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 mapvoid 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 elementvoid 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 codeint 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 elementimport java.util.*; class GFG{ // Function to find all factors of N// and keep their count in mapstatic 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 elementstatic 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 codepublic 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 elementfrom collections import defaultdictimport math # Function to find all factors of N# and keep their count in mapdef 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 elementdef 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 codeif __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 elementusing System;using System.Collections.Generic; class GFG{ // Function to find all factors of N// and keep their count in mapstatic 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 elementstatic 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 codepublic 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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