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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 : 14 Apr, 2021

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:
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:
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 cnt3
• If K is even then the result is cnt3 + cnt[K / 2]3.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the above approach``#include ``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 * cnt * cnt;` `    ``// If K is even``    ``else` `{``        ``return` `(cnt * cnt * cnt``                ``+ 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;``}`

## Java

 `// 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`

## C#

 `// 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 * cnt * cnt;``        ``}``        ``// If K is even``        ``else``        ``{``            ``return` `(cnt * cnt * cnt``                    ``+ 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`

## Python3

 `# 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`

## Javascript

 ``
Output:
`9`

Time Complexity: O(N)

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