Given two arrays a[] and b[], the task is to count the pairs (a[i], b[j]) such that (a[i] + b[j]) is a Fibonacci number.Note that (a, b) is equal to (b, a) and will be counted once.
First few Fibonacci numbers are:
0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 141, …..
Examples:
Input: a[] = {99, 1, 33, 2}, b[] = {1, 11, 2}
Output: 4
Total distinct pairs are (1, 1), (1, 2), (33, 1) and (2, 11)
Input: a[] = {5, 0, 8}, b[] = {0, 9}
Output: 3
Approach:
- Take an empty set.
- Run two nested loops to generate all possible pairs from the two arrays taking one element from first array(call it a) and one from second array(call it b).
- Apply fibonacci test on (a + b) i.e. in order for a number x to be a Fibonacci number, any one of either 5 * x2 + 4 or 5 * x2 – 4 must be a perfect square.
- If it is Fibonacci number then push (a, b) in the set, if a < b or (b, a) if b < a. This is done to avoid duplicacy.
- The size of the set in the end is the total count of valid pairs.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;// Function that returns true if// x is a perfect squarebool isPerfectSquare(long double x){ // Find floating point value of // square root of x long double sr = sqrt(x); // If square root is an integer return ((sr - floor(sr)) == 0);}// Function that returns true if// n is a Fibonacci Numberbool isFibonacci(int n){ return isPerfectSquare(5 * n * n + 4) || isPerfectSquare(5 * n * n - 4);}// Function to return the count of distinct pairs// from the given array such that the sum of the// pair elements is a Fibonacci numberint totalPairs(int a[], int b[], int n, int m){ // Set is used to avoid duplicate pairs set<pair<int, int> > s; for (int i = 0; i < n; i++) { for (int j = 0; j < m; j++) { // If sum is a Fibonacci number if (isFibonacci(a[i] + b[j]) == true) { if (a[i] < b[j]) s.insert(make_pair(a[i], b[j])); else s.insert(make_pair(b[j], a[i])); } } } // Return the size of the set return s.size();}// Driver codeint main(){ int a[] = { 99, 1, 33, 2 }; int b[] = { 1, 11, 2 }; int n = sizeof(a) / sizeof(a[0]); int m = sizeof(b) / sizeof(b[0]); cout << totalPairs(a, b, n, m); return 0;} |
Java
// Java implementation of the approachimport java.util.*;class GFG { static class pair{ int first, second; public pair(int first, int second) { this.first = first; this.second = second; } } // Function that returns true if// x is a perfect squarestatic boolean isPerfectSquare(double x){ // Find floating point value of // square root of x double sr = Math.sqrt(x); // If square root is an integer return ((sr - Math.floor(sr)) == 0);}// Function that returns true if// n is a Fibonacci Numberstatic boolean isFibonacci(int n){ return isPerfectSquare(5 * n * n + 4) || isPerfectSquare(5 * n * n - 4);}// Function to return the count of distinct pairs// from the given array such that the sum of the// pair elements is a Fibonacci numberstatic int totalPairs(int a[], int b[], int n, int m){ // Set is used to avoid duplicate pairs List<pair> s = new LinkedList<>(); for (int i = 0; i < n; i++) { for (int j = 0; j < m; j++) { // If sum is a Fibonacci number if (isFibonacci(a[i] + b[j]) == true) { if (a[i] < b[j]) { if(checkDuplicate(s, new pair(a[i], b[j]))) s.add(new pair(a[i], b[j])); } else { if(checkDuplicate(s, new pair(b[j], a[i]))) s.add(new pair(b[j], a[i])); } } } } // Return the size of the set return s.size();}static boolean checkDuplicate(List<pair> pairList, pair newPair){ for(pair p: pairList) { if(p.first == newPair.first && p.second == newPair.second) return false; } return true;} // Driver codepublic static void main(String[] args) { int a[] = { 99, 1, 33, 2 }; int b[] = { 1, 11, 2 }; int n = a.length; int m = b.length; System.out.println(totalPairs(a, b, n, m));}} // This code is contributed by Rajput-Ji |
Python3
