# 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)

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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++ 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[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; ` `} `

 `// 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# 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 `

 `# 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 `

Output:
```9
```

Time Complexity: O(N)

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Improved By : princi singh, AnkitRai01, jit_t

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