Count of triplets in an Array with odd sum
Last Updated :
13 Sep, 2021
Given an array arr[] with N integers, find the number of triplets of i, j and k such that 1<= i < j < k <= N and arr[i] + arr[j] + arr[k] is odd.
Example:
Input: arr[] = {1, 2, 3, 4, 5}
Output: 4
Explanation: The given array contains 4 triplets with an odd sum. They are {1, 2, 4}, {1, 3, 5}, {2, 3, 4} and {2, 4, 5}.
Input: arr[] ={4, 5, 6, 4, 5, 10, 1, 7}
Output: 28
Naive Approach: The given problem can be solved by iterating over all possible unordered triplets in the array and keep a track of the number of triplets such that their sum is odd.
Time Complexity: O(N3)
Efficient Approach: The above approach can be optimized using the below property of integers:
- odd + even + even = odd
- odd + odd + odd = odd
Therefore, to implement the above idea, we can count the number of even and odd integers in the array. Suppose the count of odd integers is O and the count of even integers is E. So the distinct ways to arrange one odd integer and two even integers is OC1 * EC2 -> (O * E * (E-1))/2 and the distinct ways to arrange three odd integers is OC3 -> (O * (O-1) * (O-2)) / 6. The final answer will be the sum of the above two values.
Below is the implementation of the above approach:
C++
// C++ Program for the above approach
#include <iostream>
using namespace std;
// Function to count the number of
// unordered triplets such that their
// sum is an odd integer
int countTriplets(int arr[], int n)
{
// Count the number of odd and
// even integers in the array
int odd = 0, even = 0;
for (int i = 0; i < n; i++) {
if (arr[i] & 1)
odd++;
else
even++;
}
// Number of ways to create triplets
// using one odd and two even integers
int c1 = odd * (even * (even - 1)) / 2;
// Number of ways to create triplets
// using three odd integers
int c2 = (odd * (odd - 1) * (odd - 2)) / 6;
// Return answer
return c1 + c2;
}
// Driver Code
int main()
{
int arr[] = { 4, 5, 6, 4, 5, 10, 1, 7 };
int n = sizeof(arr) / sizeof(int);
// Function Call
int ans = countTriplets(arr, n);
// Print Answer
cout << ans;
return 0;
}
// Java program for the above approach
import java.io.*;
class GFG {
// Function to count the number of
// unordered triplets such that their
// sum is an odd integer
static int countTriplets(int arr[], int n)
{
// Count the number of odd and
// even integers in the array
int odd = 0, even = 0;
for (int i = 0; i < n; i++) {
if ((arr[i] & 1) != 0)
odd++;
else
even++;
}
// Number of ways to create triplets
// using one odd and two even integers
int c1 = odd * (even * (even - 1)) / 2;
// Number of ways to create triplets
// using three odd integers
int c2 = (odd * (odd - 1) * (odd - 2)) / 6;
// Return answer
return c1 + c2;
}
// Driver code
public static void main(String[] args)
{
int arr[] = { 4, 5, 6, 4, 5, 10, 1, 7 };
int n = arr.length;
// Function Call
int ans = countTriplets(arr, n);
// Print Answer
System.out.println(ans);
}
}
// This code is contributed by code_hunt.
# Python Program for the above approach
# Function to count the number of
# unordered triplets such that their
# sum is an odd integer
def countTriplets(arr, n):
# Count the number of odd and
# even integers in the array
odd = 0
even = 0
for i in range(n):
if (arr[i] & 1):
odd+=1
else:
even+=1
# Number of ways to create triplets
# using one odd and two even integers
c1 = odd * (even * (even - 1)) // 2
# Number of ways to create triplets
# using three odd integers
c2 = (odd * (odd - 1) * (odd - 2)) // 6
# Return answer
return c1 + c2
# Driver Code
arr = [4, 5, 6, 4, 5, 10, 1, 7]
n = len(arr)
# Function Call
ans = countTriplets(arr, n)
# Print Answer
print(ans)
# This code is contributed by Shivani
// C# program for the above approach
using System;
class GFG{
// Function to count the number of unordered
// triplets such that their sum is an odd integer
static int countTriplets(int []arr, int n)
{
// Count the number of odd and
// even integers in the array
int odd = 0, even = 0;
for(int i = 0; i < n; i++)
{
if ((arr[i] & 1) != 0)
odd++;
else
even++;
}
// Number of ways to create triplets
// using one odd and two even integers
int c1 = odd * (even * (even - 1)) / 2;
// Number of ways to create triplets
// using three odd integers
int c2 = (odd * (odd - 1) * (odd - 2)) / 6;
// Return answer
return c1 + c2;
}
// Driver code
public static void Main(String[] args)
{
int []arr = { 4, 5, 6, 4, 5, 10, 1, 7 };
int n = arr.Length;
// Function Call
int ans = countTriplets(arr, n);
// Print Answer
Console.Write(ans);
}
}
// This code is contributed by shivanisinghss2110.
<script>
// JavaScript program for the above approach;
// Function to count the number of
// unordered triplets such that their
// sum is an odd integer
function countTriplets(arr, n)
{
// Count the number of odd and
// even integers in the array
let odd = 0, even = 0;
for (let i = 0; i < n; i++) {
if (arr[i] & 1)
odd++;
else
even++;
}
// Number of ways to create triplets
// using one odd and two even integers
let c1 = Math.floor(odd * (even * (even - 1)) / 2);
// Number of ways to create triplets
// using three odd integers
let c2 = Math.floor((odd * (odd - 1) * (odd - 2)) / 6);
// Return answer
return c1 + c2;
}
// Driver Code
let arr = [4, 5, 6, 4, 5, 10, 1, 7];
let n = arr.length;
// Function Call
let ans = countTriplets(arr, n);
// Print Answer
document.write(ans);
// This code is contributed by Potta Lokesh
</script>
Time Complexity: O(N)
Auxiliary Space: O(1)
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