Count the divisors or multiples present in the Array for each element - GeeksforGeeks

15 Sep, 2021
Given an array A[] with N integers, for each integer A[i] in the array, the task is to find the number of integers A[j] (j != i) in the array such that A[i] % A[j] = 0 or A[j] % A[i] = 0.
Examples:
Input: A = {2, 3, 4, 5, 6}Output: 2 1 1 0 2Explanation: For i=0, the valid indices are 2 and 4 as 4%2 = 0 and 6%2 = 0.For i=1, the only valid index is 4 as 6%3 = 0.For i=2, the only valid index is 0 as 4%2 = 0.For i=3, there are no valid indices.For i=0, the valid indices are 0 and 1 as 6%2 = 0 and 6%3 = 0.
Input: A = {6, 6, 6, 6, 6}Output: 4 4 4 4 4
 Approach: The given problem can be solved by using the observation that the number of integers that satisfies the given condition can be categorized into two cases. Suppose the current integer is P and Q is an integer that satisfies the given conditions.
Case 1 where Q is a multiple of P. Therefore, the count of integers in the given array that are divisible by P is the required answer. This case can be handled using a simple modification of Sieve of Eratosthenes which is discussed here.
Case 2 where P is a multiple of Q. Therefore, the count of integers in the given array that Q divides P is the required answer. This case can be handled similarly using sieve as that of 1st Case.
So, the required answer for any integer is the sum of resulting integers of Case 1 and Case 2. In cases where P = Q, both Case 1 and Case 2 represents the same value and should be considered only once.
Below is the implementation of the above approach:
C++
Java
Python3
C#
Javascript
// C++ Program for the above approach#include <bits/stdc++.h>using namespace std; // Function to find the count of integers// such that A[i]%A[j] = 0 or A[j]%A[i] = 0// for each index of the array A[]void countIndex(int A[], int N){     // Stores the maximum integer in A[]    int MAX = *max_element(A, A + N);     // Stores the frequency of each    // element in the array A[]    vector<int> freq(MAX + 1, 0);     for (int i = 0; i < N; i++)        freq[A[i]]++;     // Stores the valid integers in A[]    // for all integers from 1 to MAX    vector<int> res(MAX + 1, 0);     for (int i = 1; i <= MAX; ++i) {        for (int j = i; j <= MAX; j += i) {             // Case where P = Q            if (i == j) {                 // Subtract 1 because P & Q                // cannot have same index                res[i] += (freq[j] - 1);            }            else {                // Case 1                res[i] += freq[j];                 // Case 2                res[j] += freq[i];            }        }    }     // Loop to print answer for    // each index of array A[]    for (int i = 0; i < N; i++) {        cout << res[A[i]] << " ";    }} // Driver Codeint main(){    int A[] = { 2, 3, 4, 5, 6 };    int N = sizeof(A) / sizeof(int);     // Function Call    countIndex(A, N);     return 0;}
// Java Program for the above approachimport java.util.*; class GFG{ // Function to find the count of integers// such that A[i]%A[j] = 0 or A[j]%A[i] = 0// for each index of the array []Astatic void countIndex(int []A, int N){     // Stores the maximum integer in []A    int MAX = Arrays.stream(A).max().getAsInt();     // Stores the frequency of each    // element in the array []A     int []freq = new int[MAX + 1];     for (int i = 0; i < N; i++)        freq[A[i]]++;     // Stores the valid integers in []A    // for all integers from 1 to MAX    int []res = new int[MAX + 1];     for (int i = 1; i <= MAX; ++i) {        for (int j = i; j <= MAX; j += i) {             // Case where P = Q            if (i == j) {                 // Subtract 1 because P & Q                // cannot have same index                res[i] += (freq[j] - 1);            }            else {                // Case 1                res[i] += freq[j];                 // Case 2                res[j] += freq[i];            }        }    }     // Loop to print answer for    // each index of array []A    for (int i = 0; i < N; i++) {        System.out.print(res[A[i]]+ " ");    }} // Driver Codepublic static void main(String[] args){    int []A = { 2, 3, 4, 5, 6 };    int N = A.length;     // Function Call    countIndex(A, N);}} // This code is contributed by Princi Singh
# Python 3 Program for the above approach # Function to find the count of integers# such that A[i]%A[j] = 0 or A[j]%A[i] = 0# for each index of the array A[]def countIndex(A, N):       # Stores the maximum integer in A[]    MAX = max(A)     # Stores the frequency of each    # element in the array A[]    freq = [0 for i in range(MAX+1)]     for i in range(N):        if A[i] in freq:            freq[A[i]] += 1        else:            freq[A[i]] = 1     # Stores the valid integers in A[]    # for all integers from 1 to MAX    res = [0 for i in range(MAX+1)]     for i in range(1, MAX + 1, 1):        for j in range(i, MAX + 1, i):                       # Case where P = Q            if (i == j):                               # Subtract 1 because P & Q                # cannot have same index                res[i] += (freq[j] - 1)            else:                # Case 1                res[i] += freq[j]                 # Case 2                res[j] += freq[i]     # Loop to print answer for    # each index of array A[]    for i in range(N):        print(res[A[i]],end = " ") # Driver Codeif __name__ == '__main__':    A = [2, 3, 4, 5, 6]    N = len(A)     # Function Call    countIndex(A, N)         # This code is contributed by SURENDRA_GANGWAR.
