Count triplets (a, b, c) such that a + b, b + c and a + c are all divisible by K | Set 2

Count triplets (a, b, c) such that a + b, b + c and a + c are all divisible by K | Set 2

Given two positive integers N and K, the task is to count the number of triplets (a, b, c) such that 0 < a, b, c < N and (a + b), (b + c) and (c + a) are all multiples of K.

Examples:

Input: N = 2, K = 2
Output: 2
Explanation: All possible triplets that satisfy the given property are (1, 1, 1) and (2, 2, 2).
Therefore, the total count is 2.

Input: N = 3, K = 2
Output: 9

Naive Approach: Refer to the previous post for the simplest approach to solve this problem.
Time Complexity: O(N³)
Auxiliary Space: O(1)

Efficient Approach: The above approach can also be optimized based on the following observations:

• The given condition is expressed by a congruence formula:

=> a + b ≡ b + c ≡ c + a ≡ 0(mod K)
=> a+b ≡ b+c (mod K)
=> a ≡ c(mod K)

• The above relation can also be observed without using the congruence formula as:
  • As (a + b) is a multiple of K and (c + b) is a multiple of K. Therefore, (a + b) − (c + b) = a – c is also a multiple of K, i.e., a ≡ b ≡ c (mod K).
  • Therefore, the expression can be further evaluated to:

=> a + b ≡ 0 (mod K)
=> a + a ≡ 0 (mod K) (since a is congruent to b)
=> 2a ≡ 0 (mod K)

From the above observations, the result can be calculated for the following two cases:

• If K is odd, then a ≡ b ≡ c ≡ 0(mod K) since all three are congruent and the total number of triplets can be calculated as (N / K)³.
• If K is even, then K is divisible by 2 and a ≡ 0 (mod K), b ≡ 0 (mod K), and c ≡ 0 (mod K). Therefore, the total number of triplets can be calculated as (N / K)³ ((N + (K/2)) / K)³.

Below is the implementation of the above approach:

C++

#include "bits/stdc++.h" using namespace std;    int countTriplets(int N, int K) {          if (K % 2 == 0) {         long long int x = N / K;         long long int y = (N + (K / 2)) / K;            return x * x * x + y * y * y;     }             else {         long long int x = N / K;         return x * x * x;     } }    int main() {     int N = 2, K = 2;     cout << countTriplets(N, K);        return 0; }

Time Complexity: O(1)
Auxiliary Space: O(1)

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