Distinct pairs from given arrays (a[i], b[j]) such that (a[i] + b[j]) is a Fibonacci number

0 0 6 minutes read

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 square
bool 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 Number
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 number
int 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 code
int 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 approach
import 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 square
static 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 Number
static 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 number
static 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 code
public 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 approach
using 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 square
static 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 Number
static 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 number
static 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 code
public 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 square
function 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 Number
function 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 number
function 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 code
var 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:

Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 zambiatek!