Print all pairs in an unsorted array with equal sum
Last Updated :
15 Feb, 2023
Given an unsorted array A[]. The task is to print all unique pairs in the unsorted array with equal sum.
Note: Print the result in the format as shown in the below examples.
Examples:
Input: A[] = { 6, 4, 12, 10, 22, 54, 32, 42, 21, 11}
Output:
Pairs : ( 4, 12) ( 6, 10) have sum : 16
Pairs : ( 10, 22) ( 21, 11) have sum : 32
Pairs : ( 12, 21) ( 22, 11) have sum : 33
Pairs : ( 22, 21) ( 32, 11) have sum : 43
Pairs : ( 32, 21) ( 42, 11) have sum : 53
Pairs : ( 12, 42) ( 22, 32) have sum : 54
Pairs : ( 10, 54) ( 22, 42) have sum : 64
Input:A[]= { 4, 23, 65, 67, 24, 12, 86}
Output:
Pairs : ( 4, 86) ( 23, 67) have sum : 90
The idea is to use map in C++ STL for avoiding duplicate pairs of elements.
- Create a map with key as pair of integer and value as integer to store all unique pairs of elements and their corresponding sum.
- Traverse the array and generate all possible pairs and store the pairs and their corresponding sum on the first map.
- Create a second map with key as integer and value as a vector of pair to store a list of all pairs of elements with a corresponding sum.
- Finally, traverse the second map, and for a sum with more than one pair, print all pairs and then the corresponding sum in a format as shown in the above example.
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
void pairWithEqualSum(int A[], int n)
{
map<int, vector<pair<int, int> > > mp;
for (int i = 0; i < n - 1; i++) {
for (int j = i + 1; j < n; j++) {
pair<int, int> p = make_pair(A[i], A[j]);
mp[A[i] + A[j]].push_back(p);
}
}
for (auto itr = mp.begin(); itr != mp.end(); itr++) {
if (itr->second.size() > 1) {
cout << "Pairs : ";
for (int i = 0; i < itr->second.size(); i++) {
cout << "( " << itr->second[i].first << ", "
<< itr->second[i].second << ") ";
}
cout << " have sum : " << itr->first << endl;
}
}
}
int main()
{
int A[]
= { 6, 4, 12, 10, 22, 54, 32, 42, 21, 11, 8, 2 };
int n = sizeof(A) / sizeof(A[0]);
pairWithEqualSum(A, n);
return 0;
}
|
Java
import java.util.*;
class GFG {
static class pair {
int first, second;
public pair(int first, int second)
{
this.first = first;
this.second = second;
}
}
static void pairWithEqualSum(int A[], int n)
{
Map<Integer, Vector<pair> > mp = new HashMap<>();
for (int i = 0; i < n - 1; i++) {
for (int j = i + 1; j < n; j++) {
pair p = new pair(A[i], A[j]);
Vector<pair> pp = new Vector<pair>();
if (mp.containsKey(A[i] + A[j]))
pp.addAll(mp.get(A[i] + A[j]));
pp.add(p);
mp.put(A[i] + A[j], pp);
}
}
for (Map.Entry<Integer, Vector<pair> > itr :
mp.entrySet()) {
if (itr.getValue().size() > 1) {
System.out.print("Pairs : ");
for (int i = 0; i < itr.getValue().size();
i++) {
System.out.print(
"( " + itr.getValue().get(i).first
+ ", "
+ itr.getValue().get(i).second
+ ") ");
}
System.out.print(
" have sum : " + itr.getKey() + "\n");
}
}
}
public static void main(String[] args)
{
int A[] = { 6, 4, 12, 10, 22, 54,
32, 42, 21, 11, 8, 2 };
int n = A.length;
pairWithEqualSum(A, n);
}
}
|
Python3
