Maximum count of elements divisible on the left for any element

Given an array arr[] of N elements. The good value of an element arr[i] is the number of valid indices j<i such that arr[j] is divisible by arr[i].

Example:

Input: arr[] = {9, 6, 2, 3}
Output: 2
9 doesn’t has any element on its left.
6 doesn’t divide any element on its left.
2 divides 6.
3 divides 6 and 9.

Input: arr[] = {8, 1, 28, 4, 2, 6, 7}
Output: 3

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

Naive approach: For every element, find the count of numbers divisible by it on its left and print the maximum of these values.

Below is the implementation of the above approach:

C++

 // C++ implementation of the approach #include using namespace std;    // Function to return the maximum count  // of required elements  int findMax(int arr[], int n) {      int res = 0;     int i, j;            // For every element in the array starting      // from the second element      for(i = 0; i < n ; i++)     {                    // Check all the elements on the left          // of current element which are divisible          // by the current element          int count = 0;         for(j = 0; j < i; j++)         {              if (arr[j] % arr[i] == 0)                  count += 1;         }         res = max(count, res);     }     return res; }    // Driver code int main() {     int arr[] = { 8, 1, 28, 4, 2, 6, 7 };     int n = sizeof(arr) / sizeof(int);        cout << findMax(arr, n);        return 0; }    // This code is contributed by Rajput-Ji

Java

 // Java implementation of the approach  class GFG  {    // Function to return the maximum count  // of required elements  static int findMax(int arr[], int n) {      int res = 0;     int i, j;            // For every element in the array starting      // from the second element      for(i = 0; i < n ; i++)     {                    // Check all the elements on the left          // of current element which are divisible          // by the current element          int count = 0;         for(j = 0; j < i; j++)         {              if (arr[j] % arr[i] == 0)                  count += 1;         }         res = Math.max(count, res);     }     return res; }    // Driver Code public static void main (String[] args) {     int arr[] = {8, 1, 28, 4, 2, 6, 7};     int n = arr.length;      System.out.println(findMax(arr, n)); } }    // This code is contributed by AnkitRai01

Python3

 # Python3 implementation of the approach    # Function to return the maximum count  # of required elements def findMax(arr, n):     res = 0            # For every element in the array starting      # from the second element     for i in range(1, n):                    # Check all the elements on the left          # of current element which are divisible          # by the current element         count = 0         for j in range(0, i):             if arr[j] % arr[i] == 0:                 count += 1         res = max(count, res)     return res    # Driver code arr = [8, 1, 28, 4, 2, 6, 7] n = len(arr) print(findMax(arr, n))

C#

 // C# implementation of the above approach  using System;        class GFG  {    // Function to return the maximum count  // of required elements  static int findMax(int []arr, int n) {      int res = 0;     int i, j;            // For every element in the array       // starting from the second element      for(i = 0; i < n ; i++)     {                    // Check all the elements on the left          // of current element which are divisible          // by the current element          int count = 0;         for(j = 0; j < i; j++)         {              if (arr[j] % arr[i] == 0)                  count += 1;         }         res = Math.Max(count, res);     }     return res; }    // Driver Code public static void Main (String[] args) {     int []arr = {8, 1, 28, 4, 2, 6, 7};     int n = arr.Length;      Console.WriteLine(findMax(arr, n)); } }    // This code is contributed by PrinciRaj1992

Output:

3

Time Complexity: O(N2)

Efficient approach: It can be observed that for any element pair (arr[i], arr[j]) where i < j and (arr[i] % arr[j]) = 0, if the count of elements divisible by arr[i] on its left is X then the count of elements divisible by arr[j] on its left will definitely be greater than X as all the elements which are divisible by arr[i] will also be divisible by arr[j]. So, for every element which is divisible by any other element on its right, the count of elements on its left which are divisible by it doesn’t need to be calculated which will improve the time complexity of the overall program but it should be noted that for the input where no element is divisible by any other element (for example, when all the lements are prime), the worst-case time complexity would still be O(N2).

