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++
// C++ implementation of the above approach#include <bits/stdc++.h>using namespace std;// Function to return the number of tripletsint 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 Codeint main(){ int N = 3, K = 2; // Function Call cout << NoofTriplets(N, K); return 0;} |
Java
// Java implementation of the approachimport 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 approachusing 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
# 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
<script>// Javascript implementation of the above approach // Function to return the number of tripletsfunction NoofTriplets(N, K){ let cnt = Array(K); for(let i = 0; i < K; i++) cnt[i] = 0; // Storing the frequency of // each modulo class for(let 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 Codelet N = 3;let K = 2;// Function Calldocument.write(NoofTriplets(N, K));// This code is contributed by mohit kumar 29</script> |
9
Time Complexity: O(N)
Auxiliary Space: O(K)
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