def pairWithEqualSum(A, n):
mp = {}
for i in range(n):
for j in range(i+1, n):
if A[i]+A[j] in mp:
mp[A[i]+A[j]].append((A[i], A[j]))
else:
mp[A[i]+A[j]] = [(A[i], A[j])]
for itr in mp:
if len(mp[itr]) > 1:
print("Pairs : ", end="")
for i in range(0, len(mp[itr])):
print("(", mp[itr][i][0], ",",
mp[itr][i][1], ")", end=" ")
print("have sum :", itr)
if __name__ == "__main__":
A = [6, 4, 12, 10, 22, 54,
32, 42, 21, 11, 8, 2]
n = len(A)
pairWithEqualSum(A, n)
|
C#
using System;
using System.Collections.Generic;
class GFG {
class pair {
public int first, second;
public pair(int first, int second)
{
this.first = first;
this.second = second;
}
}
static void pairWithEqualSum(int[] A, int n)
{
Dictionary<int, List<pair> > mp
= new Dictionary<int, List<pair> >();
for (int i = 0; i < n - 1; i++) {
for (int j = i + 1; j < n; j++) {
pair p = new pair(A[i], A[j]);
List<pair> pp = new List<pair>();
if (mp.ContainsKey(A[i] + A[j]))
pp = AddAll(mp[A[i] + A[j]]);
pp.Add(p);
if (mp.ContainsKey(A[i] + A[j]))
mp[A[i] + A[j]] = pp;
else
mp.Add(A[i] + A[j], pp);
}
}
foreach(KeyValuePair<int, List<pair> > itr in mp)
{
if (itr.Value.Count > 1) {
Console.Write("Pairs : ");
for (int i = 0; i < itr.Value.Count; i++) {
Console.Write(
"( " + itr.Value[i].first + ", "
+ itr.Value[i].second + ") ");
}
Console.Write(" have sum : " + itr.Key
+ "\n");
}
}
}
static List<pair> AddAll(List<pair> list)
{
List<pair> l = new List<pair>();
foreach(pair p in list) l.Add(p);
return l;
}
public static void Main(String[] args)
{
int[] A = { 6, 4, 12, 10, 22, 54,
32, 42, 21, 11, 8, 2 };
int n = A.Length;
pairWithEqualSum(A, n);
}
}
|
Javascript
function pairWithEqualSum(A, n) {
let mp = new Map();
for (let i = 0; i < n - 1; i++) {
for (let j = i + 1; j < n; j++) {
let p = [A[i], A[j]];
let sum = A[i] + A[j];
if (mp.has(sum)) {
let arr = mp.get(sum);
arr.push(p);
mp.set(sum, arr);
} else {
let temp = []
temp.push(p);
mp.set(sum, temp);
}
}
}
for (let [key, value] of mp) {
if (value.length > 1) {
console.log("Pairs: ");
for (let i = 0; i < value.length; i++) {
console.log("(" + value[i][0] + ", " + value[i][1] + ") ");
}
console.log(" have sum : " + key);
}
}
}
let A = [6, 4, 12, 10, 22, 54, 32, 42, 21, 11, 8, 2];
let n = A.length;
pairWithEqualSum(A, n);
|
Output
Pairs : ( 6, 4) ( 8, 2) have sum : 10
Pairs : ( 4, 8) ( 10, 2) have sum : 12
Pairs : ( 6, 8) ( 4, 10) ( 12, 2) have sum : 14
Pairs : ( 6, 10) ( 4, 12) have sum : 16
Pairs : ( 6, 12) ( 10, 8) have sum : 18
Pairs : ( 12, 11) ( 21, 2) have sum : 23
Pairs : ( 10, 22) ( 21, 11) have sum : 32
Pairs : ( 12, 21) ( 22, 11) have sum : 33
Pairs : ( 12, 22) ( 32, 2) have sum : 34
Pairs : ( 22, 21) ( 32, 11) have sum : 43
Pairs : ( 12, 32) ( 42, 2) have sum : 44
Pairs : ( 32, 21) ( 42, 11) have sum : 53
Pairs : ( 12, 42) ( 22, 32) have sum : 54
Pairs : ( 10, 54) ( 22, 42) have sum : 64
Time Complexity: O(n2log(n)), as nested loops are used
Auxiliary Space: O(n), as extra space of size n is used to create a map
Implementation: Another implementation of the same above approach using one Single map.
