# Find all Pairs possible from the given Array

• Difficulty Level : Easy
• Last Updated : 13 Sep, 2022

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<

## Java

 `/*``    ``Java 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)``*/` `import` `java.util.*;` `class` `GFG {``  ``static` `void``    ``getPairsMerge(``int``[] arr, ``int` `l, ``int` `r, ``int` `mid,``                  ``ArrayList > 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];``        ``ArrayList a1``          ``= ``new` `ArrayList();``        ``a1.add(arr[i]);``        ``a1.add(arr[j]);` `        ``ArrayList a2``          ``= ``new` `ArrayList();``        ``a2.add(arr[j]);``        ``a2.add(arr[i]);` `        ``ArrayList a3``          ``= ``new` `ArrayList();``        ``a3.add(arr[j]);``        ``a3.add(arr[j]);` `        ``p.add(a1);``        ``p.add(a2);``        ``p.add(a3);` `        ``k++;``        ``j++;``      ``}``      ``else` `{` `        ``ArrayList a1``          ``= ``new` `ArrayList();``        ``a1.add(arr[i]);``        ``a1.add(arr[j]);` `        ``ArrayList a2``          ``= ``new` `ArrayList();``        ``a2.add(arr[j]);``        ``a2.add(arr[i]);` `        ``ArrayList a3``          ``= ``new` `ArrayList();``        ``a3.add(arr[i]);``        ``a3.add(arr[i]);` `        ``p.add(a1);``        ``p.add(a2);``        ``p.add(a3);` `        ``b[k] = arr[i];``        ``i++;``        ``k++;``      ``}``    ``}` `    ``while` `(i <= mid) {``      ``b[k] = arr[i];``      ``ArrayList a3``        ``= ``new` `ArrayList();``      ``a3.add(arr[i]);``      ``a3.add(arr[i]);` `      ``p.add(a3);``      ``i++;``      ``k++;``    ``}``    ``while` `(j <= r) {``      ``b[k] = arr[j];``      ``ArrayList a3``        ``= ``new` `ArrayList();``      ``a3.add(arr[j]);``      ``a3.add(arr[j]);` `      ``p.add(a3);``      ``j++;``      ``k++;``    ``}` `    ``for` `(``int` `x = l; x <= r; x++) {``      ``arr[x] = b[x];``    ``}``  ``}` `  ``static` `void``    ``getAllPairs(``int``[] arr, ``int` `l, ``int` `r,``                ``ArrayList > 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` `};``    ``ArrayList > p``      ``= ``new` `ArrayList >();``    ``getAllPairs(arr, ``0``, n - ``1``, p);``    ``for` `(ArrayList it : p)``      ``System.out.println(it.get(``0``) + ``" "` `+ it.get(``1``));``  ``}``}` `// This code is contributed by phasing17`

## 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<``int``[]> 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<``int``[]> 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<``int``[]> p = ``new` `List<``int``[]>();``    ``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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