# 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++

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

 // 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

 # 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#

 // 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

## Javascript



Output

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

Time Complexity: O(N2)

Auxiliary Space: O(1)

## Using Merge Sort

Approach :

During the merge operation in merge sort we can get all the pairs and we can store those pairs into a vector of pairs.

By just making few changes into the merge sort algorithm we can get all the pairs.

## C++

 /*     This code was submitted by : Chirag Mittal from IIIT Dharwad ( username : iitjeechirag )     storing all the pairs while merging     Time Complexity : O(N logN)     Space Complexity : O(N) + O(Number Of Pairs)       using Merge Sort Algorithm */   #include using namespace std; void getPairsMerge(int arr[],int l,int r,int mid,vector>&p){     int b[l+r+1],i=l,k=l,j=mid+1;     while(i<=mid && j<=r){         if(arr[i]>arr[j]){             b[k]=arr[j];             p.push_back({arr[i],arr[j]});             p.push_back({arr[j],arr[i]});             p.push_back({arr[j],arr[j]});             k++;             j++;         }         else{             p.push_back({arr[i],arr[j]});             p.push_back({arr[j],arr[i]});             p.push_back({arr[i],arr[i]});             b[k]=arr[i];             i++;             k++;         }     }       while(i<=mid){         b[k]=arr[i];         p.push_back({arr[i],arr[i]});         i++;         k++;     }     while(j<=r){         b[k]=arr[j];         p.push_back({arr[j],arr[j]});         j++;         k++;     }       for(int x=l;x<=r;x++){         arr[x]=b[x];     } }   void getAllPairs(int arr[],int l,int r,vector>&p){     if(l>p;     getAllPairs(arr,0,n-1,p);     for(auto it:p){         cout<

## Python3

 '''     Python3 program to store all the pairs while merging     using Merge Sort Algorithm       '''   # Function to perform merge sort # Time Complexity : O(N logN) # Space Complexity : O(N) + O(Number Of Pairs) def getPairsMerge(arr, l, r, mid, p):           b = [0 for _ in range(l + r + 1)]     i=l     k=l     j=mid+1;     while(i<=mid and j<=r):         if(arr[i]>arr[j]):             b[k]=arr[j];             p.append([arr[i],arr[j]]);             p.append([arr[j],arr[i]]);             p.append([arr[j],arr[j]]);             k+= 1;             j+= 1;                   else:             p.append([arr[i],arr[j]]);             p.append([arr[j],arr[i]]);             p.append([arr[i],arr[i]]);             b[k]=arr[i];             i+= 1;             k+= 1;                       while(i<=mid):         b[k]=arr[i];         p.append([arr[i],arr[i]]);         i+= 1;         k+= 1;           while(j<=r):         b[k]=arr[j];         p.append([arr[j],arr[j]]);         j+= 1;         k+= 1;           for x in range(l, r + 1):         arr[x]=b[x];           # Function to get all pairs def getAllPairs(arr, l, r, p):       if(l

## C#

 /*     C# program that     stores all the pairs while merging     using Merge Sort Algorithm     Time Complexity : O(N logN)     Space Complexity : O(N) + O(Number Of Pairs) */   using System; using System.Collections.Generic;   class GFG {   static void getPairsMerge(int[] arr, int l, int r,                             int mid, List p)   {     int[] b = new int[l + r + 1];     int i = l, k = l, j = mid + 1;     while (i <= mid && j <= r) {       if (arr[i] > arr[j]) {         b[k] = arr[j];         p.Add(new[] { arr[i], arr[j] });         p.Add(new[] { arr[j], arr[i] });         p.Add(new[] { arr[j], arr[j] });         k++;         j++;       }       else {         p.Add(new[] { arr[i], arr[j] });         p.Add(new[] { arr[j], arr[i] });         p.Add(new[] { arr[i], arr[i] });         b[k] = arr[i];         i++;         k++;       }     }       while (i <= mid) {       b[k] = arr[i];       p.Add(new[] { arr[i], arr[i] });       i++;       k++;     }     while (j <= r) {       b[k] = arr[j];       p.Add(new[] { arr[j], arr[j] });       j++;       k++;     }       for (int x = l; x <= r; x++) {       arr[x] = b[x];     }   }     static void getAllPairs(int[] arr, int l, int r,                           List p)   {     if (l < r) {       int mid = (l + r) / 2;       getAllPairs(arr, l, mid, p);       getAllPairs(arr, mid + 1, r, p);       getPairsMerge(arr, l, r, mid, p);     }   }     public static void Main(string[] args)   {     int n = 2;     int[] arr = { 1, 2 };     List p = new List();     getAllPairs(arr, 0, n - 1, p);     foreach(var it in p)       Console.WriteLine(it[0] + " " + it[1]);   } }   // This code is contributed by phasing17

## Javascript

 /*     JavaScript program to store all the pairs while merging     using Merge Sort Algorithm */   // Function to perform merge sort // Time Complexity : O(N logN) // Space Complexity : O(N) + O(Number Of Pairs) function getPairsMerge(arr, l, r, mid, p) {     let b = new Array(l+r+1).fill(0);     let i=l,k=l,j=mid+1;     while(i<=mid && j<=r){         if(arr[i]>arr[j]){             b[k]=arr[j];             p.push([arr[i],arr[j]]);             p.push([arr[j],arr[i]]);             p.push([arr[j],arr[j]]);             k++;             j++;         }         else{             p.push([arr[i],arr[j]]);             p.push([arr[j],arr[i]]);             p.push([arr[i],arr[i]]);             b[k]=arr[i];             i++;             k++;         }     }       while(i<=mid){         b[k]=arr[i];         p.push([arr[i],arr[i]]);         i++;         k++;     }     while(j<=r){         b[k]=arr[j];         p.push([arr[j],arr[j]]);         j++;         k++;     }       for(let x=l;x<=r;x++){         arr[x]=b[x];     } }   // Function to get all pairs function getAllPairs(arr, l, r, p) {     if(l

Output

1 2
2 1
1 1
2 2

Time Complexity : O (N LogN )

Auxiliary Space: O(l + r)

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