Count of pairs in an array such that the highest power of 2 that divides their product is 1
Given an array arr[] of N positive integers. The task is to find the count of pairs (arr[i], arr[j]) such that the maximum power of 2 that divides arr[i] * arr[j] is 1.
Examples:
Input: arr[] = {3, 5, 2, 8}
Output: 3
(3, 2), (5, 2) and (3, 5) are the only valid pairs.
Since the power of 2 that divides 3 * 2 = 6 is 1,
5 * 2 = 10 is 1 and 3 * 5 = 15 is 0.
Input: arr[] = {4, 2, 7, 11, 14, 15, 18}
Output: 12
Approach: As the maximum power of 2 that divides arr[i] * arr[j] is at max 1, it means that if P is the product then it must either be odd or 2 is the only even factor of P.
It implies that both arr[i] and arr[j] must be odd or exactly one of them is even and 2 is the only even factor of this number.
If odd is the count of odd numbers and even is the count of even numbers such that 2 is the only even factor of that number then the answer will be odd * even + odd * (odd – 1) / 2.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach #include <bits/stdc++.h> using namespace std; // Function to return the count of valid pairs int cntPairs( int a[], int n) { // To store the count of odd numbers and // the count of even numbers such that 2 // is the only even factor of that number int odd = 0, even = 0; for ( int i = 0; i < n; i++) { // If current number is odd if (a[i] % 2 == 1) odd++; // If current number is even and 2 // is the only even factor of it else if ((a[i] / 2) % 2 == 1) even++; } // Calculate total number of valid pairs int ans = odd * even + (odd * (odd - 1)) / 2; return ans; } // Driver code int main() { int a[] = { 4, 2, 7, 11, 14, 15, 18 }; int n = sizeof (a) / sizeof (a[0]); cout << cntPairs(a, n); return 0; } |
Java
// Java implementation of the approach class GFG { // Function to return the count of valid pairs static int cntPairs( int a[], int n) { // To store the count of odd numbers and // the count of even numbers such that 2 // is the only even factor of that number int odd = 0 , even = 0 ; for ( int i = 0 ; i < n; i++) { // If current number is odd if (a[i] % 2 == 1 ) odd++; // If current number is even and 2 // is the only even factor of it else if ((a[i] / 2 ) % 2 == 1 ) even++; } // Calculate total number of valid pairs int ans = odd * even + (odd * (odd - 1 )) / 2 ; return ans; } // Driver code public static void main(String []args) { int a[] = { 4 , 2 , 7 , 11 , 14 , 15 , 18 }; int n = a.length; System.out.println(cntPairs(a, n)); } } // This code is contributed by 29AjayKumar |
Python3
# Python3 implementation of the approach # Function to return the count of valid pairs def cntPairs(a, n) : # To store the count of odd numbers and # the count of even numbers such that 2 # is the only even factor of that number odd = 0 ; even = 0 ; for i in range (n) : # If current number is odd if (a[i] % 2 = = 1 ) : odd + = 1 ; # If current number is even and 2 # is the only even factor of it elif ((a[i] / 2 ) % 2 = = 1 ) : even + = 1 ; # Calculate total number of valid pairs ans = odd * even + (odd * (odd - 1 )) / / 2 ; return ans; # Driver code if __name__ = = "__main__" : a = [ 4 , 2 , 7 , 11 , 14 , 15 , 18 ]; n = len (a); print (cntPairs(a, n)); # This code is contributed by AnkitRai01 |
C#
// C# implementation of the approach using System; class GFG { // Function to return the count of valid pairs static int cntPairs( int []a, int n) { // To store the count of odd numbers and // the count of even numbers such that 2 // is the only even factor of that number int odd = 0, even = 0; for ( int i = 0; i < n; i++) { // If current number is odd if (a[i] % 2 == 1) odd++; // If current number is even and 2 // is the only even factor of it else if ((a[i] / 2) % 2 == 1) even++; } // Calculate total number of valid pairs int ans = odd * even + (odd * (odd - 1)) / 2; return ans; } // Driver code public static void Main(String []args) { int []a = { 4, 2, 7, 11, 14, 15, 18 }; int n = a.Length; Console.WriteLine(cntPairs(a, n)); } } // This code is contributed by Ajay KUmar |
Javascript
<script> // Javascript implementation of the approach // Function to return the count of valid pairs function cntPairs(a, n) { // To store the count of odd numbers and // the count of even numbers such that 2 // is the only even factor of that number var odd = 0, even = 0; for ( var i = 0; i < n; i++) { // If current number is odd if (a[i] % 2 == 1) odd++; // If current number is even and 2 // is the only even factor of it else if ((a[i] / 2) % 2 == 1) even++; } // Calculate total number of valid pairs var ans = odd * even + (odd * (odd - 1)) / 2; return ans; } // Driver code var a = [4, 2, 7, 11, 14, 15, 18]; var n = a.length; document.write( cntPairs(a, n)); // This code is contributed by rrrtnx. </script> |
12
Time Complexity: O(n) where n is number of elements in given array. As, we are using a loop to traverse N times so it will cost us O(N) time
Auxiliary Space: O(1), as we are not using any extra space.
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