Given an array arr[], the task is to find the count even-odd pairs in the array.
Examples:
Input: arr[] = { 1, 2, 1, 3 }
Output: 2
Explanation:
The 2 pairs of the form (even, odd) are {2, 1} and {2, 3}.
Input: arr[] = { 5, 4, 1, 2, 3}
Output: 3
Naive Approach:
- Run two nested loops to get all the possible pairs of numbers in the array.
- For each pair, check whether if a[i] is even and a[j] is odd.
- If it is, then increase the required count by one.
- When all the pairs have been checked, print the count in the end.
Below is the implementation of the above approach:
C++
// C++ program to count the pairs in array// of the form (even, odd)#include <bits/stdc++.h>using namespace std;// Function to count the pairs in array// of the form (even, odd)int findCount(int arr[], int n){ // variable to store count of such pairs int res = 0; // Iterate through all pairs for (int i = 0; i < n - 1; i++) for (int j = i + 1; j < n; j++) // Increment count // if condition is satisfied if ((arr[i] % 2 == 0) && (arr[j] % 2 == 1)) { res++; } return res;}// Driver codeint main(){ int a[] = { 5, 4, 1, 2, 3 }; int n = sizeof(a) / sizeof(a[0]); cout << findCount(a, n); return 0;} |
Java
// Java program to count the pairs in array// of the form (even, odd)import java.util.*;class GFG{// Function to count the pairs in array// of the form (even, odd)static int findCount(int arr[], int n){ // Variable to store count of such pairs int res = 0; // Iterate through all pairs for (int i = 0; i < n - 1; i++) for (int j = i + 1; j < n; j++) // Increment count // if condition is satisfied if ((arr[i] % 2 == 0) && (arr[j] % 2 == 1)) { res++; } return res;}// Driver codepublic static void main(String[] args){ int a[] = { 5, 4, 1, 2, 3 }; int n = a.length; System.out.print(findCount(a, n));}}// This code is contributed by Rohit_ranjan |
Python3
# Python3 program to count the pairs # in array of the form (even, odd) # Function to count the pairs in # array of the form (even, odd) def findCount(arr, n): # Variable to store count # of such pairs res = 0 # Iterate through all pairs for i in range(0, n - 1): for j in range(i + 1, n): # Increment count # if condition is satisfied if ((arr[i] % 2 == 0) and (arr[j] % 2 == 1)): res = res + 1 return res# Driver code a = [ 5, 4, 1, 2, 3 ] n = len(a)print(findCount(a, n))# This code is contributed by PratikBasu |
C#
// C# program to count the pairs in array// of the form (even, odd)using System;class GFG{// Function to count the pairs in array// of the form (even, odd)static int findCount(int []arr, int n){ // Variable to store count of such pairs int res = 0; // Iterate through all pairs for(int i = 0; i < n - 1; i++) for(int j = i + 1; j < n; j++) // Increment count // if condition is satisfied if ((arr[i] % 2 == 0) && (arr[j] % 2 == 1)) { res++; } return res;}// Driver codepublic static void Main(String[] args){ int []a = { 5, 4, 1, 2, 3 }; int n = a.Length; Console.Write(findCount(a, n));}}// This code is contributed by Rohit_ranjan |
Javascript
<script>// Javascript program to count the pairs in array// of the form (even, odd)// Function to count the pairs in array// of the form (even, odd)function findCount(arr, n){ // variable to store count of such pairs let res = 0; // Iterate through all pairs for (let i = 0; i < n - 1; i++) for (let j = i + 1; j < n; j++) // Increment count // if condition is satisfied if ((arr[i] % 2 == 0) && (arr[j] % 2 == 1)) { res++; } return res;}// Driver codelet a = [ 5, 4, 1, 2, 3 ];let n = a.length;document.write(findCount(a, n));</script> |
Output:
3
Time Complexity: O(n2)
Auxiliary Space: O(1)
Efficient Approach:
- For each element starting from index 0 we will check if it’s even or not.
- If it’s even we will increase the count by 1.
- Else we will increase our final answer by count.
Below is the implementation of the above approach:
C++
// C++ program to count the pairs in array// of the form (even, odd)#include <bits/stdc++.h>using namespace std;// Function to count the pairs in array// of the form (even, odd)int findCount(int arr[], int n){ int count = 0, ans = 0; for (int i = 0; i < n; i++) { // check if number is even or not if (arr[i] % 2 == 0) count++; else ans = ans + count; } return ans;}// Driver codeint main(){ int a[] = { 5, 4, 1, 2, 3 }; int n = sizeof(a) / sizeof(a[0]); cout << findCount(a, n); return 0;} |
Java
// Java program to count the pairs // in array of the form (even, odd)class GFG{// Function to count the pairs in // array of the form (even, odd)static int findCount(int arr[], int n){ int count = 0, ans = 0; for(int i = 0; i < n; i++) { // Check if number is even // or not if (arr[i] % 2 == 0) { count++; } else { ans = ans + count; } } return ans;}// Driver codepublic static void main(String[] args){ int a[] = { 5, 4, 1, 2, 3 }; int n = a.length; System.out.print(findCount(a, n));}}// This code is contributed by amal kumar choubey |
Python3
# Python3 program to count the pairs # in array of the form (even, odd) # Function to count the pairs in # array of the form (even, odd) def findCount(arr, n): count = 0 ans = 0 for i in range(0, n): # Check if number is even or not if (arr[i] % 2 == 0): count = count + 1 else: ans = ans + count return ans # Driver code a = [ 5, 4, 1, 2, 3 ]n = len(a) print(findCount(a, n))# This code is contributed by PratikBasu |
C#
// C# program to count the pairs in // array of the form (even, odd)using System;class GFG{// Function to count the pairs in // array of the form (even, odd)static int findCount(int []arr, int n){ int count = 0, ans = 0; for(int i = 0; i < n; i++) { // Check if number is even or not if (arr[i] % 2 == 0) { count++; } else { ans = ans + count; } } return ans;}// Driver codepublic static void Main(){ int []a = { 5, 4, 1, 2, 3 }; int n = a.Length; Console.WriteLine(findCount(a, n));}}// This code is contributed by Code_Mech |
Javascript
<script>// Javascript program to count the pairs in array// of the form (even, odd)// Function to count the pairs in array// of the form (even, odd)function findCount(arr, n){ let count = 0, ans = 0; for (let i = 0; i < n; i++) { // check if number is even or not if (arr[i] % 2 == 0) count++; else ans = ans + count; } return ans;}// Driver codelet a = [ 5, 4, 1, 2, 3 ];let n = a.length;document.write(findCount(a, n));// This code is contributed by subhammahato348.</script> |
Output:
3
Time Complexity: O(n)
Auxiliary Space: O(1)
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