# Count of pairs (i, j) in the array such that arr[i] is a factor of arr[j]

Given an array of integers arr, the task is to calculate the number of pairs (i, j) where i < j such that arr[j] % arr[i] = 0.

Examples:

Input: arr[] = {1, 2, 3, 4, 5}
Output: 5

Input: arr[] = {1, 1, 2, 2, 3, 3}
Output: 11

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach 1:
Iterate over all pairs of the array and keep incrementing the count of pairs that satisfy the required condition.

Below code is the implementation of the above approach:

## C++

 `// C++ Program to find ` `// the number of pairs ` `// (i, j) such that arr[i] ` `// is a factor of arr[j] ` ` `  `#include ` `using` `namespace` `std; ` ` `  `// Function to return the ` `// count of Pairs ` `int` `numPairs(``int` `arr[], ``int` `n) ` `{ ` `    ``int` `ans = 0; ` `    ``for` `(``int` `i = 0; i < n; i++) { ` `        ``for` `(``int` `j = i + 1; j < n; j++) { ` `            ``if` `(arr[j] % arr[i] == 0) ` `                ``ans++; ` `        ``} ` `    ``} ` `    ``return` `ans; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `arr[] = { 1, 1, 2, 2, 3, 3 }; ` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr); ` ` `  `    ``cout << numPairs(arr, n) << endl; ` `    ``return` `0; ` `} `

## Java

 `// Java Program to find the number of pairs ` `// (i, j) such that arr[i] is a factor of arr[j] ` `import` `java.util.*; ` `import` `java.lang.*; ` `class` `GFG{ ` ` `  `// Function to return the ` `// count of Pairs ` `static` `int` `numPairs(``int` `arr[], ``int` `n) ` `{ ` `    ``int` `ans = ``0``; ` `    ``for` `(``int` `i = ``0``; i < n; i++) ` `    ``{ ` `        ``for` `(``int` `j = i + ``1``; j < n; j++)  ` `        ``{ ` `            ``if` `(arr[j] % arr[i] == ``0``) ` `                ``ans++; ` `        ``} ` `    ``} ` `    ``return` `ans; ` `} ` ` `  `// Driver code ` `public` `static` `void` `main(String[] args) ` `{ ` `    ``int` `arr[] = { ``1``, ``1``, ``2``, ``2``, ``3``, ``3` `}; ` `    ``int` `n = arr.length; ` ` `  `    ``System.out.println(numPairs(arr, n)); ` `} ` `} ` ` `  `// This code is contributed by offbeat `

## Python3

 `# Python3 program to find the number ` `# of pairs (i, j) such that arr[i]  ` `# is a factor of arr[j]  ` ` `  `# Function to return the  ` `# count of Pairs  ` `def` `numPairs(arr, n):  ` ` `  `    ``ans ``=` `0` `    ``for` `i ``in` `range``(n):  ` `        ``for` `j ``in` `range``(i ``+` `1``, n):  ` `             `  `            ``if` `arr[j] ``%` `arr[i] ``=``=` `0``:  ` `                ``ans ``+``=` `1` ` `  `    ``return` `ans  ` ` `  `# Driver code  ` `arr ``=` `[ ``1``, ``1``, ``2``, ``2``, ``3``, ``3` `]  ` `n ``=` `len``(arr)  ` ` `  `print``(numPairs(arr, n))  ` ` `  `# This code is contributed by divyamohan123 `

## C#

 `// C# Program to find the number of pairs ` `// (i, j) such that arr[i] is a factor of arr[j] ` `using` `System; ` ` `  `class` `GFG{ ` ` `  `// Function to return the ` `// count of Pairs ` `static` `int` `numPairs(``int` `[]arr, ``int` `n) ` `{ ` `    ``int` `ans = 0; ` `    ``for``(``int` `i = 0; i < n; i++) ` `    ``{ ` `       ``for``(``int` `j = i + 1; j < n; j++)  ` `       ``{ ` `          ``if` `(arr[j] % arr[i] == 0) ` `              ``ans++; ` `       ``} ` `    ``} ` `    ``return` `ans; ` `} ` ` `  `// Driver code ` `public` `static` `void` `Main() ` `{ ` `    ``int` `[]arr = { 1, 1, 2, 2, 3, 3 }; ` `    ``int` `n = arr.Length; ` ` `  `    ``Console.Write(numPairs(arr, n)); ` `} ` `} ` ` `  `// This code is contributed by Code_Mech `

Output:

```11
```

Approach 2: Store the indices of all array elements in a map. Traverse the map and for every occurrence of an element:

• Add the occurrences of the same element after the current occurence.
• Add the occurrences of all its multiples after the current occurrence using upper_bound

Below code is the implementation of the above approach:

## C++

 `// C++ Program to find ` `// the number of pairs ` `// (i, j) such that arr[i] ` `// is a factor of arr[j] ` ` `  `#include ` `using` `namespace` `std; ` ` `  `// Function to return the ` `// count of Pairs ` `int` `numPairs(``int` `arr[], ``int` `n) ` `{ ` `    ``map<``int``, vector<``int``> > mp; ` `    ``int` `mx = 0; ` `    ``for` `(``int` `i = 0; i < n; i++) { ` ` `  `        ``// Update the maximum ` `        ``mx = max(mx, arr[i]); ` ` `  `        ``// Store the indices of ` `        ``// every element ` `        ``mp[arr[i]].push_back(i); ` `    ``} ` ` `  `    ``int` `ans = 0; ` `    ``for` `(``auto` `i : mp) { ` ` `  `        ``int` `ctr = 1; ` ` `  `        ``// Access all indices of i ` `        ``for` `(``int` `j : i.second) { ` ` `  `            ``// Add the number of ` `            ``// occurences of i ` `            ``// after j-th index ` `            ``ans += i.second.size() - ctr; ` ` `  `            ``// Traverse all multiples of i ` `            ``for` `(``int` `k = 2 * i.first; ` `                 ``k <= mx; ` `                 ``k += i.first) { ` ` `  `                ``// Find their occurrences ` `                ``// after the j-th index ` `                ``int` `numGreater = 0; ` `                ``if` `(mp.find(k) != mp.end()) ` `                    ``numGreater ` `                        ``= ``int``( ` `                            ``mp[k] ` `                                ``.end() ` `                            ``- upper_bound( ` `                                  ``mp[k].begin(), ` `                                  ``mp[k].end(), j)); ` `                ``// Add the count ` `                ``ans += numGreater; ` `            ``} ` `            ``ctr++; ` `        ``} ` `    ``} ` ` `  `    ``return` `ans; ` `} ` `// Driver code ` `int` `main() ` `{ ` `    ``int` `arr[] = { 1, 2, 3, 4, 5 }; ` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr); ` ` `  `    ``cout << numPairs(arr, n) << endl; ` `    ``return` `0; ` `} `

Output:

```5
```

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