Given a number N, the task is to find the number of quadraples such that a2 + b2 = c2 + d2 where (1 <= a, b, c, d <= N).
Example:
Input: N = 2
Output: 6
Explanation:
There are 6 valid quadraples (1, 1, 1, 1), (1, 2, 1, 2), (1, 2, 2, 1), (2, 1, 1, 2), (2, 1, 2, 1), (2, 2, 2, 2).Input: N = 4
Output 28
Explanation:
There are 28 valid quadraples for N = 4.
Naive Approach: The naive approach is to use 4 nested loops in the range [1, N] and count those quadruplets (a, b, c, d) such that a2 + b2 = c2 + d2
Time Complexity: O(N4)
Auxiliary Space: O(1)
Efficient approach than Naive approach: The above solution can be reduced in terms of time complexity using some mathematics. Find the quadruplets such that a2 + b2 = c2 + d2.
a2 + b2 = c2 + d2
a2 + b2 – c2 = d2
Taking square root on both sides
(a2 + b2 – c2)1/2 = d
By using the above formula, we can conclude that d exists only if (a2 + b2 – c2)1/2 is a positive integer.
Time Complexity: O(N3)
Efficient Approach: The idea is to use Hashing. Below are the steps:
- Traverse over all the pairs(say (a, b)) over [1, N] and store the value of a2 + b2 with its occurrence in a Map.
- Iterate over all the pairs [1, N] and if the sum exists in the map then count the sum of each pair stored in the map.
- Print the count.
Below is the implementation of the approach:
C++
// C++ program for the above approach #include <bits/stdc++.h> using namespace std; typedef long long int ll; // Function to count the quadruples ll countQuadraples(ll N) { // Counter variable ll cnt = 0; // Map to store the // sum of pair (a^2 + b^2) map<ll, ll> m; // Iterate till N for (ll a = 1; a <= N; a++) { for (ll b = 1; b <= N; b++) { // Calculate a^2 + b^2 ll x = a * a + b * b; // Increment the value in map m[x] += 1; } } for (ll c = 1; c <= N; c++) { for (ll d = 1; d <= N; d++) { ll x = c * c + d * d; // Check if this sum // was also in a^2 + b^2 if (m.find(x) != m.end()) cnt += m[x]; } } // Return the count return cnt; } // Driver Code int main() { // Given N ll N = 2; // Function Call cout << countQuadraples(N) << endl; return 0; } |
Java
// Java program for the above approach import java.util.*;class GFG{// Function to count the quadruplesstatic long countQuadraples(long N){ // Counter variable long cnt = 0; // Map to store the // sum of pair (a^2 + b^2) Map<Long, Long> m = new HashMap<>(); // Iterate till N for(long a = 1; a <= N; a++) { for(long b = 1; b <= N; b++) { // Calculate a^2 + b^2 long x = a * a + b * b; // Increment the value in map m.put(x, m.getOrDefault(x, 0l) + 1); } } for(long c = 1; c <= N; c++) { for(long d = 1; d <= N; d++) { long x = c * c + d * d; // Check if this sum // was also in a^2 + b^2 if (m.containsKey(x)) cnt += m.get(x); } } // Return the count return cnt;}// Driver codepublic static void main(String[] args){ // Given N long N = 2; // Function call System.out.println(countQuadraples(N));}}// This code is contributed by offbeat |
Python3
# Python3 program for # the above approach from collections import defaultdict# Function to count # the quadruples def countQuadraples(N): # Counter variable cnt = 0 # Map to store the # sum of pair (a^2 + b^2) m = defaultdict (int) # Iterate till N for a in range (1, N + 1): for b in range (1, N + 1): # Calculate a^2 + b^2 x = a * a + b * b # Increment the value in map m[x] += 1 for c in range (1, N + 1): for d in range (1, N + 1): x = c * c + d * d # Check if this sum # was also in a^2 + b^2 if x in m: cnt += m[x] # Return the count return cnt # Driver Code if __name__ == "__main__": # Given N N = 2 # Function Call print (countQuadraples(N)) # This code is contributed by Chitranayal |
C#
// C# program for the above approach using System;using System.Collections.Generic;class GFG{// Function to count the quadruplesstatic long countQuadraples(long N){ // Counter variable long cnt = 0; // Map to store the // sum of pair (a^2 + b^2) Dictionary<long, long> m = new Dictionary<long, long>(); // Iterate till N for(long a = 1; a <= N; a++) { for(long b = 1; b <= N; b++) { // Calculate a^2 + b^2 long x = a * a + b * b; // Increment the value in map if (m.ContainsKey(x)) m[x] = m[x] + 1; else m.Add(x, 1); } } for(long c = 1; c <= N; c++) { for(long d = 1; d <= N; d++) { long x = c * c + d * d; // Check if this sum // was also in a^2 + b^2 if (m.ContainsKey(x)) cnt += m[x]; } } // Return the count return cnt;}// Driver codepublic static void Main(String[] args){ // Given N long N = 2; // Function call Console.WriteLine(countQuadraples(N));}}// This code is contributed by Amit Katiyar |
6
Time Complexity: O(N2 log N)
Auxiliary Space: O(N)