Below is the implementation of the above approach:

C++

 // C++ implementation of the approach #include using namespace std;    // Function to return the maximum count // of required elements int findMax(int arr[], int n) {        // divisible[i] will store true     // if arr[i] is divisible by     // any element on its right     bool divisible[n] = { false };        // To store the maximum required count     int res = 0;        // For every element of the array     for (int i = n - 1; i > 0; i--) {            // If the current element is         // divisible by any element         // on its right         if (divisible[i])             continue;            // Find the count of element         // on the left which are divisible         // by the current element         int cnt = 0;         for (int j = 0; j < i; j++) {                // If arr[j] is divisible then             // set divisible[j] to true             if ((arr[j] % arr[i]) == 0) {                 divisible[j] = true;                 cnt++;             }         }            // Update the maximum required count         res = max(res, cnt);     }        return res; }    // Driver code int main() {     int arr[] = { 8, 1, 28, 4, 2, 6, 7 };     int n = sizeof(arr) / sizeof(int);        cout << findMax(arr, n);        return 0; }

Java

 // Java implementation of the approach class GFG {    // Function to return the maximum count // of required elements static int findMax(int arr[], int n) {        // divisible[i] will store true     // if arr[i] is divisible by     // any element on its right     boolean []divisible = new boolean[n];        // To store the maximum required count     int res = 0;        // For every element of the array     for (int i = n - 1; i > 0; i--)     {            // If the current element is         // divisible by any element         // on its right         if (divisible[i])             continue;            // Find the count of element         // on the left which are divisible         // by the current element         int cnt = 0;         for (int j = 0; j < i; j++)          {                // If arr[j] is divisible then             // set divisible[j] to true             if ((arr[j] % arr[i]) == 0)             {                 divisible[j] = true;                 cnt++;             }         }            // Update the maximum required count         res = Math.max(res, cnt);     }     return res; }    // Driver code public static void main(String[] args)  {     int arr[] = { 8, 1, 28, 4, 2, 6, 7 };     int n = arr.length;        System.out.println(findMax(arr, n)); } }    // This code is contributed by Rajput-Ji

Python3

 # Python3 implementation of the approach     # Function to return the maximum count  # of required elements  def findMax(arr, n) :         # divisible[i] will store true      # if arr[i] is divisible by      # any element on its right      divisible = [ False ] * n;         # To store the maximum required count      res = 0;         # For every element of the array      for i in range(n - 1, -1, -1) :             # If the current element is          # divisible by any element          # on its right          if (divisible[i]) :             continue;             # Find the count of element          # on the left which are divisible          # by the current element          cnt = 0;          for j in range(i) :                # If arr[j] is divisible then              # set divisible[j] to true              if ((arr[j] % arr[i]) == 0) :                 divisible[j] = True;                  cnt += 1;             # Update the maximum required count          res = max(res, cnt);         return res;     # Driver code  if __name__ == "__main__" :         arr = [ 8, 1, 28, 4, 2, 6, 7 ];      n = len(arr);         print(findMax(arr, n));     # This code is contributed by kanugargng

C#

 // C# implementation of the approach using System;    class GFG {    // Function to return the maximum count // of required elements static int findMax(int []arr, int n) {        // divisible[i] will store true     // if arr[i] is divisible by     // any element on its right     bool []divisible = new bool[n];        // To store the maximum required count     int res = 0;        // For every element of the array     for (int i = n - 1; i > 0; i--)     {            // If the current element is         // divisible by any element         // on its right         if (divisible[i])             continue;            // Find the count of element         // on the left which are divisible         // by the current element         int cnt = 0;         for (int j = 0; j < i; j++)          {                // If arr[j] is divisible then             // set divisible[j] to true             if ((arr[j] % arr[i]) == 0)             {                 divisible[j] = true;                 cnt++;             }         }            // Update the maximum required count         res = Math.Max(res, cnt);     }     return res; }    // Driver code public static void Main(String[] args)  {     int []arr = { 8, 1, 28, 4, 2, 6, 7 };     int n = arr.Length;        Console.WriteLine(findMax(arr, n)); } }    // This code is contributed by Rajput-Ji

Output:

3

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