Find all Pairs possible from the given Array

Given an array arr[] of N integers, the task is to find all the pairs possible from the given array. 
Note: 
 

1. (arr[i], arr[i]) is also considered as a valid pair.
2. (arr[i], arr[j]) and (arr[j], arr[i]) are considered as two different pairs.

Examples: 
 

Input: arr[] = {1, 2} 
Output: (1, 1), (1, 2), (2, 1), (2, 2).
Input: arr[] = {1, 2, 3} 
Output: (1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3)  

Approach: 
In order to find all the possible pairs from the array, we need to traverse the array and select the first element of the pair. Then we need to pair this element with all the elements in the array from index 0 to N-1. 
Below is the step by step approach: 
 

• Traverse the array and select an element in each traversal.
• For each element selected, traverse the array with help of another loop and form the pair of this element with each element in the array from the second loop.
• The array in the second loop will get executed from its first element to its last element, i.e. from index 0 to N-1.
• Print each pair formed.

Below is the implementation of the above approach:
 

// C++ implementation to find all
// Pairs possible from the given Array

#include <bits/stdc++.h>
using namespace std;

// Function to print all possible
// pairs from the array
void printPairs(int arr[], int n)
{

    // Nested loop for all possible pairs
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < n; j++) {
            cout << "(" << arr[i] << ", "
                 << arr[j] << ")"
                 << ", ";
        }
    }
}

// Driver code
int main()
{
    int arr[] = { 1, 2 };
    int n = sizeof(arr) / sizeof(arr[0]);

    printPairs(arr, n);

    return 0;
}

// Java implementation to find all
// Pairs possible from the given Array
class GFG{
 
// Function to print all possible
// pairs from the array
static void printPairs(int arr[], int n)
{
 
    // Nested loop for all possible pairs
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < n; j++) {
            System.out.print("(" +  arr[i]+ ", "
                 + arr[j]+ ")"
                + ", ");
        }
    }
}
 
// Driver code
public static void main(String[] args)
{
    int arr[] = { 1, 2 };
    int n = arr.length;
 
    printPairs(arr, n);
 
}
}

// This code is contributed by PrinciRaj1992

# Python3 implementation to find all
# Pairs possible from the given Array

# Function to print all possible
# pairs from the array
def printPairs(arr, n):

    # Nested loop for all possible pairs
    for i in range(n):
        for j in range(n):
            print("(",arr[i],",",arr[j],")",end=", ")

# Driver code

arr=[1, 2]
n = len(arr)

printPairs(arr, n)

# This code is contributed by mohit kumar 29

// C# implementation to find all
// Pairs possible from the given Array
using System;

class GFG{
 
// Function to print all possible
// pairs from the array
static void printPairs(int []arr, int n)
{
 
    // Nested loop for all possible pairs
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < n; j++) {
            Console.Write("(" +  arr[i]+ ", "
                 + arr[j]+ ")"
                + ", ");
        }
    }
}
 
// Driver code
public static void Main(string[] args)
{
    int []arr = { 1, 2 };
    int n = arr.Length;
 
    printPairs(arr, n);
}
}

// This code is contributed by AnkitRai01

<script>

// Javascript implementation to find all
// Pairs possible from the given Array

// Function to print all possible
// pairs from the array
function printPairs(arr, n)
{
    // Nested loop for all possible pairs
    for (var i = 0; i < n; i++) {
        for (var j = 0; j < n; j++) {
            document.write("(" + arr[i] + ", "
                 + arr[j] + ")"
                 + ", ");
        }
    }
}

// Driver code
var arr = [ 1, 2 ];
var n = arr.length;
printPairs(arr, n);

// This code is contributed by rutvik_56.
</script>

(1, 1), (1, 2), (2, 1), (2, 2), 

Time Complexity: O(N²)

Auxiliary Space: O(1)