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Number of triplets such that each value is less than N and each pair sum is a multiple of K
  • Last Updated : 08 Jul, 2019

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++

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// 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;
}

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Java

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// 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

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C#

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// 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

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Python3

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# 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

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Output:

9

Time Complexity: O(N)

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