# Python3 implementation of the approach from math import sqrt,floor# Function that returns true if # x is a perfect square def isPerfectSquare(x) : # Find floating point value of # square root of x sr = sqrt(x) # If square root is an integer return ((sr - floor(sr)) == 0)# Function that returns true if # n is a Fibonacci Number def isFibonacci(n ) : return (isPerfectSquare(5 * n * n + 4) or isPerfectSquare(5 * n * n - 4))# Function to return the count of distinct pairs # from the given array such that the sum of the # pair elements is a Fibonacci number def totalPairs(a, b, n, m) : # Set is used to avoid duplicate pairs s = set(); for i in range(n) : for j in range(m) : # If sum is a Fibonacci number if (isFibonacci(a[i] + b[j]) == True) : if (a[i] < b[j]) : s.add((a[i], b[j])); else : s.add((b[j], a[i])); # Return the size of the set return len(s); # Driver code if __name__ == "__main__" : a = [ 99, 1, 33, 2 ]; b = [ 1, 11, 2 ]; n = len(a); m = len(b); print(totalPairs(a, b, n, m)); # This code is contributed by Ryuga |
C#
// C# implementation of the approachusing System;using System.Collections.Generic; class GFG {public class pair{ public int first, second; public pair(int first, int second) { this.first = first; this.second = second; } } // Function that returns true if// x is a perfect squarestatic bool isPerfectSquare(double x){ // Find floating point value of // square root of x double sr = Math.Sqrt(x); // If square root is an integer return ((sr - Math.Floor(sr)) == 0);}// Function that returns true if// n is a Fibonacci Numberstatic bool isFibonacci(int n){ return isPerfectSquare(5 * n * n + 4) || isPerfectSquare(5 * n * n - 4);}// Function to return the count of distinct pairs// from the given array such that the sum of the// pair elements is a Fibonacci numberstatic int totalPairs(int []a, int []b, int n, int m){ // Set is used to avoid duplicate pairs List<pair> s = new List<pair>(); for (int i = 0; i < n; i++) { for (int j = 0; j < m; j++) { // If sum is a Fibonacci number if (isFibonacci(a[i] + b[j]) == true) { if (a[i] < b[j]) { if(checkDuplicate(s, new pair(a[i], b[j]))) s.Add(new pair(a[i], b[j])); } else { if(checkDuplicate(s, new pair(b[j], a[i]))) s.Add(new pair(b[j], a[i])); } } } } // Return the size of the set return s.Count;}static bool checkDuplicate(List<pair> pairList, pair newPair){ foreach(pair p in pairList) { if(p.first == newPair.first && p.second == newPair.second) return false; } return true;} // Driver codepublic static void Main(String[] args) { int []a = { 99, 1, 33, 2 }; int []b = { 1, 11, 2 }; int n = a.Length; int m = b.Length; Console.WriteLine(totalPairs(a, b, n, m));}} // This code is contributed by Rajput-Ji |
Javascript
<script>// Javascript implementation of the approach// Function that returns true if// x is a perfect squarefunction isPerfectSquare(x){ // Find floating point value of // square root of x var sr = Math.sqrt(x); // If square root is an integer return ((sr - Math.floor(sr)) == 0);}// Function that returns true if// n is a Fibonacci Numberfunction isFibonacci(n){ return isPerfectSquare(5 * n * n + 4) || isPerfectSquare(5 * n * n - 4);}// Function to return the count of distinct pairs// from the given array such that the sum of the// pair elements is a Fibonacci numberfunction totalPairs(a, b, n, m){ // Set is used to avoid duplicate pairs var s = new Set(); for (var i = 0; i < n; i++) { for (var j = 0; j < m; j++) { // If sum is a Fibonacci number if (isFibonacci(a[i] + b[j])) { if (a[i] < b[j]) { var tmp = a[i]+" "+b[j]; s.add(tmp); } else { var tmp = b[j]+" "+a[i]; s.add(tmp); } } } } // Return the size of the set return s.size;}// Driver codevar a = [99, 1, 33, 2 ];var b = [1, 11, 2 ];var n = a.length;var m = b.length;document.write( totalPairs(a, b, n, m));</script> |
Output:
4
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