Given an array arr[] of positive integers of size N, the task is to count number of triplets in the array such that a[i] divides a[j] and a[j] divides a[k] and i < j < k.
Examples:
Input: arr[] = {1, 2, 3, 4, 5, 6}
Output: 3
Explanation:
The triplets are: (1, 2, 4), (1, 2, 6), (1, 3, 6).
Input: arr[] = {1, 2, 2}
Output: 1
Explanation:
The triplet is (1, 2, 2)
Naive Approach: To solve the problem mentioned above, we will try to implement brute force solution. Traverse the array for all three numbers a[i], a[j] and a[k] and count the number of triplets which satisfies the given condition.
Time complexity: O(N3)
Auxiliary Space complexity: O(1)
Efficient Approach: To optimize the above method we can traverse the array for the middle element from index 1 to n-2 and for every middle element we can traverse the left array for a[i] and count number of possible a[i]’s such that a[i] divides a[j]. Similarly, we can traverse in the right array and do the same thing for a[k].
Below is the implementation of the above approach:
C++
// C++ program to find count of triplets // (a, b, c) in the Array such that // a divides b and b divides c #include <bits/stdc++.h> using namespace std; // Function to count triplets int getCount(int arr[], int n) { int count = 0; // Iterate for middle element for (int j = 1; j < n - 1; j++) { int p = 0, q = 0; // Iterate left array for a[i] for (int i = 0; i < j; i++) { if (arr[j] % arr[i] == 0) p++; } // Iterate right array for a[k] for (int k = j + 1; k < n; k++) { if (arr[k] % arr[j] == 0) q++; } count += p * q; } // return the final result return count; } // Driver code int main() { int arr[] = { 1, 2, 2 }; int N = sizeof(arr) / sizeof(arr[0]); cout << getCount(arr, N) << endl; return 0; } |
Java
// Java program to find count of triplets // (a, b, c) in the Array such that // a divides b and b divides c import java.io.*; import java.util.*; class GFG { // Function to count triplets static int getCount(int arr[], int n) { int count = 0; // Iterate for middle element for(int j = 1; j < n - 1; j++) { int p = 0, q = 0; // Iterate left array for a[i] for(int i = 0; i < j; i++) { if (arr[j] % arr[i] == 0) p++; } // Iterate right array for a[k] for(int k = j + 1; k < n; k++) { if (arr[k] % arr[j] == 0) q++; } count += p * q; } // return the final result return count; } // Driver code public static void main(String[] args) { int arr[] = { 1, 2, 2 }; int N = arr.length; System.out.println(getCount(arr, N)); } } // This code is contributed by coder001 |
Python3
# Python3 program to find the count of # triplets (a, b, c) in the Array such # that a divides b and b divides c # Function to count triplets def getCount(arr, n): count = 0 # Iterate for middle element for j in range(1, n - 1): p, q = 0, 0 # Iterate left array for a[i] for i in range(j): if (arr[j] % arr[i] == 0): p += 1 # Iterate right array for a[k] for k in range(j + 1, n): if (arr[k] % arr[j] == 0): q += 1 count += p * q # Return the final result return count # Driver code if __name__ == '__main__': arr = [ 1, 2, 2 ] N = len(arr) print(getCount(arr, N)) # This code is contributed by mohit kumar 29 |
C#
// C# program to find count of triplets // (a, b, c) in the Array such that // a divides b and b divides c using System; class GFG{ // Function to count triplets public static int getCount(int[] arr, int n) { int count = 0; // Iterate for middle element for(int j = 1; j < n - 1; j++) { int p = 0, q = 0; // Iterate left array for a[i] for(int i = 0; i < j; i++) { if (arr[j] % arr[i] == 0) p++; } // Iterate right array for a[k] for(int k = j + 1; k < n; k++) { if (arr[k] % arr[j] == 0) q++; } count += p * q; } // return the final result return count; } // Driver code public static void Main() { int[] arr = { 1, 2, 2 }; int N = arr.Length; Console.WriteLine(getCount(arr, N)); } } // This code is contributed by jrishabh99 |
1
Time complexity: O(N2)
Auxiliary Space complexity: O(1)
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