// C# program of above approachusing System; public class GFG {    static void countIndex(int[] A, int N)    {         // Stores the maximum integer in []A        int MAX = A[0];        for (int i = 1; i < N; i++) {            if (A[i] > MAX) {                MAX = A[i];            }        }         // Stores the frequency of each        // element in the array []A        int[] freq = new int[MAX + 1];         for (int i = 0; i < N; i++)            freq[A[i]]++;         // Stores the valid integers in []A        // for all integers from 1 to MAX        int[] res = new int[MAX + 1];         for (int i = 1; i <= MAX; ++i) {            for (int j = i; j <= MAX; j += i) {                 // Case where P = Q                if (i == j) {                     // Subtract 1 because P & Q                    // cannot have same index                    res[i] += (freq[j] - 1);                }                else {                    // Case 1                    res[i] += freq[j];                     // Case 2                    res[j] += freq[i];                }            }        }         // Loop to print answer for        // each index of array []A        for (int i = 0; i < N; i++) {            Console.Write(res[A[i]] + " ");        }    }     // Driver Code    static public void Main()    {        int[] A = { 2, 3, 4, 5, 6 };        int N = A.Length;         // Function Call        countIndex(A, N);    }} // This code is contributed by maddler.
<script>         // JavaScript program for the above approach;         // Function to find the count of integers        // such that A[i]%A[j] = 0 or A[j]%A[i] = 0        // for each index of the array A[]        function max_element(A, N) {            let MAX = Number.MIN_VALUE;            for (let i = 0; i < A.length; i++) {                if (A[i] > MAX) {                    MAX = A[i];                }            }            return MAX;        }        function countIndex(A, N) {             // Stores the maximum integer in A[]            let MAX = max_element(A, A + N);             // Stores the frequency of each            // element in the array A[]            let freq = new Array(MAX + 1).fill(0);             for (let i = 0; i < N; i++)                freq[A[i]]++;             // Stores the valid integers in A[]            // for all integers from 1 to MAX            let res = new Array(MAX + 1).fill(0);             for (let i = 1; i <= MAX; ++i) {                for (let j = i; j <= MAX; j += i) {                     // Case where P = Q                    if (i == j) {                         // Subtract 1 because P & Q                        // cannot have same index                        res[i] += (freq[j] - 1);                    }                    else {                        // Case 1                        res[i] += freq[j];                         // Case 2                        res[j] += freq[i];                    }                }            }             // Loop to print answer for            // each index of array A[]            for (let i = 0; i < N; i++) {                document.write(res[A[i]] + " ");            }        }         // Driver Code        let A = [2, 3, 4, 5, 6];        let N = A.length;         // Function Call        countIndex(A, N);    // This code is contributed by Potta Lokesh    </script>
2 1 1 0 2 
Time Complexity: O(N*log N)Auxiliary Space: O(MAX) where MAX represents the maximum integer in the given array.
lokeshpotta20
princi singh
maddler
SURENDRA_GANGWAR
singghakshay
khushboogoyal499
divisors
sieve
Arrays
Mathematical
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Mathematical
sieve
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