Data Structure
Networking
RDBMS
Operating System
Java
MS Excel
iOS
HTML
CSS
Android
Python
C Programming
C++
C#
MongoDB
MySQL
Javascript
PHP
Physics
Chemistry
Biology
Mathematics
English
Economics
Psychology
Social Studies
Fashion Studies
Legal Studies
- Selected Reading
- UPSC IAS Exams Notes
- Developer's Best Practices
- Questions and Answers
- Effective Resume Writing
- HR Interview Questions
- Computer Glossary
- Who is Who
Count number of ordered pairs with Even and Odd Product in C++
We are given an array of n positive numbers.The goal is to count the ordered pairs (arr[x], arr[y]) with the product of arr[x] and arr[y] is even or odd. Pair ( arr[i],arr[j] ) and ( arr[j],arr[i] are counted as separate.
We will traverse the array using two for loops for each number of pairs. Now calculate product, if it is even increment count by 2 for even products else increment count by 2 for odd products.
Let’s understand with examples.
Input− Arr[]= { 1,1,2,3 } N=4
Output − Count of even product pairs: 6 Count of odd product pairs: 6
Explanation − Valid odd product pairs are −
Arr[0] & Arr[1] → (1,1) Arr[1] & Arr[0] → (1,1) count=2 Arr[0] & Arr[3] → (1,3) Arr[3] & Arr[0] → (3,1) count=2 Arr[1] & Arr[3] → (1,3) Arr[3] & Arr[1] → (3,1) count=2 Total=6 Valid even product pairs are: Arr[0] & Arr[2] → (1,2) Arr[2] & Arr[0] → (2,1) count=2 Arr[1] & Arr[2] → (1,2) Arr[2] & Arr[1] → (2,1) count=2 Arr[2] & Arr[3] → (2,3) Arr[3] & Arr[2] → (3,2) count=2 Total=6
Input − Arr[]= { 2,2,2 } N=3
Output − Count of even product pairs − 6 Count of odd product pairs − 0
Explanation − Valid even product pairs are −
Arr[0] & Arr[1] → (2,2) Arr[1] & Arr[0] → (2,2) count=2 Arr[1] & Arr[2] → (2,2) Arr[2] & Arr[1] → (2,2) count=2 Arr[2] & Arr[3] → (2,2) Arr[3] & Arr[2] → (2,2) count=2 Total=6 No odd product as all elements are even.
Approach used in the below program is as follows
We take an integer array arr[] initialized with random numbers.
Take a variable n which stores the length of Arr[].
Function countPairs(int arr[], int n) takes an array, its length as input and prints the count of pairs with even and odd products.
Traverse array using two for loops for each element of the pair.
Outer Loop from 0<=i<n-1, inner loop i<j<n
Check if arr[i]*arr[j]%2==0. Increment count1 for count of even product pairs by 2 as arr[i],arr[j] and arr[j],arr[i] will be two pairs.
If the above condition is false increment count2 for odd product pairs by 2.
At the end of all loops count1 will have a total number of pairs that have even product and count2 will have a total number of pairs that have odd product
Print the count1 and count2 as result.
Example
#include <bits/stdc++.h>
using namespace std;
void countPairs(int arr[], int n){
int count1=0; //even product pairs
int count2=0; //odd product pairs
int prod=1;
for(int i=0;i<n-1;i++){
for(int j=i+1;j<n;j++){
prod=arr[i]*arr[j];
if(prod%2==0) //product is even
{ count1+=2; } //(a,b) and (b,a) as two pairs
else
{ count2+=2; }
}
}
cout<<"Even Product pairs: "<<count1;
cout<<endl<<"Odd Product pairs: "<<count2;
}
int main(){
int arr[] = { 1,2,7,3 };
int n = sizeof(arr) / sizeof(int);
countPairs(arr, n);
return 0;
}Output
If we run the above code it will generate the following output −
Even Product pairs: 6 Odd Product pairs: 6
123 Views
