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# Count of multiples in an Array before every element

• Difficulty Level : Expert
• Last Updated : 27 May, 2021

Given an array arr of size N, the task is to count the number of indices j (j<i) such that a[i] divides a[j], for all valid indexes i
Examples:

Input: arr[] = {8, 1, 28, 4, 2, 6, 7}
Output: 0, 1, 0, 2, 3, 0, 1
No of multiples for each element before itself –
N(8) = 0 ()
N(1) = 1 (8)
N(28) = 0 ()
N(4) = 2 (28, 8)
N(2) = 3 (4, 28, 8)
N(6) = 0 ()
N(7) = 1 (28)
Input: arr[] = {1, 1, 1, 1}
Output: 0, 1, 2, 3

Naive Approach: Traverse through all valid indices j, in range [0, i-1], for each index i; and count the divisors for each indexes.
Time Complexity: O(N2)
Space Complexity: O(1)
Efficient Approach: This approach is to use map. Increment the count of factors in the map while traversing the array and lookup for that count in the map to find all valid j (< i) without traversing back.
Below is the implementation of the above approach.

## C++

 `// C++ program to count of multiples``// in an Array before every element` `#include ``using` `namespace` `std;` `// Function to find all factors of N``// and keep their count in map``void` `add_factors(``int` `n,``                 ``unordered_map<``int``, ``int``>& mp)``{``    ``// Traverse from 1 to sqrt(N)``    ``// if i divides N,``    ``// increment i and N/i in map``    ``for` `(``int` `i = 1; i <= ``int``(``sqrt``(n)); i++) {``        ``if` `(n % i == 0) {``            ``if` `(n / i == i)``                ``mp[i]++;``            ``else` `{``                ``mp[i]++;``                ``mp[n / i]++;``            ``}``        ``}``    ``}``}` `// Function to count of multiples``// in an Array before every element``void` `count_divisors(``int` `a[], ``int` `n)``{` `    ``// To store factors all of all numbers``    ``unordered_map<``int``, ``int``> mp;` `    ``// Traverse for all possible i's``    ``for` `(``int` `i = 0; i < n; i++) {``        ``// Printing value of a[i] in map``        ``cout << mp[a[i]] << ``" "``;` `        ``// Now updating the factors``        ``// of a[i] in the map``        ``add_factors(a[i], mp);``    ``}``}` `// Driver code``int` `main()``{``    ``int` `arr[] = { 8, 1, 28, 4, 2, 6, 7 };``    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr[0]);` `    ``// Function call``    ``count_divisors(arr, n);` `    ``return` `0;``}`

## Java

 `// Java program to count of multiples``// in an Array before every element``import` `java.util.*;` `class` `GFG{`` ` `// Function to find all factors of N``// and keep their count in map``static` `void` `add_factors(``int` `n,``                 ``HashMap mp)``{``    ``// Traverse from 1 to Math.sqrt(N)``    ``// if i divides N,``    ``// increment i and N/i in map``    ``for` `(``int` `i = ``1``; i <= (Math.sqrt(n)); i++) {``        ``if` `(n % i == ``0``) {``            ``if` `(n / i == i) {``                ``if``(mp.containsKey(i))``                    ``mp.put(i, mp.get(i) + ``1``);``                ``else``                    ``mp.put(i, ``1``);``            ``}``            ``else` `{``                ``if``(mp.containsKey(i))``                    ``mp.put(i, mp.get(i) + ``1``);``                ``else``                    ``mp.put(i, ``1``);``                ``if``(mp.containsKey(n / i))``                    ``mp.put(n / i, mp.get(n / i) + ``1``);``                ``else``                    ``mp.put(n / i, ``1``);``            ``}``        ``}``    ``}``}`` ` `// Function to count of multiples``// in an Array before every element``static` `void` `count_divisors(``int` `a[], ``int` `n)``{`` ` `    ``// To store factors all of all numbers``    ``HashMap mp = ``new` `HashMap();`` ` `    ``// Traverse for all possible i's``    ``for` `(``int` `i = ``0``; i < n; i++) {``        ``// Printing value of a[i] in map``        ``System.out.print(mp.get(a[i]) == ``null` `? ``0` `+ ``" "` `: mp.get(a[i]) + ``" "``);`` ` `        ``// Now updating the factors``        ``// of a[i] in the map``        ``add_factors(a[i], mp);``    ``}``}`` ` `// Driver code``public` `static` `void` `main(String[] args)``{``    ``int` `arr[] = { ``8``, ``1``, ``28``, ``4``, ``2``, ``6``, ``7` `};``    ``int` `n = arr.length;`` ` `    ``// Function call``    ``count_divisors(arr, n);`` ` `}``}` `// This code is contributed by 29AjayKumar`

