code
a = [ 4, 5, 2 ]
n = len(a)
k = 2
 
print(countPairs(a, n, k))
 
# This code is contributed by chitranayal

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C#

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// C# program to count pairs
// of array elements which are
// divisible by K when concatenated
using System;
class GFG{
      
static int [,]rem = new int[11, 11];
  
// Function to calculate and
// return the count of pairs
static int countPairs(int[] a,
                      int n, int k)
{
  int[] len = new int[n];
 
  // Compute power of 10 modulo k
  int[] p = new int[11];
  p[0] = 1;
 
  for(int i = 1; i <= 10; i++)
  {
    p[i] = (p[i - 1] * 10) % k;
  }
 
  for(int i = 0; i < n; i++)
  {
    int x = a[i];
 
    // Calculate length of a[i]
    while (x > 0)
    {
      len[i]++;
      x /= 10;
    }
 
    // Increase count of remainder
    rem[len[i], a[i] % k]++;
  }
 
  int ans = 0;
 
  for(int i = 0; i < n; i++)
  {
    for(int j = 1; j <= 10; j++)
    {
      // Calculate (a[i]* 10^lenj) % k
      int r = (a[i] * p[j]) % k;
 
      // Calculate (k - (a[i]* 10^lenj)% k) % k
      int xr = (k - r) % k;
 
      // Increase answer by count
      ans += rem[j, xr];
 
      // If a pair (a[i], a[i]) is counted
      if (len[i] == j &&
         (r + a[i] % k) % k == 0)
        ans--;
    }
  }
 
  // Return the count of pairs
  return ans;
}  
  
// Driver code
public static void Main(string[] args)
{
  int[] a = {4, 5, 2};
  int n = a.Length, k = 2;
  Console.Write(countPairs(a, n, k));
}
}
 
// This code is contributed by rutvik_56

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Output

4


Time Complexity: O(N * len(maxm)) 
Auxiliary Space: O(N)

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Improved By : chitranayal, offbeat, rutvik_56