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 <bits/stdc++.h> 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<Integer,Integer> 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<Integer,Integer> mp = new HashMap<Integer,Integer>(); // 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 |
0 1 0 2 3 0 1
Time Complexity: O(N * sqrt(N))
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