## Python3

 `# Python 3 program to count of multiples``# in an Array before every element``from` `collections ``import` `defaultdict``import` `math`` ` `# Function to find all factors of N``# and keep their count in map``def` `add_factors(n, mp):` `    ``# Traverse from 1 to sqrt(N)``    ``# if i divides N,``    ``# increment i and N/i in map``    ``for` `i ``in` `range``(``1``, ``int``(math.sqrt(n)) ``+` `1``,):``        ``if` `(n ``%` `i ``=``=` `0``):``            ``if` `(n ``/``/` `i ``=``=` `i):``                ``mp[i] ``+``=` `1``            ``else` `:``                ``mp[i] ``+``=` `1``                ``mp[n ``/``/` `i] ``+``=` `1`` ` `# Function to count of multiples``# in an Array before every element``def` `count_divisors(a, n):`` ` `    ``# To store factors all of all numbers``    ``mp ``=` `defaultdict(``int``)`` ` `    ``# Traverse for all possible i's``    ``for` `i ``in` `range``(n) :``        ``# Printing value of a[i] in map``        ``print``(mp[a[i]], end``=``" "``)`` ` `        ``# Now updating the factors``        ``# of a[i] in the map``        ``add_factors(a[i], mp)`` ` `# Driver code``if` `__name__ ``=``=` `"__main__"``:``    ` `    ``arr ``=` `[ ``8``, ``1``, ``28``, ``4``, ``2``, ``6``, ``7` `]``    ``n ``=` `len``(arr)`` ` `    ``# Function call``    ``count_divisors(arr, n)`` ` `# This code is contributed by chitranayal`

## C#

 `// C# program to count of multiples``// in an Array before every element``using` `System;``using` `System.Collections.Generic;` `class` `GFG{``  ` `// Function to find all factors of N``// and keep their count in map``static` `void` `add_factors(``int` `n,``                 ``Dictionary<``int``,``int``> mp)``{``    ``// Traverse from 1 to Math.Sqrt(N)``    ``// if i divides N,``    ``// increment i and N/i in map``    ``for` `(``int` `i = 1; i <= (Math.Sqrt(n)); i++) {``        ``if` `(n % i == 0) {``            ``if` `(n / i == i) {``                ``if``(mp.ContainsKey(i))``                    ``mp[i] = mp[i] + 1;``                ``else``                    ``mp.Add(i, 1);``            ``}``            ``else` `{``                ``if``(mp.ContainsKey(i))``                    ``mp[i] = mp[i] + 1;``                ``else``                    ``mp.Add(i, 1);``                ``if``(mp.ContainsKey(n / i))``                    ``mp[n / i] = mp[n / i] + 1;``                ``else``                    ``mp.Add(n / i, 1);``            ``}``        ``}``    ``}``}``  ` `// Function to count of multiples``// in an Array before every element``static` `void` `count_divisors(``int` `[]a, ``int` `n)``{``  ` `    ``// To store factors all of all numbers``    ``Dictionary<``int``,``int``> mp = ``new` `Dictionary<``int``,``int``>();``  ` `    ``// Traverse for all possible i's``    ``for` `(``int` `i = 0; i < n; i++) {``        ``// Printing value of a[i] in map``        ``Console.Write(!mp.ContainsKey(a[i]) ? 0 + ``" "` `: mp[a[i]] + ``" "``);``  ` `        ``// Now updating the factors``        ``// of a[i] in the map``        ``add_factors(a[i], mp);``    ``}``}``  ` `// Driver code``public` `static` `void` `Main(String[] args)``{``    ``int` `[]arr = { 8, 1, 28, 4, 2, 6, 7 };``    ``int` `n = arr.Length;``  ` `    ``// Function call``    ``count_divisors(arr, n);``  ` `}``}` `// This code is contributed by sapnasingh4991`

## Javascript

 ``
Output:

`0 1 0 2 3 0 1`

Time Complexity: O(N * sqrt(N))

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