C++
#include <bits/stdc++.h>
using namespace std;
void findPairs(vector<int> arr)
{
map<int, int> mp;
int sum = 0, element = 0;
for (int i = 0; i < arr.size(); i++) {
for (int j = i + 1; j < arr.size(); j++) {
sum = arr[i] + arr[j];
if (mp.count(sum)) {
element = mp[sum];
cout << "Pairs:(" << arr[i] << "," << arr[j]
<< ") (" << element << ","
<< "(" << sum - element << ")"
<< ") have sum:" << sum << endl;
}
else {
mp[sum] = arr[i];
}
}
}
}
int main()
{
vector<int> arr
= { 6, 4, 12, 10, 22, 54, 32, 42, 21, 11, 8, 2 };
findPairs(arr);
}
|
Java
import java.io.*;
import java.util.HashMap;
import java.util.Map;
class GFG{
public void findPairs(int arr[]){
Map<Integer,
Integer> map = new HashMap<Integer,
Integer>();
int sum = 0, element = 0;
for(int i = 0; i < arr.length; i++)
{
for(int j = i + 1; j < arr.length; j++)
{
sum = arr[i] + arr[j];
if(map.get(sum) != null)
{
element = map.get(sum);
System.out.println("Pairs:(" + arr[i] +
"," + arr[j] + ") (" +
element + "," +
(sum - element) +
") have sum:" + sum);
}
else
{
map.put(sum, arr[i]);
}
}
}
}
public static void main(String[] args)
{
int arr[] = {6, 4, 12, 10, 22, 54,
32, 42, 21, 11, 8, 2};
GFG obj = new GFG();
obj.findPairs(arr);
}
}
|
Python3
class GFG :
def findPairs(self, arr) :
map = dict()
sum = 0
element = 0
i = 0
while (i < len(arr)) :
j = i + 1
while (j < len(arr)) :
sum = arr[i] + arr[j]
if (map.get(sum) != None) :
element = map.get(sum)
print("Pairs:(" + str(arr[i]) + "," + str(arr[j]) + ") (" + str(element) + "," + str((sum - element)) + ") have sum:" + str(sum))
else :
map[sum] = arr[i]
j += 1
i += 1
@staticmethod
def main( args) :
arr = [6, 4, 12, 10, 22, 54, 32, 42, 21, 11, 8, 2]
obj = GFG()
obj.findPairs(arr)
if __name__=="__main__":
GFG.main([])
|
C#
using System;
using System.Collections.Generic;
class GFG
{
static void findPairs(int[] arr){
Dictionary<int,int> map = new Dictionary<int,int>();
int sum = 0, element = 0;
for(int i = 0; i < arr.Length; i++)
{
for(int j = i + 1; j < arr.Length; j++)
{
sum = arr[i] + arr[j];
if(map.ContainsKey(sum))
{
element = map[sum];
Console.WriteLine("Pairs:(" + arr[i] +
"," + arr[j] + ") (" +
element + "," +
(sum - element) +
") have sum:" + sum);
}
else
{
map[sum] = arr[i];
}
}
}
}
public static void Main(String[] args)
{
int[] arr = {6, 4, 12, 10, 22, 54,
32, 42, 21, 11, 8, 2};
findPairs(arr);
}
}
|
Javascript
function findPairs(arr) {
let mp = {};
let sum = 0, element = 0;
for (let i = 0; i < arr.length; i++) {
for (let j = i + 1; j < arr.length; j++) {
sum = arr[i] + arr[j];
if (mp[sum]) {
element = mp[sum];
document.write("Pairs:(" + arr[i] + "," + arr[j]
+ ") (" + (element) + ","
+ "(" + (sum - element) + ")"
+ ") have sum:" + sum);
}
else {
mp[sum] = arr[i];
}
}
}
}
let arr = [6, 4, 12, 10, 22, 54, 32, 42, 21, 11, 8, 2];
findPairs(arr);
|
Output
Pairs:(4,12) (6,(10)) have sum:16
Pairs:(4,10) (6,(8)) have sum:14
Pairs:(12,2) (6,(8)) have sum:14
Pairs:(10,8) (6,(12)) have sum:18
Pairs:(10,2) (4,(8)) have sum:12
Pairs:(22,32) (12,(42)) have sum:54
Pairs:(22,42) (10,(54)) have sum:64
Pairs:(22,11) (12,(21)) have sum:33
Pairs:(32,11) (22,(21)) have sum:43
Pairs:(32,2) (12,(22)) have sum:34
Pairs:(42,11) (32,(21)) have sum:53
Pairs:(42,2) (12,(32)) have sum:44
Pairs:(21,11) (10,(22)) have sum:32
Pairs:(21,2) (12,(11)) have sum:23
Pairs:(8,2) (6,(4)) have sum:10
Complexity Analysis:
- Time Complexity: O(n2logn), as nested loops are used
- Auxiliary Space: O(n), as extra space of size n is used to create a Hashmap
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