Count of elements in given Array divisible by all elements in their prefix

• Last Updated : 24 Jan, 2022

Given an array arr[] containing N positive integers, the task is to find the total number of elements in the array that are divisible by all the elements present before them.

Examples:

Input: arr[] = {10, 6, 60, 120, 30, 360}
Output: 3
Explanation: 60, 120 and 360 are the required elements.

Input: arr[] = {2, 6, 5, 60}
Output: 2
Explanation: 6 and 60 are the elements.

Approach: As known, that any number X is divided by {X1, X2, X3, X4, . . ., Xn}, if X is divided by LCM of {X1, X2, X3, X4, …, Xn). And LCM of any number A, B is [(A*B)/gcd(A, B)]. Now to solve this problem, follow the below steps:

1. Create a variable ans which stores the final answer and initialize it with 0.
2. Create another variable lcm which stores LCM up to the ith element while iterating through the array. Initialize lcm with arr[0].
3. Iterate over the array from i = 1 to i = N, and in each iteration, check if arr[i] is divided by lcm. If yes increment ans by 1. Also, update lcm with lcm up to ith element.
4. Print ans as the final answer to this problem.

Below is the implementation of the above approach:

C++

 `// C++ code to implement the above approach``#include ``using` `namespace` `std;` `// Function to return total number of``// elements which are divisible by``// all their previous elements``int` `countElements(``int` `arr[], ``int` `N)``{``    ``int` `ans = 0;``    ``int` `lcm = arr[0];``    ``for` `(``int` `i = 1; i < N; i++) {` `        ``// To check if number is divisible``        ``// by lcm of all previous elements``        ``if` `(arr[i] % lcm == 0) {``            ``ans++;``        ``}` `        ``// Updating LCM``        ``lcm = (lcm * arr[i]) / __gcd(lcm, arr[i]);``    ``}``    ``return` `ans;``}` `// Driver code``int` `main()``{``    ``int` `arr[] = { 10, 6, 60, 120, 30, 360 };` `    ``int` `N = ``sizeof``(arr) / ``sizeof``(``int``);``    ``cout << countElements(arr, N);``    ``return` `0;``}`

Java

 `// Java program for the above approach``import` `java.util.*;``public` `class` `GFG {` `  ``// Recursive function to return gcd of a and b``  ``static` `int` `__gcd(``int` `a, ``int` `b)``  ``{` `    ``// Everything divides 0``    ``if` `(a == ``0``)``      ``return` `b;``    ``if` `(b == ``0``)``      ``return` `a;` `    ``// base case``    ``if` `(a == b)``      ``return` `a;` `    ``// a is greater``    ``if` `(a > b)``      ``return` `__gcd(a - b, b);``    ``return` `__gcd(a, b - a);``  ``}` `  ``// Function to return total number of``  ``// elements which are divisible by``  ``// all their previous elements``  ``static` `int` `countElements(``int` `arr[], ``int` `N)``  ``{``    ``int` `ans = ``0``;``    ``int` `lcm = arr[``0``];``    ``for` `(``int` `i = ``1``; i < N; i++) {` `      ``// To check if number is divisible``      ``// by lcm of all previous elements``      ``if` `(arr[i] % lcm == ``0``) {``        ``ans++;``      ``}` `      ``// Updating LCM``      ``lcm = (lcm * arr[i]) / __gcd(lcm, arr[i]);``    ``}``    ``return` `ans;``  ``}` `  ``public` `static` `void` `main(String args[])``  ``{``    ``int` `arr[] = { ``10``, ``6``, ``60``, ``120``, ``30``, ``360` `};``    ``int` `N = arr.length;``    ``System.out.print(countElements(arr, N));``  ``}``}` `// This code is contributed by Samim Hossain Mondal.`

Python3

 `# Python code for the above approach` `# Recursive function to return gcd of a and b``def` `__gcd(a, b):``  ` `    ``# Everything divides 0``    ``if` `(a ``=``=` `0``):``        ``return` `b;``    ``if` `(b ``=``=` `0``):``        ``return` `a;` `    ``# base case``    ``if` `(a ``=``=` `b):``        ``return` `a;` `    ``# a is greater``    ``if` `(a > b):``        ``return` `__gcd(a ``-` `b, b);``    ``return` `__gcd(a, b ``-` `a);` `# Function to return total number of``# elements which are divisible by``# all their previous elements``def` `countElements(arr, N):``    ``ans ``=` `0``;``    ``lcm ``=` `arr[``0``];``    ``for` `i ``in` `range``(``1``, N):` `        ``# To check if number is divisible``        ``# by lcm of all previous elements``        ``if` `(arr[i] ``%` `lcm ``=``=` `0``):``            ``ans ``+``=` `1` `        ``# Updating LCM``        ``lcm ``=` `(lcm ``*` `arr[i]) ``/` `__gcd(lcm, arr[i]);``    ``return` `ans;` `# Driver code``arr ``=` `[``10``, ``6``, ``60``, ``120``, ``30``, ``360``];` `N ``=` `len``(arr)``print``(countElements(arr, N));` `# This code is contributed by Saurabh Jaiswal`

C#

 `// C#program for the above approach``using` `System;` `public` `class` `GFG {` `  ``// Recursive function to return gcd of a and b``  ``static` `int` `__gcd(``int` `a, ``int` `b)``  ``{` `    ``// Everything divides 0``    ``if` `(a == 0)``      ``return` `b;``    ``if` `(b == 0)``      ``return` `a;` `    ``// base case``    ``if` `(a == b)``      ``return` `a;` `    ``// a is greater``    ``if` `(a > b)``      ``return` `__gcd(a - b, b);``    ``return` `__gcd(a, b - a);``  ``}` `  ``// Function to return total number of``  ``// elements which are divisible by``  ``// all their previous elements``  ``static` `int` `countElements(``int``[] arr, ``int` `N)``  ``{``    ``int` `ans = 0;``    ``int` `lcm = arr[0];``    ``for` `(``int` `i = 1; i < N; i++) {` `      ``// To check if number is divisible``      ``// by lcm of all previous elements``      ``if` `(arr[i] % lcm == 0) {``        ``ans++;``      ``}` `      ``// Updating LCM``      ``lcm = (lcm * arr[i]) / __gcd(lcm, arr[i]);``    ``}``    ``return` `ans;``  ``}` `  ``public` `static` `void` `Main()``  ``{``    ``int``[] arr = { 10, 6, 60, 120, 30, 360 };``    ``int` `N = arr.Length;``    ``Console.Write(countElements(arr, N));``  ``}``}` `// This code is contributed by ukasp.`

Javascript

 ``

Output
`3`

Time complexity: O(N * logD) where D is the maximum array element
Auxiliary Space: O(N)

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