Number of triplets such that each value is less than N and each pair sum is a multiple of K

Given two integers N and K. Find the numbers of triplets (a, b, c) such that 0 ≤ a, b, c ≤ N and (a + b), (b + c) and (c + a) are multiples of K.

Examples:

Input: N = 3, K = 2
Output: 9
Triplets possible are:
{(1, 1, 1), (1, 1, 3), (1, 3, 1)
(1, 3, 3), (2, 2, 2), (3, 1, 1)
(3, 1, 1), (3, 1, 3), (3, 3, 3)}

Input: N = 5, K = 3
Output: 1
Only possible triplet is (3, 3, 3)

Approach: Given that (a + b), (b + c) and (c + a) are multiples of K. Hence, we can say that (a + b) % K = 0, (b + c) % K = 0 and (c + a) % K = 0.
If a belongs to the x modulo class of K then b should be in the (K – x)th modulo class using the first condition.
From the second condition, it can be seen that c belongs to the x modulo class of K. Now as both a and c belong to the same modulo class and they have to satisfy the third relation which is (a + c) % K = 0. It could be only possible if x = 0 or x = K / 2.
When K is an odd integer, x = K / 2 is not valid.
Hence to solve the problem, count the number of elements from 0 to N in the 0th modulo class and the (K / 2)th modulo class of K.

  • If K is odd then the result is cnt[0]3
  • If K is even then the result is cnt[0]3 + cnt[K / 2]3.

Below is the implementation of the above approach:

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ implementation of the above approach
#include <bits/stdc++.h>
using namespace std;
  
// Function to return the number of triplets
int NoofTriplets(int N, int K)
{
    int cnt[K];
  
    // Initializing the count array
    memset(cnt, 0, sizeof(cnt));
  
    // Storing the frequency of each modulo class
    for (int i = 1; i <= N; i += 1) {
        cnt[i % K] += 1;
    }
  
    // If K is odd
    if (K & 1)
        return cnt[0] * cnt[0] * cnt[0];
  
    // If K is even
    else {
        return (cnt[0] * cnt[0] * cnt[0]
                + cnt[K / 2] * cnt[K / 2] * cnt[K / 2]);
    }
}
  
// Driver Code
int main()
{
    int N = 3, K = 2;
  
    // Function Call
    cout << NoofTriplets(N, K);
  
    return 0;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// Java implementation of the approach
import java.util.Arrays;
  
class GFG 
{
  
    // Function to return the number of triplets
    static int NoofTriplets(int N, int K) 
    {
        int[] cnt = new int[K];
  
        // Initializing the count array
        Arrays.fill(cnt, 0, cnt.length, 0);
  
        // Storing the frequency of each modulo class
        for (int i = 1; i <= N; i += 1)
        {
            cnt[i % K] += 1;
        }
  
        // If K is odd
        if ((K & 1) != 0
        {
            return cnt[0] * cnt[0] * cnt[0];
        
        // If K is even
        else 
        {
            return (cnt[0] * cnt[0] * cnt[0]
                    + cnt[K / 2] * cnt[K / 2] * cnt[K / 2]);
        }
    }
  
    // Driver Code
    public static void main(String[] args) 
    {
  
        int N = 3, K = 2;
  
        // Function Call
        System.out.println(NoofTriplets(N, K));
    }
}
  
// This code is contributed by Princi Singh

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// C# implementation of the approach
using System;
  
class GFG
{
      
    // Function to return the number of triplets
    static int NoofTriplets(int N, int K) 
    {
        int[] cnt = new int[K];
  
        // Initializing the count array
        Array.Fill(cnt, 0, cnt.Length, 0);
  
        // Storing the frequency of each modulo class
        for (int i = 1; i <= N; i += 1)
        {
            cnt[i % K] += 1;
        }
  
        // If K is odd
        if ((K & 1) != 0) 
        {
            return cnt[0] * cnt[0] * cnt[0];
        
        // If K is even
        else
        {
            return (cnt[0] * cnt[0] * cnt[0]
                    + cnt[K / 2] * cnt[K / 2] * cnt[K / 2]);
        }
    }
  
    // Driver Code
    static public void Main ()
    {
            int N = 3, K = 2;
  
        // Function Call
        Console.Write(NoofTriplets(N, K));
    }
}
  
// This code is contributed by ajit

chevron_right


Python3

filter_none

edit
close

play_arrow

link
brightness_4
code

# Python3 implementation of the above approach 
  
# Function to return the number of triplets 
def NoofTriplets(N, K) : 
      
    # Initializing the count array
    cnt = [0]*K; 
  
    # Storing the frequency of each modulo class 
    for i in range(1, N + 1) :
        cnt[i % K] += 1
  
    # If K is odd 
    if (K & 1) :
        rslt = cnt[0] * cnt[0] * cnt[0]; 
        return rslt
  
    # If K is even 
    else :
        rslt = (cnt[0] * cnt[0] * cnt[0] +
                cnt[K // 2] * cnt[K // 2] * cnt[K // 2]); 
        return rslt
  
# Driver Code 
if __name__ == "__main__"
  
    N = 3; K = 2
  
    # Function Call 
    print(NoofTriplets(N, K)); 
  
# This code is contributed by AnkitRai01

chevron_right


Output:

9

Time Complexity: O(N)



My Personal Notes arrow_drop_up

Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.



Improved By : princi singh, AnkitRai01, jit_t