Given an array arr[] of N integers, the task is to find all the pairs possible from the given array.
Note:
- (arr[i], arr[i]) is also considered as a valid pair.
- (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 <bits/stdc++.h>using namespace std;// Function to print all possible// pairs from the arrayvoid 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 codeint 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 Arrayclass GFG{ // Function to print all possible// pairs from the arraystatic 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 codepublic 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 arraydef 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 codearr=[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 Arrayusing System;class GFG{ // Function to print all possible// pairs from the arraystatic 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 codepublic static void Main(string[] args){ int []arr = { 1, 2 }; int n = arr.Length; printPairs(arr, n);}}// This code is contributed by AnkitRai01 |
Javascript
<script>// Javascript implementation to find all// Pairs possible from the given Array// Function to print all possible// pairs from the arrayfunction 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 codevar arr = [ 1, 2 ];var n = arr.length;printPairs(arr, n);// This code is contributed by rutvik_56.</script> |
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<bits/stdc++.h>using namespace std;void getPairsMerge(int arr[],int l,int r,int mid,vector<pair<int,int>>&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<pair<int,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); }}int main(){ int n=2; int arr[n]={1,2}; vector<pair<int,int>>p; getAllPairs(arr,0,n-1,p); for(auto it:p){ cout<<it.first<<" "<<it.second<<endl; }} |
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<ArrayList<Integer> > 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<Integer> a1 = new ArrayList<Integer>(); a1.add(arr[i]); a1.add(arr[j]); ArrayList<Integer> a2 = new ArrayList<Integer>(); a2.add(arr[j]); a2.add(arr[i]); ArrayList<Integer> a3 = new ArrayList<Integer>(); a3.add(arr[j]); a3.add(arr[j]); p.add(a1); p.add(a2); p.add(a3); k++; j++; } else { ArrayList<Integer> a1 = new ArrayList<Integer>(); a1.add(arr[i]); a1.add(arr[j]); ArrayList<Integer> a2 = new ArrayList<Integer>(); a2.add(arr[j]); a2.add(arr[i]); ArrayList<Integer> a3 = new ArrayList<Integer>(); 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<Integer> a3 = new ArrayList<Integer>(); a3.add(arr[i]); a3.add(arr[i]); p.add(a3); i++; k++; } while (j <= r) { b[k] = arr[j]; ArrayList<Integer> a3 = new ArrayList<Integer>(); 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<ArrayList<Integer> > 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<ArrayList<Integer> > p = new ArrayList<ArrayList<Integer> >(); getAllPairs(arr, 0, n - 1, p); for (ArrayList<Integer> 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 pairsdef getAllPairs(arr, l, r, p): if(l<r): mid=int((l+r)/2); getAllPairs(arr,l,mid,p); getAllPairs(arr,mid+1,r,p); getPairsMerge(arr,l,r,mid,p);# Driver coden=2;arr = [0 for _ in range(n)]arr[0] = 1;arr[1] = 2;p = [];getAllPairs(arr,0,n-1,p);# Displaying the sorted pairsfor it in p: print(it[0], it[1])# This code is contributed by phasing17 |
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 pairsfunction getAllPairs(arr, l, r, p){ if(l<r){ let mid= Math.floor((l+r)/2); getAllPairs(arr,l,mid,p); getAllPairs(arr,mid+1,r,p); getPairsMerge(arr,l,r,mid,p); }}// Driver codelet n=2;let arr = new Array(n).fill(0);arr[0] = 1;arr[1] = 2;let p = [];getAllPairs(arr,0,n-1,p);// Displaying the sorted pairsfor(var it of p) console.log(it[0], it[1])// This code is contributed by phasing17 |
Output
1 2 2 1 1 1 2 2
Time Complexity : O (N LogN )
Auxiliary Space: O(l + r)
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