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

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

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 ` `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 > 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 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 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 s = ``new` `List(); ` ` `  `    ``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 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 `

Output:

```4
```

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