Maximum count of elements divisible on the left for any element

• Last Updated : 02 Jun, 2021

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:
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:

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 elementsint 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 codeint 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 approachclass GFG{ // Function to return the maximum count// of required elementsstatic 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 Codepublic 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 elementsdef 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 codearr = [8, 1, 28, 4, 2, 6, 7]n = len(arr)print(findMax(arr, n))

C#

 // C# implementation of the above approachusing System;     class GFG{ // Function to return the maximum count// of required elementsstatic 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 Codepublic 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

Javascript


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 elements 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 elementsint 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 codeint 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 approachclass GFG{ // Function to return the maximum count// of required elementsstatic 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 codepublic 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 elementsdef 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 codeif __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 approachusing System; class GFG{ // Function to return the maximum count// of required elementsstatic 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 codepublic 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

Javascript


Output:
3

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