Skip to content
Related Articles

Related Articles

Table of Contents

Improve Article
Save Article
Like Article

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:

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.

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



// 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.
Python3



# 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#



// 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.
Javascript



<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>
Output
28
Time Complexity: O(N)
Auxiliary Space: O(1)



My Personal Notes
arrow_drop_up
Previous
Generate N-size Array with equal difference of adjacent elements containing given numbers A and B
Next

Find the shortest Binary string containing one or more occurrences of given strings
Recommended Articles
Page :
Make all the elements of array odd by incrementing odd-indexed elements of odd-length subarrays
16, Dec 20
Check if a number has an odd count of odd divisors and even count of even divisors
20, Aug 20
Count of triplets from the given Array such that sum of any two elements is the third element
27, Feb 20
Count of even sum triplets in the array for Q range queries
24, Aug 21
Maximize sum of array by reducing array elements to contain no triplets (i, j, k) where a[i] < a[j] and a[i] < a[k] and j <i < k
09, Sep 20
Absolute difference between sum of even elements at even indices & odd elements at odd indices in given Array
04, Dec 21
Count triplets with sum smaller than a given value
10, Nov 15
Count Triplets such that one of the numbers can be written as sum of the other two
26, Nov 18
Count non-negative triplets with sum equal to N
19, Jul 19
Count triplets such that sum of any two number is equal to third | Set 2
30, Mar 20
Count triplets from a given range having sum of two numbers of a triplet equal to the third number
09, Mar 21
Count of triplets having sum of product of any two numbers with the third number equal to N
26, Jul 21
Count distinct prime triplets up to N such that sum of two primes is equal to the third prime
09, Feb 21
Count prime triplets upto N having sum of first two elements equal to the third number
31, Jan 21
Count triplets in an array such that i<j<k and a[j] - a[i] = a[k] - a[j] = D
19, Apr 20
Count of triplets (a, b, c) in the Array such that a divides b and b divides c
13, May 20
Count of triplets in an Array (i, j, k) such that i < j < k and a[k] < a[i] < a[j]
29, Jun 20
Count of triplets in a given Array having GCD K
19, Jun 20
Rearrange array to maximize count of triplets (i, j, k) such that arr[i] > arr[j] < arr[k] and i < j < k
30, Sep 20
Count triplets having product 0 from a given array
16, Oct 20
Count non-equidistant triplets of distinct array elements having indices in increasing order
12, Jan 21
Count triplets from an array which can form quadratic equations with real roots
29, Jan 21
Count triplets (i, j, k) in an array of distinct elements such that a[i] < a[j] > a[k] and i < j < k
12, Apr 21
Count triplets from a sorted array having difference between adjacent elements equal to D
12, May 21
Article Contributed By :
https://media.geeksforgeeks.org/auth/avatar.png
madhav_mohan
@madhav_mohan
Vote for difficulty





Improved By :
Article Tags :
Practice Tags :
Report Issue

We use cookies to ensure you have the best browsing experience on our website. By using our site, you
acknowledge that you have read and understood our
Cookie Policy &
        Privacy Policy

Lightbox

Start Your Coding Journey Now!