# 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++

 `// C++ Program to count pairs` `// of array elements which are` `// divisible by K when concatenated` `#include ` `using` `namespace` `std;`   `map<``int``, ``int``> rem;`   `// 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 = 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);` `}`

## Java

 `// 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`

## Python3

 `# 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`

## C#

 `// 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``;` `  ``p = 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`

Output

```4

```

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

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Improved By : chitranayal, offbeat, rutvik_56